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Pakistani History/Islamic Republic of Pakistan. The Islamic Republic of Pakistan, is a country in South Asia. It is the fifth-most populous country, with a population of over 241.5 million, having the second-largest Muslim population as of 2023. Islamabad is the nation's capital, while Karachi is its largest city and financial centre. Pakistan is the 33rd-largest country by area and the ninth-largest in Asia. Bounded by the Arabian Sea on the south, the Gulf of Oman on the southwest, and the Sir Creek on the southeast, it shares land borders with India to the east; Afghanistan to the west; Iran to the southwest; and China to the northeast. It shares a maritime border with Oman in the Gulf of Oman, and is separated from Tajikistan in the northwest by Afghanistan's narrow Wakhan Corridor. Pakistan is the site of several ancient cultures, including the 8,500-year-old Neolithic site of Mehrgarh in Balochistan, the Indus Valley civilisation of the Bronze Age, and the ancient Gandhara civilisation. The regions that compose the modern state of Pakistan were the realm of multiple empires and dynasties, including the Achaemenid, the Maurya, the Kushan, the Gupta; the Umayyad Caliphate in its southern regions, the Samma, the Hindu Shahis, the Shah Miris, the Ghaznavids, the Delhi Sultanate, the Mughals, and most recently, the British Raj from 1858 to 1947. Spurred by the Pakistan Movement, which sought a homeland for the Muslims of British India, and election victories in 1946 by the All-India Muslim League, Pakistan gained independence in 1947 after the Partition of the British Indian Empire, which awarded separate statehood to its Muslim-majority regions and was accompanied by an unparalleled mass migration and loss of life. Initially a Dominion of the British Commonwealth, Pakistan officially drafted its constitution in 1956, and emerged as a declared Islamic republic. In 1971, the exclave of East Pakistan seceded as the new country of Bangladesh after a nine-month-long civil war. In the following four decades, Pakistan has been ruled by governments whose descriptions, although complex, commonly alternated between civilian and military, democratic and authoritarian, relatively secular and Islamist. Pakistan is considered a middle power nation, with the world's sixth-largest standing armed forces. It is a declared nuclear-weapons state, and is ranked amongst the emerging and growth-leading economies, with a large and rapidly-growing middle class. Pakistan's political history since independence has been characterized by periods of significant economic and military growth as well as those of political and economic instability. It is an ethnically and linguistically diverse country, with similarly diverse geography and wildlife. The country continues to face challenges, including poverty, illiteracy, corruption, and terrorism. Pakistan is a member of the United Nations, the Shanghai Cooperation Organisation, the Organisation of Islamic Cooperation, the Commonwealth of Nations, the South Asian Association for Regional Cooperation, and the Islamic Military Counter-Terrorism Coalition, and is designated as a major non-NATO ally by the United States.

Chess Opening Theory/1. e4/1...Nf6/2. e5/2...Nd5/3. c4/3...Nf4. = Alekhine's Defence: The Squirrel = Black saves their Knight, which was attacked by the pawn on c4. Even though the knight moves forward, this move is not seen as best because the Knight is unstable on f4 and will likely have to move again to a worse square to save itself.

AI Art Generation Handbook/List of AI Image Model. This page is to show the available AI models that are available to public that can be used to run in local PC to generate images

Mouse Trap/Revised Rules. Sometime in the mid-2020s Hasbro, for whatever reason, decided to completely overhaul the rules of "Mouse Trap". The objective of the revised game is to be the first player to collect six pieces of cheese. Game Equipment. The revised "Mouse Trap" game comes with the following equipment: Setup. To prepare the game: One player is selected to take the first turn, then play proceeds clockwise. Basic Gameplay. A player's turn consists of the player rolling the die and moving their mouse clockwise that many spaces along the board's track. They then take an action corresponding to the type of space they landed on. After this is resolved, that player's turn is over and play proceeds to the player left from them. Types of Spaces. There are several different types of spaces, each with different actions a player must take when landing on them: Start. Appearance: A circular wooden-looking space with the word "Start" and two dotted cheese slices. This space is where the players begin the game. At game start two cheese pieces are placed on this space - the first player to make it around the board and back to this space will collect these cheese pieces. Normal. Appearance: A purple space. Nothing happens. The player will move from here on their next turn. Launch the Trap. Appearance: A blue space with an illustration of the Mouse Trap's cage on it. If a player lands here, they are allowed to select one other player and move that player's mouse to the normal space immediately underneath the Mouse Trap's cage, denoted by the wheel of cheese. Then the player may pull back on the stop sign and release it to set the Trap in motion. If the Trap succeeds in capturing the other mouse, the player who was trapped must turn over one of their cheese pieces to the player who trapped them. If the Trap fails for whatever reason, they don't turn over any cheese. Unlike with the original rules, the player caught by the Mouse Trap is not out of the game. Regardless of the outcome of an attempted capture, the player who operated the Mouse Trap must reset it by restoring the metal marble, diver and cage to their original positions and rotating the broom down, using the wrench to secure it. If when a player lands on this space there is already a mouse on the trap space, the player may still select another player's mouse to move to the trap space. If the Mouse Trap succeeds in capturing multiple players, all of the trapped players must turn over one cheese piece. Cheese. Appearance: A yellow space with an illustration of a piece of cheese on it. If a player lands here, they may take one piece of cheese from the pile. Steal. Appearance: A green space with the silhouette of a mouse and the word "STEAL!" on it. If a player lands here, they may steal one piece of cheese from another player. Lose Cheese. Appearance: A red space with the dotted line silhouette of a cheese piece bearing a sad face. If a player lands here, they must return one piece of cheese to the pile. Mouse-Hole Tunnel. Appearance: The entrance to a tunnel (arch-shaped in some versions, circular in others) with a mouse in front of it. If a player lands here, they may, if they want to, travel through the mouse-hole tunnel to the space on the other side. Winning the Game. The winner of the game is the first player to collect six cheese pieces, forming a complete cheese wheel.

Palliative Pharmacotherapy/Miscellaneous/Helpful Dot Phrases/New Consult Inpatient Pain Management Note Template. The following is an EXAMPLE note template that could be used for comprehensive medication management for an inpatient pain management consult service. This is the author's own personal template and therefore has information in place to remind this scatterbrained author to ask questions that may be very obvious to other practitioners. Please feel free to use and edit this template as you wish. Beware that this note template has been created for CPRS (VA EMR) and is not suitable in its raw form for use in Epic, Cerner, Meditech, etc. "For those unfamiliar with CPRS, the items / text between bars" (bars = | |) "are called TIU (text integration utility) data objects and these pull data from the chart automatically. For example, in this template the first TIU data object encountered is" |ADMITTING DIAGNOSIS|, "which when used in a template will automatically populate with the selected patient's admitting diagnosis for the current inpatient admission. CPRS TIU data objects vary across VA facilities, though some may be similar. That is to say - if you would like to use this template within CPRS at your facility, some of the TIU data objects may not work properly."

AI Art Generation Handbook/Node Prompting in Stable Diffusion. This type of "programming language" used in ComfyUI may overwhelm especially to the user whom is not familiar to before compared to other local UI (such as Auto1111, Foooocus, Oomost and such...) but if we bite the bullet and get through it , this may become easy . As you can see, there are square blocks that are connected to one another using multiple coloured lines. In this context, the square blocks are known as "node" or "block diagram" and the multiple coloured lines are known as "wires" or "pipelines" Control. Once you successfully installing ComfyUI, we learn how to control inside ComfyUI Anatomy of the nodes. Every nodes have this 3 characteristic :<br> Input Data Types (Red): The data flows into the nodes through this pipelines. Take note that the wire colours are matching to the input data types<br> Output Data Types (Blue): The data flows out of nodes through this pipelines. Take note that the wire colours are matching to the output data types<br> Control Parameter (Green): The results of the calculations performed within this node is largely influenced by the setting values inside control parameters<br> Basically, how this works is the whole nodes (as shown on the left) is basically a big "black box" where it takes data flow into the nodes and perform abstractions as well as the complex calculations (based on the control parameters within its nodes). After the calculations is done, the data is outputted to the next nodes for other processing. Node Programming. To learn node programming, we can clear the whole workflow by clicking "Clear" button on the right. As seen on the diagram of the Text-2-Image of how it works, we can recreate the workflow . Node Programming - Add in new nodes. To add new nodes, we can double click on any empty spaces in canvas and using search functions to search required block diagrams easier: We add this few important block diagram Node Programming - Connecting the nodes. We can start connecting the block diagrams by the pipelines. "Note: Either way is OK but if you wanted to connect many to one nodes, the recommended method is to connect from the block diagrams input (left side) to the block diagrams output (right side) to minimises potential bugs and issues later on. See" "Connecting CLIP to K-Sampler parts" Let's start with the Load Checkpoint block diagrams and K-Sampler block diagrams . Click on the "MODEL" output (with lavender circle) , you can notices that only "model" input is highlighted meanwhile the rest is grayed out (Meaning it is not connectable). Another point to take note that , pay attention to the colour of input and output data types (Only both of same colours are able to be connected together) Now , Load Check Point should be connected to K-Sampler through the MODEL pipelines. There are CLIP to be connected. As mentioned above, the CLIP is related to Text Encoder, therefore , we can connect both CLIP Block Diagrams . But However, before we started to connect those , rename the Block Diagrams first into more appropriate name to prevent confusions. Right click on the block diagrams and click on "Properties Panel" . There should be Title, Mode , Color and Node name for S&R, change the both "Title" and "Node Name for S&R" to more self explanatory names likme codice_1 as examples. "Note : We want to distinguish which block diagrams is for positive and for negative prompts." Connect the Load Checkpoints "CLIP" Output to both Positive and Negative Block Diagrams "clip" input (See yellow pipelines). Lastly connect the CLIP "CONDITIONING" output to "positive" and "negative" K-Sampler input (See orange pipelines)respective to the block diagrams names You should have the connections as seen in the left screenshot by now "Note: There are potential bugs of connecting output data types CONDITIONING to input data types positive or negative , therefore reverse the sequence to connect from input positive / negative to output CONDITIONING." As we can see now in Load Checkpoints block diagram, "VAE" block diagram is still unconnected . We can click on "VAE" output datatypes (red pipelines) to see which it can connect to, only VAE Decode Block Diagram "vae" input is highlighted. Repeat the same in VAE Decode diagram block for "sampler" input datatypes (pink pipelines) to be connected to K-SAMPLER's "latent" output data type Finally , repeat the same for VAE Decode "IMAGE" output datatypes (blue pipelines) to be connected to "images" input datatypes in Save Images Block Diagrams. You should get the similar diagrams as below and left only one connections between the K-Sampler and Empty Latent Image block diagrams to make it functional. Connect the "LATENT" in Empty Latent Image to KSampler's "latent_image" to finish the workflow. Congrats , you finished the basic text to image workflow and have slightly better understanding on how this all works. Node Programming - Generate Image. Try to click the "Queue Prompt" on the right button on the sidebar. You may not notice at first but the KSampler node is suddenly highlighted in green colour and then, there is green colour bar that starting to go across. It meant that the AI Art Models is working. To know what does each of the setting meant , go to AI Art Generation Handbook/Stable Diffusion settings to learn more Improving the ComfyUI. Save Prompt Names as Image Ouput File Name. As we generate the images, we get the generic names of "ComfyUI_0001.png" as default , but do you know that with some additional steps , we can save prompt as filenames in ComfyUI ? Here is how we do it very easily. On the Positive CLIP block diagram, right click on the block diagrams to search for the name of codice_2 Use the values that you key in earlier (in this case "Positive CLIP") and type the following into filename_prefix inside Save Image block diagram: %Positive_CLIP.text% , refer to picture on the left for clarity . From now onwards, you can see the filename changes to the saved images if the Save Image block diagram is not altered in anyway. For more customisation on the name , here are the page where we can get to change the custom names (especially prompt names) for the ComfyUI output files. Improve Image Generations Quality. SDXL and above are trained on 1024 x 1024 px images , therefore anything less than that will results in low quality of images as shown per this experiments here. Therefore, on Empty Latent Image block diagram you can adjust both width and height up to 1024px to get more aesthetic image generations. References. [1] https://www.youtube.com/watch?v=RVwIz63bxN4 - KSamplers [2] https://www.reddit.com/r/StableDiffusion/comments/174g0s9/prompt_in_file_name_in_comfyui/ - Added filenames into prompt

Moving objects in retarded gravitational potentials of an expanding spherical shell/Brief historical review. Brief historical review. The finiteness of the propagation speed of gravity and its influence on gravitational forces was originally published by the Austrian astronomer Josef von Hepperger (1855–1928) in 1888 in Vienna. The German physicist Paul Gerber (1854–1909) published 1898 his paper on "The spatial and temporal expansion of gravity", where he established that the perihelion precession of the planet Mercury is related to the propagation speed of gravity, which is quite close to that of electromagnetic radiation. His formula for the perihelion precession of the planet Mercury wasn't known to Albert Einstein (1879–1955), but six years after Paul Gerber's death, Einstein found an identical formula in his publication "Erklärung der Perihelbewegung des Merkur aus der allgemeinen Relativitätstheorie" (English: "Explanation of Mercury's perihelion motion from the general theory of relativity") by applying the laws of General Relativity. However, the contemporary scientists couldn't reproduce the derivation of Paul Gerber for the formula, and furthermore, they stated that some of the prerequisites used by him were wrong. Shortly before his early death the German astronomer and physicist Karl Schwarzschild (1873–1916) published a paper on "Über das Gravitationsfeld einer Kugel aus inkompressibler Flüssigkeit nach der Einsteinschen Theorie" (English: "On the gravitational field of a sphere of incompressible fluid according to Einstein's theory"), where he described how to compute the smallest possible radius for a sphere with a given mass. He found the radius for a sphere with the mass of the sun to be three kilometres. In recognition of his achievement, the corresponding radius is now called the Schwarzschild radius. It is interesting to notice that Schwarzschild also made important contributions to retarded potentials in electrodynamics already in 1903. According to the electrokinetic potential he claimed: Es sind in jedem Raumelement die Werte der Dichte und der Geschwindigkeit zu benutzen, welche dort zu einer um die Lichtzeit zurückliegenden Epoche galten. In each spatial element, the values of density and velocity are to be used, which were valid there at an epoch around light time ago. The Belgian theologian and physicist George Lemaître SJ (1894–1966) is regarded as the founder of the Big Bang theory. In 1931, he introduced the term "atome primitif" (English: "primordial atom") to describe the hot initial state of the universe. Already in 1927 he wrote as the second conclusion of his publication about "a homogeneous universe of constant mass and increasing radius, accounting for the radial velocity of extra-galactic nebulae" with reference to the investigations of Edwin Hubble (1889–1953) in 1926: Le rayon de l'univers croit sans cesse depuis une valeur asymptotique formula_1 pour formula_2 The radius of the universe increases without limit from an asymptotic value formula_1 for formula_2 In 1933 the Swiss astronomer Fritz Zwicky (1898–1974) obeserved a gravitational anomaly in the Coma galaxy cluster, and he coined the term dark matter (in German: "dunkle Materie") for the cause of this anomaly. In his book "Relativity, Gravitation and World Structure" of the year 1935 the British astrophysicist and mathematician Edward Arthur Milne (1896–1950) introduced the special-relativistic cosmological Milne model. This model assumes an isotropic and for all time linearly expanding, but not homogeneous universe. It is independent of general theory of relativity, but nevertheless equivalent to a special case of the Friedmann–Lemaître–Robertson–Walker metric in the limit of zero energy density. The redshift is explained by the velocity caused by a hypothetical initial explosion of matter. The density of such a universe must be small in comparison to the critical density of the Friedmann equations that were published by Russian physicist and mathematician Alexander Friedmann (1888–1925) in 1924. Decades later, the Milne model fell almost completely into oblivion, whereas the cosmological models based on the general theory of relativity gained more and more attention. In 1953 the German astrophysicist Erwin Finlay-Freundlich (1885–1964) derived a blackbody temperature for intergalactic space of 2.3 Kelvin according to his theory of tired light. The German-British mathematician and physicist Max Born (1882–1970) immediately recommended taking the problem seriously and pursuing it further. In 1965 the cosmic microwave background (CMB) was discovered by US-American radio astronomers Arno Allan Penzias (1933–2024) and Robert Woodrow Wilson (born 1936). It has a thermal black body spectrum at a very low temperature of about 2.7 Kelvin. Since the cosmic microwave background was mainly created in the visible area of the electromagnetic spectrum, it must have undergone a strong wavelength extension on its way to the observers. This can happen because of two main reasons: In addition, it is discussed that an expansion of the spacetime during the propagation time of a light wave also would increase its wavelength to the same extent. The redshift factor formula_5 is defined as a relation between an emitted wavelength formula_6 and an observed wavelength formula_7 of electromagnetic radiation: The redshift of the cosmic microwave background was found to be formula_9 in 2003, which is an extremely high value. The age of the universe at the time this background radiation was created by hydrogen atoms has been estimated at around 379,000 years. Observations of distant type Ia supernovae published by both the Supernova Cosmology Project as well as the High-Z Supernova Search Team in 1998 show that the relative expansion of the universe is accelerating. For the analysis the astronomers Saul Perlmutter, Brian P. Schmidt and Adam Riess were awarded the Nobel Prize in Physics in 2011. From 2001 to 2010 the NASA spacecraft Wilkinson Microwave Anisotropy Probe (WMAP) was investigating the cosmic microwave background. Its measurements led to the current Standard Model of Cosmology. According to this model the universe currently consists of less than 5 percent ordinary baryonic matter; about 24 percent cold dark matter (CDM) that interacts only weakly with ordinary matter and electromagnetic radiation; and more than 70 percent of dark energy that is used to explain the accelerated expansion of the universe. These data were more or less confirmed by the Planck space observatory that was operated by the European Space Agency (ESA) from 2009 to 2013. In 2013 the Indian researcher Chandrakant Raju proposed to apply the retarded gravitation theory (RGT) to explain the flyby anomaly of spacecrafts in the graviational field of the earth as well as the acceleration of masses in the retarded gravitational fields of spriral galaxies. In 2016, researchers from the gravitational-wave observatory LIGO reported the first direct measurement of gravitational waves generated by the collision of two black holes (gravitational wave event GW150914). Already in 2017 the Nobel Prize in Physics was awarded for the first direct observation of gravitational waves. On 17th August 2017 a merging binary neutron star was observed independently and simultanuously by the Advanced LIGO and Virgo detectors (gravitational wave event GW170817) and the Fermi Gamma-ray Burst Monitor as well as the Anticoincidence Shield for the Spectrometer for the International Gamma-Ray Astrophysics Laboratory (gamma ray burst event GRB 170817A). These observations prove that the propagation velocity of gravitational waves must be extremly close to that of electromagnetic waves. In January 2024 the very young and very far galaxis JADES-GS-z14-0 was found with the Near-Infrared Spectrograph (NIRSpec) of the James Webb Space Telescope (JWST). This galaxy was observed in a state 290 million years after the big bang. Its redshift measured with the well-known Lyman alpha break at a wavelength about 1.8 micometres has a very high value of a good 14, which is a very high value for an astronomical object, but much lower than that of the cosmic microwave background.

Moving objects in retarded gravitational potentials of an expanding spherical shell/Classical approach. Classical approach. This Wikibook assumes a huge amount of dark matter that is in a huge shell behind the visible part of the universe. This matter is rather cold, and it can be assumed that it is in a in thermal equilibrium. Therefore, the electromagnetic radiation emitted by this shell can be represented by the Planck radiation law. The emission of such a cold source of electromagnetic radiation is not visible and additionally the radiation has experienced a strong redshift due to the relative velocity of the fast expanding emitting shell as well as due to the gravitation of the mass of this shell. The gravitational force formula_1 between two masses formula_2 and formula_3 in a distance of formula_4 is given by Newton's law of universal gravitation with the gravitational constant formula_5: In a thought experiment it is possible to investigate the behaviour of a mass formula_2 that is located within a spherical shell with the mass formula_3. The approach described in 1687 by Isaak Newton (1643–1727 in Gregorian dates) is called Newton's shell theorem. According to this theorem a mass formula_2 within a homogeneous and spherically symmetric shell with a mass formula_3 experiences no net gravitational force. Additionally, this is regardless of the location of the mass formula_2 within the shell as well as of its velocity. This behaviour can be easily explained with the following consideration: let a mass formula_2 be within a homogeneous and spherically symmetric shell with the radius formula_13 and the areal density formula_14. Then the area formula_15 and the mass formula_3 of the shell can be expressed as: If we take an axis symmetrical double cone where the tips of the two cones are located at the position of the mass formula_2 within a homogeneous and spherically symmetric shell, we get the situation shown in the adjacent figure. For the two infinitesimally small area elements formula_20 and formula_21 in the distances formula_22 and formula_23 of a mass formula_2 with the same infinitesimally small angle elements formula_25: For the appropriate masses formula_28 and formula_29 of the two area elements: With Newton's law of universal gravitation, we can express the two infinitesimally small gravitational forces formula_32 and formula_33 due to the masses of the two area elements: It is obvious that both forces have the same value, and since they must be exactly directed to opposite directions, they add to zero. In other words: a mass within a spherical shell does not experience any net gravitational force at all. It is worth noting that this classical approach assumes the infinite propagation speed of gravitational waves, which is applicable for static but not for dynamic situations as well as huge systems.

Moving objects in retarded gravitational potentials of an expanding spherical shell/Retarded gravitational potentials. Retarded gravitational potentials. Already the German physicist and astronomer Karl Schwarzschild (1873–1916) described retarded potentials for elecrodynamic fields (he still used the term "electrokinetic potential") in 1903. These potentials with a delayed effect were adopted one year later with reference to Schwarzschild by the German mathematician Alexander Wilhelm von Brill (1842–1935) in "Über zyklische Bewegung" (English: "About cyclic movement"), where he coined the term "retardiertes Potential" (English: "retarded potential") in a footnote. Furthermore, he emphasized that these retarded potentials would cause a non-zero divergence in space, i.e. that there would be sources and sinks or that the medium would have the possibility of storing or releasing potential energy. Retarded potentials are a mathematical description of potentials in a field theory in which a field quantity propagates at a finite speed (speed of light) and not instantaneously. They occur in the investigation of time-dependent problems, such as the radiation of electromagnetic waves, but also in the propagation of gravitational waves. Objects that are moving towards the outer rim of the universe will experience retarded gravitational potentials by the expanding dark matter that can be assumed at the very edge of the universe. Therefore, the appropriate gravitational forces will have delayed effects, and due to the large distances and the finite velocity of gravitational wave propagation they also will be weaker in the direction of the former location of the objects. As a result, the net gravitational force is directed in the direction of movement of these objects, and therefore, all objects that move outwards will be accelerated in the direction of their own movement, which would become observable as an accelerated expansion of the visible universe. The gravitational potential formula_1 due to a mass formula_2 in the distance formula_3 is given by: The gravitational force formula_5 to another mass formula_6 results as follows: The time-depending potential formula_8 is the solution of the inhomogeneous wave equation, where formula_9 is the inhomogeneity, and formula_10 is the speed of the propagation of the waves. For gravitational forces we can consider it as equal to the vacuum speed of electromagnetic waves: where formula_12 is the Laplace operator, formula_13 is the D’Alembert operator. The solution of the inhomogeneous wave equation is called retarded potential, and in three dimensions it can be given as: The retardation is to be interpreted in such a way that a source element formula_15 at the point formula_16 and at the time formula_17 only influences the potential at the distant point of impact at formula_18 at a later time formula_19: formula_17 is called the retarded time. At the location formula_18 and at the time formula_19 the retarded potential only depends on the inhomogeneity formula_25 in the retarding back cone of the location. This inhomogeneity has a retarded effect to the solution, and it is delayed with the wave velocity formula_10. In a shell. In the simplified example in the adjacent figure the source term is the linear mass density formula_27 that is not time-dependant and only exists in the circle of the outer rim with the constant radius formula_28 and its centre at formula_29: In all other locations within the plane the linear mass density formula_27 is zero: All mass elements outside the regarded plane have an symmetrical effect to the mass, and therefore, in these locations the contribution of the mass elements to the potential can be neglected for the determination of inhomogeneity: Furthermore, the mass element formula_36 on the homogenous circumference formula_37 of a circle with the radius formula_28 is given by: The cosine formula gives us the relation between the location of the mass point formula_6 in the horizontal distance formula_42 of the origin formula_29 of the circle with the radius formula_28, when the mass element on the circle formula_36 is in the direction of the angle formula_46 and the distance formula_3: In the normalised standard form of the quadratic equation, we get: The solution for the distance formula_3 is: It is obvious that the following simplifications are valid: The origin of the coordinate system can be also shifted to the mass formula_6: The retarded time formula_17 and the retarded potential formula_60 are given as follows, where formula_10 represents the propagation speed of the potentials: Illustration. In a simplified example we only consider the infinitesimally small angles formula_64 in the origin of a spherical shell with the radius formula_28 and the areal density formula_66. Symmetrical geometry. If the mass formula_6 is in the centre of a spherical shell with the radius formula_68 we have the following situation: In this situation the mass formula_6 does not experience any acceleration in the classical approach (see above) or if its velocity is zero. Moving masses. The situation is changed, if the masses are moving starting at formula_76 within a time span of formula_77. The mass formula_6 moves with the velocity formula_79 to the right and the two mass elements formula_80 and formula_81 move with the radial velocity formula_82: At the time formula_83 the mass has moved with the velocity formula_79 the distance formula_85 to the right: The spherical shell has expanded with the velocity formula_82 and gained an increased radius: Therefore: This means that the distance of mass element formula_93 to the mass formula_6 is always greater than the distance of mass element formula_95 to the mass formula_6. For the two retarded times for these distances to the location of formula_6 we get: With formula_100 and therefore formula_101: Therefore: This means that the retarded time for the mass element formula_95 is always later than the retarded time for the mass element formula_93. Common case. In the adjacent diagram there are three mass points that move in space. Their speed is given as follows: Their time-depending location is given by these three functions for their x-coordinates: The time-depending effective distances formula_113 and formula_114 between the outer mass elements and the mass point in between them is the difference of the corresponding x-coordinates and linked to the propagation velocity of the interacting waves formula_115 as follows: As a result, the corresponding retarded times formula_118 and formula_119 for the two outer mass elemens are: And therefore: And: These are the time-depending effective distances for the gravitational potentials of the outer mass elements at their retarded times that have a simultaneous effect to the mass in between them at the time formula_19. For formula_127 and formula_128 and according to the diagram we can make the following assumption for the comparison of the effective distances: This means that the retarded time of the upper mass element is always later than the retarded time of the lower mass element. As well as: Since formula_137 is always greater than formula_107, the assumption is proven. This means that the effective distance of the moving mass in between them to the lower mass element is always greater than to the moving upper mass element. Finally, it can be stated that the absolute value of the retarded gravitational potential at the location of the moving mass in between them is always greater for the upper mass element than for the lower mass element, if both mass elements have the same value formula_2: The moving mass formula_6 experiences the corresponding retarded forces: The net force to the mass formula_6 is the sum of both: For the acceleration formula_148 of the mass formula_6 we find: Since formula_151 the net force formula_152 as well as the acceleration formula_148 are positive, and the mass experiences an acceleration to positive x-values, i.e. in the direction of its movement. Special case. Let us have a look at the following special case: Their time-depending location is given by these three functions for their x-coordinates: The time-depending effective distances formula_160 and formula_161 between the outer mass elements and the mass point in between them is the difference of the corresponding x-coordinates: As a result, the corresponding retarded times formula_164 and formula_165 for the two outer mass elemens are: Therefore: Thought experiment. In a thought experiment we look at the following situation, where formula_10 is the propagation speed of the gravitational waves. The time line starts at formula_173, and the effect of the retarded potentials is synchronised with the occurence of the moving mass. The left moving mass element is regarded at formula_173 and formula_175, the right moving mass element is regarded at formula_173 and formula_177, whilst the moving mass formula_6 is regarded at formula_179 and formula_180, when the retarded gravitational potentials have their effect to the mass. In the following diagram the velocity of the waves is normalised and used without unit: And therefore, only for simplification and without any units, too: The effective distances formula_70 for the gravitational potential at formula_187 are both equal: At the time formula_189 the mass experiences the retarded potentials on the left-hand side (distance to mass element is formula_190) and at the right-hand side (distance to mass element is formula_191): All inhomogeneities contribute to the retarded potential at the location of the mass with the value they had at the retarded times formula_194, formula_195 and formula_196: For a mass formula_6 moving from the centre of a shell to the right the angle element formula_204 to the left becomes smaller than the original angle element formula_205, and the angle element formula_206 to the right becomes greater than the original angle element formula_205: The following applies to the appropriate mass elements: For the net force formula_210 to the mass formula_6: For the acceleration formula_213 of the mass formula_6 we find: Since formula_216 the net force formula_210 is positive, and the mass experiences an acceleration formula_213 to the right, i.e. in the direction of its movement. This result is absolutely inline with the findings above in the section "Common case" above. Furthermore, it is noteworthy to state that the acceleration formula_213 is proportional to the areal density formula_220 of the expanding shell: Nevertheless it should be noted that the areal density is decreasing with the expansion and the increasing radius of the outer shell.

Moving objects in retarded gravitational potentials of an expanding spherical shell/Gravitational redshift. Gravitational redshift. Schwarzschild radius. The Schwarzschild radius formula_1 of any mass formula_2 in a Schwarzschild sphere (typically a black hole) is given by: Apart from its mass, the Schwarzschild radius is only depending on two natural constants: The gravitational redshift z of photons emitted to the opposite direction of the Schwarzschild sphere and with a distance to the centre of the sphere formula_6can be computed by the following formula: The redshift of very far objects such as the galaxy JADES.GS.z14-0 has a value of more than 14, and this is much larger than expected. The possible high speed of such a galaxy is not sufficient for such a high value. The distance of this galaxy is given by 13.5 billion light-years, and its age is assumed to be 290 million years after Big Bang. The cosmic microwave background even has a redshift value of 1089, which is extremely high. It is associated with the first hydrogen atoms that occurred some 380,000 years after the Big Bang. In a shell. However, a gravitational redshift can not only occur outside of spheres, but also within in a hollow spherical cap. To estimate its gravitational redshift, the effective mass formula_8 of such a cap can be integrated for any point within the cap. The corresponding effect can be described by a Schwarzschild sphere with the Schwarzschild radius. For very fast-moving objects we can assume that they only experience the retarded gravitational potentials of the mass elements in front of them, since the backward potentials are even much more retarded, and therefore, contribute only weakly to the net gravity. In the following section only the arc of the shell cap is considered. The gravitational forces rectangular to the direction of movement are very small and can be neglected. This also holds in first approximation for the gravitational forces behind the very fast-moving moving mass formula_9. If the distance formula_10 between the mass formula_9 and the shell is given, we can carry out an estimation for the fraction of the retarded forces from the opposite part of the shell in the distance formula_12, where formula_13 is radius of the universe (cf. previous section "Retarded gravitational potentials"): If we assume that the distance is a fraction of the radius then the ratio of the two forces has the following relation: This means if the distance formula_10 is a tenth of the radius formula_13 of the universe, the error by neglecting the left force would be less than 0.3 percent. This distance is corresponding to a redshift of about 7.5. In the diagram on the left the ratio of the two forces is plotted against the redshift which is observed at corresponding distances. The high values of redshift at far distant objects seem to be dominated by the gravitaional redshift, whereas at shorter distances the redshift is dominated by the Doppler effect. Computation. The bold arc on the right-hand side of the diagram is representing all mass elements of the outer shell of the universe with the linear mass density formula_20 in front of the mass formula_9 that is moving to the right with high velocity. The outer shell is consisting of dark matter (mainly hydrogen), and in the central area of the universe this dark matter might only be visible due to the cosmic microwave background. We use the following constant values for the estimates derived from these premises: The mass of this arc formula_24 can be computed by integrating the arc between the angles formula_25 and formula_26 with its linear mass density formula_20: The mass of the whole shell formula_30 is given by integrating a complete circle: The Pythagorean theorem gives the following result for the half chord of the circle: If the distance formula_10 between the mass formula_9 and the outer shell is given, we can compute the distances of the mass formula_9 to any infinitesimal mass element formula_36 on the arc depending on the angle formula_37: The angle formula_40 as seen from the moving mass m to the infinitesimal mass element formula_41 on the arc can be derived by applying the law of sines and is given by the following expression: With the auxiliary sagitta formula_43 we get: The vertical components of the gravitational forces are symetrical, and therefore, their net effect is zero. The net horizontal force of this arc formula_46 that is accelerating the mass formula_9 in horizontal direction can be achieved by integrating a semi-circle with the correction factor formula_48: With the effective gravitational force formula_46 that acts on the mass formula_9 by the gravitational force of the virtual effective mass formula_8 that is located in the intersection point of the trajectory of the mass formula_9 with the outer shell, we have to consider the varying distances formula_54 between formula_9 and the infinitesimal mass elements formula_36 along the arc: We finally get the effective mass formula_8 that is acting via gravitational forces on the mass formula_9 in the distance formula_10: Schwarzschild distance. The Schwarzschild distance formula_62 of this effective mass formula_8 is equal to the Schwarzschild radius of a sphere with the effective mass: The gravitational redshift formula_65 of photons emitted to the centre of the universe that is caused by the effective mass in the distance formula_10 of the outer shell is: The equations above can be solved for the Schwarzschild distance formula_62 of every shell with the mass formula_30. With the following condition we can get the solutions with formula_70, where formula_71: Therefore formula_62 can be determined for a given formula_20: The Schwarzschild distance formula_62 can be interpreted as the distance between the surface of the Hubble sphere with the Hubble radius formula_13 and the inner Schwarzschild sphere of the black shell, which is the visible limit of the universe. The Hubble radius formula_13 would be the distance between an observer in the centre of the universe and all objects that are receding from him at the speed of light. It can be expressed by the Hubble time formula_79: Results. It is assumed that the outer shell is expanding, spherical and consists of dark matter. The effective mass of the outer shell is computed only considering the geometrical region in front of the object. This assumption is based on the fact that the retarded gravitational potentials from the opposite border of the black shell can be neglected. The following diagrams show the solution in two different representations over the relation of the mass of the invisible universe surrounding the visible universe in a spherical shell in units of the mass of the visible universe formula_81: The overall mass of the invisible black shell formula_30 can be expressed in relation to the mass of the visible universe formula_86: → See appendix: table with results for different values of formula_20 numerically computed by a Java program. In this model the mass formula_30 of the invisible outer black shell continuously increases by absorbing mass from the visible universe, due to the net forces of the retarded gravitational potentials and the corresponding acceleration in direction of the black shell. More and more matter is moving from the visible universe behind the event horizon, where it becomes invisible and unaccessible, but leaves its gravitational action to the visible part of the universe. Furthermore, the mass of the black shell is in the same magnitude as the mass of the visible universe even in the earlier cosmical ages. It is very noteworthy to recapitulate that the age of the cosmic microwave background with its afterglow light pattern is about 380,000 years (the corresponding Schwarzschild distance belongs to a shell mass formula_90, Schwarzschild distance formula_91), exactly where the radius of the visible universe had significantly begun to decrease in relation to the Hubble radius (see right diagram). The so-called "dark ages" begun, and they lasted for several hundreds of million years. The age of the oldest known galaxy JADES.GS.z14-0 represents the youngest stars that emit light, and therefore, the end of the dark ages. Its age is about 290 million years (the corresponding Schwarzschild distance belongs to a shell mass formula_92, Schwarzschild distance formula_93), where the conversion from dark matter to "dark energy" during the so-called "dark ages" was finished. At the break-even point, where the masses of the visible universe and the black shell have the same value, the Schwarzschild distance is about 860 million light-years (it belongs to a shell mass formula_94). Schwarzschild radius of the universe. The Schwarzschild radius of the visible universe formula_95 can be computed by its mass, too: This value fits very well with the radius of the invisible particle horizon at the end of inflationary period of the universe today. The question is whether this coincidence is the case by accident or due to varying values of the mass of the visible universe formula_97 or the gravitational constant formula_98. Of course it is disputable, whether the mass of the visible universe and the gravitational constant are constant on large time scales. However, the British mathematician and astrophysicist Edward Arthur Milne (1896–1950) published already in 1935 in his book on "Relativity, Gravitation and World-Structure" the possibility that the gravitational constant could be proportional to the time since the creation of the universe formula_99 and reciprocal to the mass of the visible universe formula_97. According to Edward Arthur Milne the cosmological constant formula_102 is in this model: Today this value is approximately formula_104, which is about 6.6 times smaller than the current value of the standard model of cosmology known as the ΛCDM model (CDM = cold dark matter). The time-dependent relation for the gravitational constant would mean that the gravitational constant is grewing with the following rate, if the mass of the universe keeps constant: As a result the gravitational constant would change in the following magnitudes, which are much smaller than the uncertainty of the measurements of the gravitational constant of around formula_106: Only after half a million of years the corresponding change would be in the magnitude of the uncertainty. If we substitute the expression of Milne for the gravitational constant in the formula for the Schwarzschild radius of the universe, we get: Assuming that the whole universe is visible, and therefore, formula_109, this value would be two times larger than the value of the Schwarzschild radius computed by the mass of the universe: It is noteworthy to mention that according to the results the following estimates apply for Schwarzschild distances of around 858 million light-years, where the ratio of the mass of black shell to the mass of visible universe was equal to one: This Schwarzschild distance is similar to the differences of the particle horizon and the plane of the cosmic microwave background (CMB) that is determined by the ΛCDM model that is preferred by most cosmologists today: Therefore: This assumption is supported by the fact that the mass of ordinary matter within radius of the visible universe formula_115 has the following value that can be computed by its density formula_116 and the corresponding volume formula_117 yielded by the Planck Survey of 2013: However, if the gravitational constant would be given by the following relation using the total mass of the universe formula_122 including the black shell, the values would match much better: This value is identical to the radius of the particle horizon of our universe. In this case the change of the gravitational constant would be in the magnitude of the uncertainty only after a million years. The total mass of the universe would not change or influence the gravitational constant even if any mass flow from the visible part of the universe to the invisible outer black shell would decrease the mass of the visible part of the universe. The following applies for the hypothetical case that the Schwarzschild radius of the universe is only determined by the ordinary mass of the visible part of the universe, which is only half of the total mass of the universe. The Schwarzschild radius of the visible universe with the mass formula_125 that is completely visible to an observer in the centre of the universe, causes any light originating from the centre of the universe being invisible to any observer outside of this sphere, i.e. in the outer black shell with the similar mass formula_126. Vice versa, any light originating from the outer black shell is invisible to any observer in the centre of the universe. If the these two regions share the very same spherical boundary surface, this surface would separate two worlds that obviously cannot exchange any information. According to this assumption, the Schwarzschild radius of our universe would define this spherical boundary surface and would be around 860 million light-years smaller than the radius of the particle horizon respectively the outer radius of the black shell.

Moving objects in retarded gravitational potentials of an expanding spherical shell/Conclusion. Conclusion. The considerations described above show that retarded gravitational potentials can contribute to the explanation of the accelerated expansion of the visible part of the universe, if the outer spherical black shell of the universe would consist of expanding matter with a huge mass. This would also hold in a steady state universe and even if the black shell would not move at all. In the latter case any moving masses within the black shell would reach the black shell in finite time. It can also be considered that the effect of the retarded gravitational potentials will accelerate moving objects the more, the faster they move and the closer they are to the black shell. Furthermore, also objects within smaller spherical shells that not only are surrounded by the black shell, but also by other visible objects, will experience the retarded gravitational potentials of these objects, and therefore, they will experience an additional acceleration which is directed outwards. This all leads to an inhomogeneous evolution of the mass density in particular during the early development stages of an inflating universe: Computing the corresponding and permanently increasing Schwarzschild distances gives three remarkable points: Cosmologists call the period in between the creation of the cosmic microwave background and the first stars as the dark ages, where dark matter is supposed to be transformed into "dark energy". So far, the nature of "dark energy" is unclear, but the increase of the mass of an invisible black shell behind the visible universe could contribute to find explanations. Due to Newton's third law (actio = reactio) the black shell experiences the same retarded gravitational forces as the moving masses within the shell. These forces will cause a deceleration of the shell which is the stronger the closer the masses within the visible universe are to it. The high velocity of visible objects in the vincity of the black shell as well as the huge mass of the matter in front of them seem to be the reasons for the extreme redshift of their light that can be observed by us. The observation that this redshift is greater than expected for distant objects led to the assumption that the further away these objects are, the more they are accelerated. Another reason for this could be found in the effects of the retarded gravitational potentials of a massive black shell. Furthermore, the additional gravitational redshift leads to a higher value than expected only by the relativistic Doppler effect. This is in accordance with the observations of very far and young galaxies such as JADES-GS-z14-0 or of the cosmic microwave background (CMB). However, the computation of the equations of motion based on retarded gravitational forces becomes very expensive, if the black shell is neither homogeneous, nor spherical or symmetrical, or if the concept of spacetime has a structure based on a non-Euclidean geometry. Furthermore, relativistic effects (including the transverse Doppler effect) could cause time dilatation or mass increase that would have to be considered, too. Finally, is would be necessary to consider the loss of mass in the visible universe over the time, since a significant amount of visible matter may have traversed the event horizon that is caused and built by the huge mass in the black shell of the universe. If matter and antimatter were created in equal parts during the Big Bang, it would be conceivable that the visible universe consists of the remaining matter and the black shell of the remaining antimatter. This could be a reason for the obeserved baryon asymmetry. Since we cannot obtain any information from the black shell, antimatter would not be observable for us. Furthermore, any electromagnetic radiation from any source including the annihilation of matter and antimatter that leaves the visible universe by passing the limit of the Schwarzschild distance into the black shell cannot return to us.

Moving objects in retarded gravitational potentials of an expanding spherical shell/Conclusion/. The considerations described above show that retarded gravitational potentials can contribute to the explanation of the accelerated expansion of the visible part of the universe, if the outer spherical shell of the universe would consist of expanding dark matter with a huge mass. However, this would also hold in a steady state universe and even if the outer dark shell of the universe would not move at all. But in the latter case any moving mass in that shell would reach the shell in a finite time. It can also be considered that the effect of the retarded gravitational potentials will accelerate moving objects the more, the faster they move, and therefore, the closer they are to the spherical shell of dark matter. Furthermore, also objects within smaller spherical shells that not only are surrounded by expanding dark matter, but also by expanding visible objects will experience the retarded gravitational potentials of these visible objects, and therefore, will experience an additional acceleration which is directed outwards. Computing the corresponding Schwarzschild distances results in two remarable points in time where we today can observe the cosmic microwave background at a cosmic age of 380,000 years as well as the farest known galaxy at a cosmic age of 290 million years. Due to Newton's third law (actio = reactio) the shell experiences the same retarded gravitational forces as the moving masses within the shell. These forces will cause a deceleration of the shell which is the stronger the closer the masses within the shell are to it. The high velocity of visible objects near the outer dark rim of the universe as well as the huge mass of the expanding dark matter in front of them seem to be the reason for the extreme redshift of their light that can be observed. The observation that this redshift is greater than expected for distant objects led to the assumption that the further away these objects are, the more they are accelerated. One reason for this could be found in the effects of the retarded gravitational potentials of a massive dark outer shell. Furthermore, the additional gravitational redshift leads to a higher value than expected only by the relativistic Doppler effect, which is in accordance with the observations of very far and young galaxies such as JADES-GS-z14-0. However, the computation of the accelerations of moving objects due to retarded forces becomes extremely expensive, if the outer shell is neither homogeneous, nor spherical or symmetrical. Furthermore, relativistic effects could cause time dilatation and mass increase that would have to be considered, too. Finally, is would be necessary to consider the loss of mass in the visible universe over the time, since visible matter may traverse the event horizon that is caused by the huge mass of the outer shell of the universe.

Moving objects in retarded gravitational potentials of an expanding spherical shell/Preface. Preface. Abstract. The principles of retarded gravitational potentials are presented and explained. The application of retarded gravitational potentials to an expanding spherical shell of matter with a large mass (a "black shell") leads to acceleration forces on moving objects within such a huge shell. The concept of a Schwarzschild distance from the inner surface of a hollow spherical cap is introduced. It can be used to determine equivalent spherical masses which can provide information about the evolution of the universe. The unexpected high values of the cosmological redshift that are observed at extremely far astronomical objects including the cosmic microwave background could be related to the strong asymmetrical effect of the gravitation of the outer black shell of the universe. Due to the integrative effect of the retarded gravitational potentials of a concave shaped spherical black shell the gravitational forces of masses within the Schwarzschild distance are so enormous that the light from any atoms can no longer reach us. Exactly at the boundary to our visible universe, the gravitational redshift is infinite, just like on the surface of a convex shaped spherical black hole. The evolvement of the Schwarzschild distance due to the continuously increasing mass of the outer invisible shell of our universe has two inflection points around the very points in time and space, where the cosmic microwave background as well as the youngest galaxies can be observed today. The strongly increasing Schwarzschild distance between these two points in time goes hand in hand with the postulated massive transformation of dark matter to "dark energy". Therefore, the effect of retarded gravitational potentials could also contribute to the explanation of the accelerated expansion of the universe, which often is related to the effect of "dark energy". It seems that the hypothesis of relevant dark matter that continuously moves beyond the event horizon of our visible universe could open the gates for further interesting investigations and findings.

Moving objects in retarded gravitational potentials of an expanding spherical shell/Printable version. =Appendix=

Simple Guide to Minecraft/Enchanting. The enchanting process in Minecraft is one that can be seen as complicated. However, this page will help you crush the confusion that stands in between the clear window of level 30 enchanting. Enchanting is a simple thing that takes little practice (though practice does help) to master, even small youtube videos on this subject are very helpful and could answer many question that this article might fail to bring to subject. To start you will need a few late game items. One of the most noticeable on this list being the use of a Diamond Pickaxe. See the full list below: Gather your materials (listed above) to begin, if you don't have 15 bookshelves, skip the second step, if you do, continue as normal. The first step is to place down you're enchanting table/enchantment table on a flat space, it can be inside a house, on a flat turf, of floating in the middle of the End. The second step is to place your 15 bookshelves in the configuration shown below (Important note: make sure there are no blocks obscuring the view between the bookshelves and the enchanting table): And the third and final step is to place your Lapis Lazuli and your item in the enchanting table and click the enchant level you want. And there you go, you have enchanted for your first time.

AI Art Generation Handbook/IPAdapter. IP Adapter (Image Prompt Adapter) is Stable Diffusion extension for using images as prompts. They functions as to copy the style, composition, or a face from the reference image. It is pioneered by Tencent (WeChat developer) AI Lab as a way to achieve image prompt capability for the pre-trained text-to-image diffusion models. Here are some of the Local UI that already utilizes the IP Adapter References. [1] https://github.com/tencent-ailab/IP-Adapter

AI Art Generation Handbook/IPAdapter/ComfyUI. Note: We are focusing more on IPAdapter for SDXL models here: GO to Your_Installed_Directory/ComfyUI/custom_nodes/ and on the address bar , type codice_1 and inside terminal , type codice_2 codice_3 Go to the official huggingface SDXL to download the IP Adaptor Encoder (clip_vision) first at link below : https://huggingface.co/h94/IP-Adapter/tree/main/sdxl_models/image_encoder You should find a model clip_vision file named: model.safetensors and download it to directory Your_Installed_Directory/ComfyUI/models/clip_vision and renamed it exactly to codice_4 Go back up one level https://huggingface.co/h94/IP-Adapter/tree/main/sdxl_models and download all the models with .safertensors and .bins file type into this directory Your_Installed_Directory/ComfyUI/models/ipadapter (Create folder ipadapter if not present) It should be: 1) ip-adapter-plus-face_sdxl_vit-h.safetensors 2) ip-adapter-plus_sdxl_vit-h.safetensors 3) ip-adapter_sdxl.safetensors 4) ip-adapter_sdxl_vit-h.safetensors For the Unified Loader to works, the .bins files (Take note of FaceID not Plus version) are also needed to be downloaded from here https://huggingface.co/h94/IP-Adapter-FaceID/tree/main into this directory Your_Installed_Directory/ComfyUI/models/ipadapter 5) ip-adapter-faceid-portrait_sdxl.bin 6) ip-adapter-faceid_sdxl.bin 7) ip-adapter-faceid-plusv2_sdxl.bin 8) ip-adapter-faceid-portrait_sdxl_unnorm.bin Finally, we need to download the Lora for the FaceID to work Your_Installed_Directory/ComfyUI/models/loras 9) ip-adapter-faceid_sdxl_lora.safetensors 10) ip-adapter-faceid-plusv2_sdxl_lora.safetensors References. [1] https://github.com/cubiq/ComfyUI_IPAdapter_plus

Yoruba/Vocabulary. Roots. Most verbal roots are monosyllabic of the phonological shape CV(N), for example: dá (to create), dán (to polish), pọ́n (to be red). Verbal roots that do not seem to follow this pattern are mostly former compounds in which a syllable has been elided. For example: nlá (to be large), originally a compound of ní (to have) + lá (to be big) and súfèé (to whistle), originally a compound of sú (to eject wind) + òfé or ìfé (a blowing). Vowels serve as nominalizing prefixes that turn a verb into a noun form. Nominal roots are mostly disyllabic, for example: abà (crib, barn), ara (body), ibà (fever). Monosyllabic and even trisyllabic roots do occur but they are less common.

Bodybuilding and Weight Training/Dumbbell Flys. Dumbbell Flys Main Muscles Used: Chest Description: Dumbbell Flys, also known as Dumbbell Flyes, are a strength training exercise specifically designed to target the pectoral muscles, which are the primary muscles of the chest. This exercise is typically performed lying on a flat bench, though variations can include an incline or decline bench to target different areas of the chest. To perform Dumbbell Flys, you start by lying flat on the bench with your feet firmly planted on the ground for stability. Hold a dumbbell in each hand, with your palms facing each other and your arms extended straight above your chest. A crucial aspect of this exercise is maintaining a slight bend in the elbows throughout the movement to reduce strain on the joints. Safety Tips Mistakes to Avoid

Dialect/Central Yoruba (CY)/Igbomina. The Ìgbómìnà (also colloquially Igboona or Ogboona) are a subgroup of the Yoruba ethnic group, which originates from the north central and southwest Nigeria. They speak a dialect called Ìgbómìnà or Igbonna, classified among the Central Yoruba of the three major Yoruba dialectical areas. The Ìgbómìnà spread across what is now southern Kwara State and northern Osun State. Peripheral areas of the dialectical region have some similarities to the adjoining Ekiti, Ijesha and Oyo dialects. Idoba is on Owode Ofaro Land

Dialect/Central Yoruba (CY)/Ekiti. The Ekiti people are one of the largest historical subgroups of the larger Yoruba people of West Africa, located in Nigeria. They are classified as a Central Yoruba group, alongside the Ijesha, Igbomina, Yagba and Ifes. Ekiti State is populated exclusively by Ekiti people; however, it is but a segment of the historic territorial domain of Ekiti-speaking groups, which historically included towns in Ondo State such as Akure (the current capital and largest city of Ondo State), Ilara-Mokin, Ijare, and Igbara-oke. Ogbagi, Irun, Ese, Oyin, Igasi, Afin and Eriti in the Akoko region, as well as some towns in Kwara State, are also culturally Ekiti, although belong in other states today.

Dialect/Central Yoruba (CY)/Akoko. The Akoko are a large Yoruba cultural sub-group in the Northeastern part of Yorubaland. The area spans from Ondo State to Edo State in southwest Nigeria. The Akokos as a subgroup make up 20.3% of the population of Ondo State, and 5.7% of the population of Edo State. Out of the present 18 Local Government Councils it constitutes four; Akoko North-East, Akoko North-West, Akoko South-East and Akoko South-West, as well as the Akoko Edo LGA of Edo State. The Adekunle Ajasin University, a state owned university with a capacity for about 20,000 tertiary education students and more than 50 departments in seven faculties is located in Akungba-Akoko. A state specialist hospital is situated at Ikare Akoko, while community general hospitals are located in Oka-Akoko and Ipe-Akoko.

Dialect/South-West Yoruba (SWY)/Ketu. Ketu is the name of a Yoruba subgroup, historical kingdom and region straddling parts of what is now southeastern Republic of Benin and parts of southwest Nigeria. The chief town and traditional capital of the area was the town of Kétou (Kétu), which is considered to be one of the oldest capitals of the Yoruba-speaking people, tracing its establishment to a settlement founded by a descendant of Oduduwa (also known as; Odùduwà, Oòduà and Eleduwa) called Sopasan or Soipasan.

Dialect/South-West Yoruba (SWY)/Ifè (Togo). Ifè (or Ifɛ) is a Niger–Congo language spoken by some 180,000 people in Togo, Benin and Ghana. It is also known as Ana, Ana-Ifé, Anago, Baate and Ede Ife. It has a lexical similarity of 87%–91% with Ede Nago. Written works began to be produced in the language in the 1980s, published by the Comité Provisoire de Langue Ifɛ̀ and SIL. An Ifè–French dictionary (Oŋù-afɔ ŋa nfɛ̀ òŋu òkpi-ŋà ŋa nfãrãsé), edited by Mary Gardner and Elizabeth Graveling, was produced in 2000.

Dialect/South-East Yoruba (SEY)/Ikale. Okitipupa is in Nigeria and part of the Ikale-speaking nation in Ondo State. Okitipupa Government headquarter is located in Okitipupa Major town with a university, Olusegun Agagu University of Science and Technology (OAUSTECH) which commenced academic sessions in 2010–11. Ìkálè or Old Ìkálè is part of the Yoruba tribe of Ondo state in Nigeria which was originally Ikaleland and combination of the present Okitipupa Local Government and Irele Local Government before the two local governments were split into two namely: Ìrèlè local government and Okitipupa local government.

Dialect/South-East Yoruba (SEY)/Akoko (Akoko, Ào). The Akoko are a large Yoruba cultural sub-group in the Northeastern part of Yorubaland. The area spans from Ondo State to Edo State in southwest Nigeria. The Akokos as a subgroup make up 20.3% of the population of Ondo State, and 5.7% of the population of Edo State. Out of the present 18 Local Government Councils it constitutes four; Akoko North-East, Akoko North-West, Akoko South-East and Akoko South-West, as well as the Akoko Edo LGA of Edo State. The Adekunle Ajasin University, a state owned university with a capacity for about 20,000 tertiary education students and more than 50 departments in seven faculties is located in Akungba-Akoko. A state specialist hospital is situated at Ikare Akoko, while community general hospitals are located in Oka-Akoko and Ipe-Akoko.

Dialect/South-East Yoruba (SEY)/Ilaje. Ìlàje is a Local Government Area in Ondo State, South-West Nigeria. Its headquarters are in the town of Igbokoda. The Ilajes are a distinct migratory coastal linguistic group of Yoruba peoples spread along the coastal belts of Ondo, Ogun, Lagos and Delta states and originally made up of four geo-political entities, namely: Ode Ugbo, Ode Mahin, Ode Etikan and Aheri. While most towns and villages in the Mahin kingdom (Ode Mahin) are distributed on arable lands, the towns and villages in the other three geo-politics of Ugbo, Aheri and Etikan kingdoms are spread out along the beaches and swampy terrain of the Atlantic Ocean coast.

Dialect/South-East Yoruba (SEY)/Usẹn. Ode Usen, also known as Ufe kekere and Ode Awure is the name of a small town in Edo state, Nigeria. It also doubles as the name of a Yoruba subgroup consisting of culturally related villages situated between Ofosu in the west and Ogbese in the east. Usen is surrounded on all sides by smaller villages and farmsteads known in the local Yoruba dialect as Egunre. some of the villages under the authority of Usen include; Arere, Oladaro, Arekpa, Ogunweyin, Ogidigbo, Ilorin (Ulorin), Ukankan, Ajegunle, Obome, Aghakpo, Leleji, Ofaran, Okeodo, Adeyanba. These villages were all founded by people from Usen and speak the same dialect of the Yoruba language. There are also other communities that speak the same dialect of the Yoruba language as Usen, such as; Egbeta, Utese and Igue Ogho between Ekiadọlọ and Usen.

Bodybuilding and Weight Training/Deadlift. Deadlift Main Muscles Used: Lower back, Hamstrings Description: The deadlift is a fundamental compound exercise that targets multiple muscle groups, primarily focusing on the lower back, glutes, hamstrings, and core. It is highly regarded for its ability to build strength, improve posture, and enhance overall athletic performance. It is performed by lifting a loaded barbell from the ground to hip level while maintaining a neutral spine throughout the movement. The deadlift is highly effective for building overall strength and muscle mass due to its multi-joint nature and ability to handle heavy loads. It enhances functional strength, improving performance in everyday activities and athletic endeavors. Safety Tips Mistakes to Avoid

Bodybuilding and Weight Training/Squats. Squats Main Muscles Used: Quadriceps, Hamstrings, Glutes, and Calves Description: Squats are a foundational exercise in strength training, highly regarded for their ability to build lower body strength, enhance muscle hypertrophy, and improve overall functional fitness. They primarily target the quadriceps, hamstrings, glutes, and calves while also engaging the core, lower back, and even the upper body to a certain extent, depending on the variation performed. Safety Tips Mistakes to Avoid

Character Encodings/Code Tables/EBCDIC/EBCDIC 340. IBM code page 340 (CCSID 340) is an EBCDIC code page used in IBM mainframes for optical character recognition. It has been superseded by EBCDIC 892 and EBCDIC 893. Codepage layout. Characters are shown with their equivalent Unicode codes. Note that the grave, acute, circumflex (at 0xDF), tilde, diaeresis, and cedilla con be added over (in the case of the cedilla, under) letters to form accented letters. Characters not in Unicode:

Chess Opening Theory/1. e4/1...e5/2. Nf3/2...Nc6/3. Nxe5. Irish Gambit / Chicago Gambit. 3. Nxe5?? The Irish Gambit is an unsound opening. It trades a knight for a pawn, since Black can just recapture with 3...Nxe5. While this move does let White have a stronger center, the cost of a knight is too much to pay. Still, like other dubious openings, it can be used to surprise or unsettle the opponent.

Chess/Bowdler Attack. This is a mistake that allows clear equality. Black's best answer is 2. ...e6 followed by d5 at some point which easily removes White's bishop and gives Black a tempo and clear centre. For example: 1. e4 c5 2.Bc4 e6 3. Nc3 Nc6 4. Nf3 Nf6 5. O-O d5 6. exd5 exd5 7.Bb3 Black needs to play Nc6 before d5, otherwise the king will be exposed on the a4-e8 diagonal, and therefore inviting a check from White's light square bishop. It is analogous to some lines in the Italian Game but Black has played the move c5 which helps prevent d4 and makes b3 a less attractive place for the White Bishop (because of c4). Interestingly, 1. e4 c5 2. Nf3 d6 3. Bc4 (as opposed to the usual 1. e4 c5 2. Nf3 d6 3. d4 cxd4 4. Nxd4 Nf6 5. Nc3 XXX 6. Bc4 where XXX is e6, Nc6, g6 or a6), does not allow for such shenanigans by Black who has no better option but to transpose to a main line open Sicilian where White plays Bc4. In contrast to 2. Bc4 where Black obtains immediate equality, the only real advantage of 3. Bc4 for Black is that he gets to choose his defence (Dragon, Najdorf, Scheveningen, Classical) with the knowledge that White has committed his bishop to c4. However it may confuse some unwary Black players who feel they should "prove" that White has made an "inaccurate" move.

Saluto, Jonathan! (Ido)/Chapitro 14. Dek-e-quaresma chapitro. Jonathan eniras sua chambro, sideskas sur la lito, e pensas. Esas nokto, e lu vidas la luno. La luno hodie esas tre bela, e lu vidas exter la fenestro. Lu vidas la monti, e la vali. La kozi esas bela, e lia beleso (beleso es esar bela) felicigas lu. Lu pensas: “Me esas en karcero, yes, me es karcerano, ma kande me vidas la monti e la vali me esas felica.” Pose lu vidas stranja kozo, e dicas: “Oh! Quo es to?” Jonathan vidas infre, e lu vidas Dracula. “Haha! Anke Dracula vidas exter la fenestro, quale me! Ka lu volas vidar la luno tra la fenestro?” lu pensas. “Ma… mea Deo! Quon lu facas?” Jonathan vidas Dracula e lua manui. La manui di Dracula prenas la muro, e lu komencas decensar. Yes, forta viri povas klimar muro, ma Dracula klimas lo kun sua kapo infre! Dracula vidas adinfre, decensas la muro, ed iras plu e plu fore. Nun Jonathan ne povas vidar la Komto Dracula. “Mea Deo! Ta Dracula ne es viro! Quo lu es, takaze?” lu pensas. Sequantadie, Jonathan vidas Dracula ankorfoye, e lu vidas ke Dracula decensas la muro. Jonathan savas ke lu ekiris la kastelo, e savas ke lu es sola: do lu komencas serchar kozi. Jonathan serchas tra la kastelo. Lu vidas omna pordi: klozita e klefagita, klozita e klefagita, klozita e klefagita. Omni klozita e klefagita! Jonathan pensas ke Dracula devas havar klefo, nam onu klefagas e apertas pordi per klefo. Jonathan trovis kelka mikra pordi klozita ma ne klefagita, e tin lu povis apertar sen klefo: lu uzis sua manui por apertar li. Ma dop la mikra pordi esis nulo. Pos to lu vidis pordo qua esis klozita, e semblis klefagita, ma ne esis klefagita – ol esis nure anciena! Lu apertis ol per la forteso di sua manui, ne per klefo. Nova loko por vidar! Quo es dop la pordo? Dop la pordo esas multa chambri, kun multa granda fenestri. Jonathan pensas ke ca chambri esas plu granda e plu bela, e ke dum antiqua tempi, mulieri certe vivis en li. Hike, Jonathan povas bonege vidar la luno. Lu pensas pri la antiqua tempi en ca kastelo, kande mulieri vidis la suno e la luno tra la granda fenestri. Dum kelka tempo, Jonathan es felica, nam lu trovis altra chambri e sentas su kelke plu libera. Jonathan rimemoras to quo Dracula dicis, ke Jonathan devas ne dormar en altra chambro. Jonathan memoras lo; lu ne oblivias lo (obliviar = ne memorar). Ma… Jonathan es felica en ca chambri e felica vidante la luno, e lu pensas ke forsan to quo Dracula dicis ne es importanta. Lu pensas pri esar libera, e pensas pri dormar en ca chambri ube la kozi esas plu bela, ube lu povas pensar pri bona kozi: mulieri, beleso, monti e vali… Jonathan klozas sua okuli e dormeskas. Gramatiko. ci, ti. Ci, ti = plu kam un ‘co’ o ‘to’ infre, adinfre. infre, adinfre = ⬇️, ma on uzas ‘adinfre’ se esas movo. p.ex:

Saluto, Jonathan! (Ido)/Chapitro 15. Dek-e-kinesma chapitro. Dormante, Jonathan apertas sua okuli. Ka lu ankore dormas? Lu ne savas. Ma lu savas ke lu ne esas sola. En la chambro esas tri altra personi: tri mulieri. La tri mulieri en la chambro vidas Jonathan, e parolas inter su. Du mulieri havas nigra hararo, e un muliero havas orea hararo. Eli es tre granda, e belega. Jonathan vidas la triesma muliero e pensas: “Ka me konocas ta muliero?” (konocar es savar qua es persono) Eli es bela, ma Jonathan ne esas felica; lu timas. Ma lu volas ke eli kisez lu. Lu pensas pri su amorato Mina e ne es felica, nam lu pensas ke lu volas kisar altra muliero. Ma lo esas vera… La tri mulieri parolas pluse inter su. Pose eli ridas. Elia ridado es tre bela; ol es belega. Ma ol es anke ridado tre mala, ridado tro forta por persono. Ka eli es vere personi? La unesma muliero dicas al triesma muliero: “Irez! Irez! Tu esos la unesma. Ni sequos tu.” La duesma muliero dicas al triesma muliero: “Il es yuna, e forta. Tu povas kisar il, ma anke ni povas kisar il, pro ke il es yuna e forta. Forta viro povas kisar omna ni.” La triesma muliero venas adproxim Jonathan, e Jonathan apertas sua okuli kelke. Lu volas ke la muliero kisez lu… “No! Pensez pri Mina!” Ma lu volas lo. Ma samatempe lu timas el. Lu volas kisar el, e ne volas kisar el, e lu ne savas quon facar. Ka lu konocas el? Ka lu vidis el, ulatempe dormante? La muliero venas plu e plu adproxim, e nun elua boko es tre proxim al kolo di Jonathan. Lu sentas la denti dil muliero an sua kolo. Ma ye la sequanta instanto lu sentas altra kozo: Dracula venis! La manuo di Dracula prenas la kolo dil muliero e jetas el. Jonathan vidas Dracula e timas: lu nulatempe vidis Dracula tante forta. La okuli di Dracula es quale flami, e lua vizajo es mem plu blanka. Nun il jetas la altra mulieri kontre la muro. Jonathan nun rimemoras la duktisto kontre la volfi, e la gesto quan lu facis per sua manui. La forteso di lua manui! Lo esas grandega. Dracula dicas: “Mulieri, me dicis ke vi ne povas prenar il per via manui! Ta viro es mea!” La triesma muliero ridas e dicas: “Vu savas nulo pri amoro! Vu nulatempe amorabas!” E la tri mulieri ridas e ridas. La ridado dil tri mulieri es, segun Jonathan, quale la ridado di demoni. Dracula vidas Jonathan, e vidas la mulieri, e dicas: “Yes, me povas amorar. Anke me havas historio. Ka vi ne savas lo? Bone, pos ke me facabas to quo me volas kun ca viro, vi povas kisar il. Ma ne hodie! Il dormas, e me devas vekigar il. Ni havas kozi por facar.” Un ek la mulieri dicas: “Takaze, canokte? Quon ni manjas canokte? Me vidas sako ibe; takaze, ka ni darfas havar ol?” Dracula dicas nulo, ma facas gesto per sua kapo. La tri mulieri, felice, prenas la sako e apertas ol. Jonathan kredas ke lu audas voco ekiranta la sako. "Mea Deo, ka esas vivanta persono interne di la sako?" pensas Jonathan. Ma nun la mulieri e la sako esas fore, e Jonathan savas nulo plusa. Omno es nigra e lu ne savas pluse. Gramatiko. an. an = tre proxima; tante proxima ke la du kozi tushas. p.ex: On sidas an tablo. eli, ili, oli. On povas sempre dicar ‘li’ vice eli/ili/oli, exakte quale on povas dicar ‘lu’ vice el/il/ol (o elu/ilu/olu). -ar-. -ar- = kolektajo de kozi qui es la radiko. p.ex: "haro" es un haro, ma "hararo" es omna hari del kapo. -e-. -e- = havanta la koloro di. p.ex: -ez. Quale vu ja vidis, -ez uzesas por donar instrucioni (p.ex. "Vartez!"). -ez anke uzesas en frazi quale "Me volas ke vu kisez me."

AI Art Generation Handbook/Archive/Settings. This is another examples of generating realistic human figures : Correlations between settings. The picture shows the correlations between CFG and Sampling Steps. It is note worthy to look that steps that are too low and with high CFG values will almost caused the images to become distorted. The CFG that are too low will make the picture appeared washed out. References. https://www.felixsanz.dev/articles/complete-guide-to-samplers-in-stable-diffusion#what-is-the-best-sampler-in-stable-diffusion

AI Art Generation Handbook/Node Prompting in Stable Diffusion/ComfyUI Custom Output Name. To create a custom output name is easy , the format is usually goes like this codice_1 Here are 3 most common ways to customise the prompt name (1) K-Sampler. Original Node name for S&R : codice_2 (2) Text Prompt Original Node name for S&R : codice_3 (3) Empty Latent Image Original Node name for S&R : codice_4

Cherokee/Structural Nuggets. These "structural nuggets" refer to elements in the Cherokee language categorized as units of "structure," as opposed to content words like nouns, for example. As such, the words or word parts found here do not solely belong to one part of speech. These "nuggets" are like the glue holding much of the language together! Each nugget is presented in a boldface header, followed by a line defining it. Uppercase abbreviations such as ADJ for "adjective" are used to represent variable words or word parts, while lowercase Cherokee text represents invariable elements within the nugget. Following the definition, you may find examples. At the bottom of each entry you may find questions whose answers are unclear to the writers of this book. We empower you to ask these questions to native speakers of Cherokee and to document the answers in the Discussion section for this page, found at the top of the screen to the right of the tab that says "Book." You are always encouraged to ask for and document "other" ways to express the same idea for "every" nugget. Abbreviations. ADJ = adjective S = subject S1 = subject number 1 V = verb V1 = verb number 1 CSP = citation for "Cherokee Speech Patterns" by JW Webster Related to Adjectives. ADJ + jigi. the (possibly also "a") ADJ onegiígágẽ j-igi red REL-be "the red one" CSP 22Questions: How do you say "the red ones" or "some red ones"? ADJ -yú. -ly One way to form adverbs from adjectives, equivalent to the English suffix -ly, is to add -yú to the end of the adjective. If the adjective ends in -i, the /i/ lengthens and gets a rising tone in anticipation of the high tone on the suffix -yú.uusganõli > uusganõliíyú slow > slowly CSP 23 Related to Clauses. S1 agvvyĩ V1 si ni- V2 -ṽna(wú) S2. S1 V1 before S2 V2 One way to express that an action happens before another action is to use this structure.maági agvvyĩ uùhnigíisééqi si núúluhjṽnawú uweej ageéhya Mark first 3SG.B-left still without-3SG.B-arriving 3SG.B-child woman "Mark left before his daughter arrived." CSP24

Ada Programming/Input Output/Text Example. Introduction. Those of us with American friends often have to deal with the Fahrenheit temperature scale. While the rest of the world has adopted the Celsius temperature scale, America and a few other countries have stayed true to Fahrenheit. While this is both a small and insignificant problem when dealing with friends, it is nevertheless a problem worthy of having its own program: A Fahrenheit to Celsius converter. The necessary files and directories. Before we start, we need to do a little preparation. First create the following directory/file structure somewhere on your computer: $ mkdir FtoC $ cd FtoC $ mkdir exe objects $ touch ftoc.adb ftoc.gpr The above should leave you with the following contents in the codice_1 directory: exe/ objects/ ftoc.adb ftoc.gpr We will use these two directories and files in the following manner: The Project File. Add this to the codice_3 file: project ftoc is for Source_Dirs use ("."); for Main use ("ftoc.adb"); for Exec_Dir use "exe"; for Object_Dir use "objects"; package Ide is for Compiler_Command ("ada") use "/usr/gnat/bin/gnatmake"; end Ide; package Compiler is Common_Options := ("-gnatwa", "-gnaty3abcdefhiklmnoprstux", "-Wall", "-O2"); for Default_Switches ("Ada") use Common_Options; end Compiler; end ftoc; Please go here for an explanation on the different parts of the above project file. The Actual Program. With the project file out of the way, we turn our attention to the actual codice_2 code. Add this to the codice_5 file: with Ada.Text_IO; use Ada.Text_IO; with Ada.Integer_Text_IO; use Ada.Integer_Text_IO; with Ada.Float_Text_IO; use Ada.Float_Text_IO; procedure FtoC is subtype Fahrenheit_Degree_Range is Natural range 0 .. 212; -- Now, that is a Natural Range for Ada, but I can tell you that Americans -- are not nearly so happy with "0" and not even an Icelander likes it at "212"! Fahr : Fahrenheit_Degree_Range := Fahrenheit_Degree_Range'First; Factor : constant := 5.0 / 9.0; Offset : constant := 32; Step : constant := 1; begin loop Put (Item => Fahr, Width => Fahrenheit_Degree_Range'Width); Put (Item => Factor* Float (Fahr - Offset), Fore => 4, Aft => 2, Exp => 0); New_Line; exit when Fahr = Fahrenheit_Degree_Range'Last; Fahr := Fahr + Step; end loop; end FtoC; Save the file, and lets move on to the final step: Compiling. Compile And Run FtoC. If you use the excellent GNAT Studio IDE, the compiling is a simple matter of pressing codice_6 to compile the project, and codice_7 to execute it. If you're not using this IDE, then do this instead: $ gnatmake -P ftoc.gpr You should see some output scroll by: gcc -c -gnatwa -gnaty3abcdefhiklmnoprstux -Wall -O2 -I- -gnatA /home/thomas/FtoC/ftoc.adb gnatbind -I- -x /home/thomas/FtoC/objects/ftoc.ali gnatlink /home/thomas/FtoC/objects/ftoc.ali -o /home/thomas/FtoC/exe/ftoc You now have an executable in the codice_8/ directory. When you run it, you get this (somewhat abbreviated here): 0 -17.78 1 -17.22 2 -16.67 26 -3.33 27 -2.78 28 -2.22 29 -1.67 30 -1.11 31 -0.56 32 0.00 33 0.56 34 1.11 35 1.67 48 8.89 49 9.44 50 10.00 51 10.56 52 11.11 67 19.44 68 20.00 69 20.56 70 21.11 71 21.67 84 28.89 85 29.44 86 30.00 87 30.56 210 98.89 211 99.44 212 100.00 Now we know that 0 degrees Fahrenheit equals -17.78 Celsius and at 212 degrees Fahrenheit water boils, which to us Celsius users is known as 100 degrees Celsius. The program works! Lets go over it in detail, to figure out how it works. Note: All output in the following examples will be heavily abbreviated, because printing 213 lines of output for every little example is madness. How Does The FtoC Program Work? Lets start with the first three lines: with Ada.Text_IO; use Ada.Text_IO; with Ada.Integer_Text_IO; use Ada.Integer_Text_IO; with Ada.Float_Text_IO; use Ada.Float_Text_IO; These are called codice_9 clauses and codice_10 declarations. An Ada codice_9 clause can be likened to a C codice_12. When you see codice_13, it means the utilities of the codice_14 package are being made available to the program. When next you see the declaration codice_15, it means that the utilities of codice_14 are directly visible to the program, ie. you do not have to write Ada.Text_IO.Put ("No use clause"); to call the codice_17 procedure. Instead you simply do Put ("With use clause"); In the FtoC program we utilize, and make visible, three packages: codice_14, codice_19 and codice_20. These packages provide exactly what their names imply: Text IO capabilities for different types. Looking at the source, you might wonder where the need for codice_14 comes in, as we're only outputting numbers. We will get to that in a minute, so stay tuned. The FtoC declarations. Next we have this code: procedure FtoC is subtype Fahrenheit_Degree_Range is Natural range 0 .. 212; Fahr : Fahrenheit_Degree_Range := Fahrenheit_Degree_Range'First; Factor : constant := 5.0 / 9.0; Offset : constant := 32; Step : constant := 1; begin This is called the declarative part of an Ada program. It is here we declare our variables, constants, types and whatnot. In the case of FtoC we have 5 declarations. Lets talk a bit about the first two: subtype Fahrenheit_Degree_Range is Natural range 0 .. 212; Fahr : Fahrenheit_Degree_Range := Fahrenheit_Degree_Range'First; That right there is one of the biggest selling points of Ada: The ability to create your own types, with your own constraints. It might not seem like a big deal but, believe me, it is. In this case we've created a new subtype of the built-in subtype Natural (). We've constrained the type's range to codice_22, meaning that objects declared as a codice_23 can never go below 0 or above 212. If a value outside this range is assigned to an object of the codice_23 type, a codice_25 exception is raised. With the codice_23 type in place, we direct our attention to the declaration and assignment of the codice_27 variable. If you've never seen this syntax before or it makes no sense to you, please read the Constants article on this Wiki, and then return here. The codice_27 variable is of the codice_23 type and it's initial value is 0. Why 0? Because we assign it the value codice_30, and the codice_31 part equals the first value in the range constraint of the type, in this case 0. Consequently the value for codice_32 is 212. Let's take a look at the final three object declarations: Factor : constant := 5.0 / 9.0; Offset : constant := 32; Step : constant := 1; What's going on here? Why are there no codice_33 associated with any of those declarations? Well, if you've read the Constants article, you will know that these constants are called named numbers and their codice_33 is called an universal type (). Named numbers can be of any size and precision. The codice_35 and codice_36 constants are used in the Fahrenheit-to-Celsius calculation, and codice_37 define how many conversions we do in the program. The FtoC body. With the declarations out of the way, we turn our attention to the body of the program: begin loop Put (Item => Fahr, Width => Fahrenheit_Degree_Range'Width); Put (Item => Factor* Float (Fahr - Offset), Fore => 4, Aft => 2, Exp => 0); New_Line; exit when Fahr = Fahrenheit_Degree_Range'Last; Fahr := Fahr + Step; end loop; end FtoC; The reserved word codice_38 signifies the beginning of the codice_39 and that same body ends with the final codice_40. Between those two, we have a bunch of statements. The loop. The first one is the codice_41 statement. Loops come in many different shapes in Ada, each of which obeys the basic premise of loop -- some statements end loop; Loops can be terminated using the codice_42 keyword: loop -- some statements if X = Y then exit; end if; end loop; This construction is so common that a shorter version has been made available: loop -- some statements exit when X = Y; end loop; It is this last version we use in the FtoC program and we exit the loop when codice_27 equals the codice_44 value of the codice_23 type. FtoC output - putting integers on the screen. Immediately after the codice_41 statement we encounter the codice_17 procedure: Put (Item => Fahr, Width => Fahrenheit_Degree_Range'Width); We know from the declaration that codice_27 is a subtype of codice_49, which in turn is a subtype of codice_50, so the call to codice_17 on this line actually calls codice_52. As you can see, we give codice_17 two parameters: codice_54 and codice_55. The meaning of codice_54 should be obvious: It's the integer we want to output, in our case the codice_27 variable. codice_55 on the other hand does not make as much sense. The codice_55 parameter gives the minimum number of characters required to output the integer type or, in our case, subtype, as a literal, plus one for a possible negative sign. If the codice_55 parameter is set too high, the integer literal is padded with whitespace. If it's set too low, it is automatically expanded as necessary. Setting the codice_55 parameter to 0, results in the field being the minimum width required to contain the integer. The codice_62 attribute returns the maximum width of the type, so if we change the codice_23 later on, we wouldn't have to do a single thing about this call to codice_17; it would simply adjust itself accordingly. Let's do some tests with various codice_55 parameters: Put (Item => Fahr); -- Default Width parameter The output now looks like this: 0 -17.78 1 -17.22 2 -16.67 3 -16.11 4 -15.56 Lots of wasted space there. This is because codice_17 now sets aside space for the codice_50, which is the type codice_23 is derived from. Let's try with 0: Put (Item => Fahr, Width => 0); -- Minimum required characters for the integer And the output: 0 -17.78 1 -17.22 2 -16.67 3 -16.11 9 -12.78 10 -12.22 11 -11.67 99 37.22 100 37.78 101 38.33 Unfortunately for readability, the codice_27 integer literal is no longer right-justified. Each number is given the exact width necessary to hold it and no more. Let's try it with a codice_55 that is wide enough for some of the codice_27 values, but not all of them: Put (Item => Fahr, Width => 2); -- Minimum width of 2. Expands if necessary This outputs: 0 -17.78 1 -17.22 2 -16.67 9 -12.78 10 -12.22 11 -11.67 98 36.67 99 37.22 100 37.78 101 38.33 102 38.89 103 39.44 As you can see, the single digit values are right-justified and padded with 1 space, the two-digit values come out even, but the rest of the results are expanded to hold the third character. FtoC output - now with floats. With codice_17 for integer types out of the way, we move on to the next codice_17> Put (Item => Factor* Float (Fahr - Offset), Fore => 4, Aft => 2, Exp => 0); The codice_54 parameter for this call to codice_17 is a float, because codice_35 is a named number that contains a decimal point and the expression codice_77 is converted to a float using the codice_78 expression. So when calling this codice_17, we're actually calling codice_80. The codice_81 parameter gives the minimum character count necessary to output the value preceding the decimal point. As with codice_55 for the integer types, codice_81 will automatically expand if necessary. codice_84 sets the precision after the decimal point, in this case 2. And finally codice_85 sets the exponent field size. A value of codice_86 signifies that no exponent will be output. Anything other than zero will output the exponent symbol "E", a +/-, and the digit(s) of the exponent. Note: The value of codice_85 should not be less than zero! Let's try a few different combinations: Put (Item => Factor* Float (Fahr - Offset), Fore => 10, Aft => 4, Exp => 0); The output: 0 -17.7778 1 -17.2222 15 -9.4444 16 -8.8889 17 -8.3333 26 -3.3333 27 -2.7778 100 37.7778 101 38.3333 102 38.8889 103 39.4444 104 40.0000 Or how about this: Put (Item => Factor* Float (Fahr - Offset), Fore => 4, Aft => 2, Exp => 1); And the output: 0 -1.78E+1 1 -1.72E+1 2 -1.67E+1 8 -1.33E+1 9 -1.28E+1 10 -1.22E+1 11 -1.17E+1 18 -7.78E+0 37 2.78E+0 98 3.67E+1 99 3.72E+1 100 3.78E+1 101 3.83E+1 And finally: Put (Item => Factor* Float (Fahr - Offset), Fore => 0, Aft => 1, Exp => 0); This outputs: 0-17.8 8-13.3 9-12.8 10-12.2 11-11.7 31-0.6 320.0 499.4 9836.7 9937.2 10037.8 10138.3 10238.9 Which obviously isn't very pretty to look at. A new line, an exit strategy, and a step. The last four lines of the FtoC program finish our formatting, get us out of here if we are done, and, if not, set the next value of codice_27 to be converted: New_Line; exit when Fahr = Fahrenheit_Degree_Range'Last; Fahr := Fahr + Step; end FtoC; The single call to codice_89 is the reason we have to make codice_14 available to the FtoC program using codice_13. What codice_89 does is output a single line feed. Running the program without this call to codice_89 would result in output looking like this: 0 -17.78 1 -17.22 2 -16.67 3 -16.11 4 -15.56 5 -15.00 6 -14.44 7 -13.89 8 -13.33 9 -12.78 10 -12.22 11 -11.67 ... The codice_89 procedure accepts a codice_95 parameter, meaning you can do this to output consecutive line feeds: New_Line (Spacing => 5); Or simply New_Line (5); We've already discussed the codice_96 method of terminating a loop, and the final statement is merely a simple counter. The value of codice_27 is incremented with codice_37 on each iteration of the loop. When codice_27 equals codice_32, the loop is terminated. The final line, codice_101, signifies the end of the program. There's nothing more to do and nothing more to see. Control is handed back to whatever called the program in the first place, and life goes on. Conclusion. I hope you've enjoyed this little tutorial on building a Fahrenheit to Celsius conversion table. It is left to the reader to figure out how to add Kelvin to the mix. Have fun!

Ada Programming/Libraries/GNAT.String Split. Introduction. Exploding a string into several components based on a set of separators can be done in many different ways. In this article we're going to focus on a solution involving the codice_1 package. Caveat. If you use the following example in a program of your own, the result will be a less portable program. The GNAT packages are only found in the [GPL] and in the [GCC GNAT] compilers, meaning that your program probably won't compile with other Ada compilers. The Problem. You want to split a string into a set of individual components, such as This is a string into This is a string And this is exactly what you can do with the codice_1 package. The GNAT.String_Split Solution. Let's dive straight into the code necessary to solve our string split problem. Create a file named codice_3 and add this to it: -- A procedure to illustrate the use of the GNAT.String_Split package. This -- is just the simplest, most basic usage; the package can do a lot more, like -- splitting on a char set, re-split the string with new separators, and -- return the separators found before and after each substring. Left as an -- exercise for the reader. ;) with Ada.Characters.Latin_1; with Ada.Text_IO; with GNAT.String_Split; procedure Explode is use Ada.Characters; use Ada.Text_IO; use GNAT; Data : constant String := "This becomes a " & Latin_1.HT & " bunch of substrings"; -- The input data would normally be read from some external source or -- whatever. Latin_1.HT is a horizontal tab. Subs : String_Split.Slice_Set; -- Subs is populated by the actual substrings. Seps : constant String := " " & Latin_1.HT; -- just an arbitrary simple set of whitespace. begin Put_Line ("Splitting '" & Data & "' at whitespace."); -- Introduce our job. String_Split.Create (S => Subs, From => Data, Separators => Seps, Mode => String_Split.Multiple); -- Create the split, using Multiple mode to treat strings of multiple -- whitespace characters as a single separator. -- This populates the Subs object. Put_Line ("Got" & String_Split.Slice_Number'Image (String_Split.Slice_Count (Subs)) & " substrings:"); -- Report results, starting with the count of substrings created. for I in 1 .. String_Split.Slice_Count (Subs) loop -- Loop though the substrings. declare Sub : constant String := String_Split.Slice (Subs, I); -- Pull the next substring out into a string object for easy handling. begin Put_Line (String_Split.Slice_Number'Image (I) & Sub & " (length" & Positive'Image (Sub'Length) & -- Output the individual substrings, and their length. end; end loop; end Explode; You compile and execute the codice_4 program like this: $ gnatmake explode.adb $ ./explode You should see output similar to this: Splitting 'This becomes a bunch of substrings' at whitespace. Got 6 substrings: 1 -> This (length 4) 2 -> becomes (length 7) 3 -> a (length 1) 4 -> bunch (length 5) 5 -> of (length 2) 6 -> substrings (length 10) The comments in the example should more or less explain what's going on, but for the sake of clarity, we're going to do a step-by-step walk-through of the code, starting with the dependencies and codice_5 clauses: with Ada.Characters.Latin_1; with Ada.Text_IO; with GNAT.String_Split; procedure Explode is use Ada.Characters; use Ada.Text_IO; use GNAT; The three codice_6 lines list the packages on which our program depends. When the compiler encounters these, it retrieves those packages from its library. The "//Procedure Explode is//" line marks the start of our program, specifically the declarative part, where we declare/initialize our constants and variables. It also names our program codice_4. Note the codice_5 clauses. Adding these enables us to do this: Put_Line ("Some text"); instead of this Ada.Text_IO.Put_Line ("Some text"); in the program. Very handy. As an exercise, try commenting the three codice_5 clauses, and prefix the actual package names to all types and procedures in the program. Next up we have this: Data : constant String := "This becomes a " & Latin_1.HT & " bunch of substrings"; This is the codice_10 we're going to split into individual components. codice_11 is a constant declared in codice_12. It inserts a horizontal tab in the string. Since we don't change the value of codice_13 throughout the program, we've initialized it as a constant. Subs : String_Split.Slice_Set; The codice_14 variable is the container for the individual components, or "slices". Seps : constant String := " " & Latin_1.HT; These are our separators. In this case we want to split the string on space (" ") and horizontal tabs (//Latin_1.HT//). Note that the separators are NOT included as part of the resulting codice_15. Try experimenting with different separators. begin Put_Line ("Splitting '" & Data & "' at whitespace."); codice_16 marks the beginning of the body of our program. Immediately after codice_16 we output a short message. String_Split.Create (S => Subs, From => Data, Separators => Seps, Mode => String_Split.Multiple); This is the meat of the program. In this one statement the codice_13 codice_10 is split into individual slices based on the codice_20 separators, and the resulting slices are placed in the codice_21. Note the codice_22 parameter. When using codice_23 mode, codice_24 will treat consecutive whitespace and horizontal tabs as one separator. As an exercise, try changing codice_23 to codice_26 and see what happens. Put_Line ("Got" & String_Split.Slice_Number'Image (String_Split.Slice_Count (Subs)) & " substrings:"); This is the line that's responsible for the output: Got 6 substrings: Yes, it looks like an awfully long line for very little output, but there's method to the madness: String_Split.Slice_Number'Image (String_Split.Slice_Count (Subs)) That line is responsible for the "6" part of the output. What it does is transform the codice_27 value codice_28 into the codice_10 value "6", and it does so using the codice_30 []. codice_31 return a codice_32 type, which is basically just an codice_27 with a value >=0, and codice_30 then convert this to a codice_10 suitable for output. for I in 1 .. String_Split.Slice_Count (Subs) loop -- Loop though the substrings. declare Sub : constant String := String_Split.Slice (Subs, I); -- Pull the next substring out into a string object for easy handling. begin Put_Line (String_Split.Slice_Number'Image (I) & Sub & " (length" & Positive'Image (Sub'Length) & -- Output the individual substrings, and their length. end; end loop; Here we start a loop that repeats codice_31 times, which in our case is 6. So on the first loop codice_37 is 1 and on the final loop codice_37 is 6. Inside the loop we codice_39 a new block. This enables us to locally initialize the codice_40 constant, which on each repeat of the loop is initialized anew with the next slice from our split. This is done using the codice_41 function which takes our codice_40 constant and the codice_37 loop counter as parameters, and return a codice_10. In the body of the block we output each slice, along with its index in the codice_21 and its length. As you can see, we once again make use of the codice_30 attribute to convert numeric values to codice_47. You can get rid of the block inside the loop like this: for I in 1 .. String_Split.Slice_Count (Subs) loop -- Loop though the substrings. Put_Line (String_Split.Slice_Number'Image (I) & String_Split.Slice (Subs, I) & " (length" & Positive'Image (String_Split.Slice (Subs, I)'Length) & -- Output the individual substrings, and their length. end loop; As you can see, we're no longer using the codice_40 constant. Instead we call codice_49 directly. It works just the same, but it is perhaps a bit less readable. Another option is to use an codice_50. You can see a possible solution here: Finally we have: end Explode; Which simply ends the program. And with that, we've concluded this small tutorial on how to split a string into individual parts (slices) based on a set of separators. I hope you enjoyed reading it, as much as I enjoyed writing it. See also. Wikibook. GNAT.String_Split

AI Art Generation Handbook/GUI/SD.Next. SD.Next is one of the most up to date of text prompting local Web UI (at the rate faster than Auto1111) . Don't be overwhelmed if you are majority of the users as you are probably be using just the 20% of the features listed here Here, you can do pretty much the following

World War II/Fall of Denmark and Norway. Fall of Denmark and Norway. Northern Europe in German and British strategic considerations. One of the few books that Adolf Hitler read on the topic of naval strategy was "The Sea Strategy of the World War" (1929) by Wolfgang Wegener, a former Vice Admiral of the German Imperial Navy of World War I. In it, the author confidently asserted that the German navy, which in the world war had proven an uninspired besieged force under constant naval blockade, could have drastically improved its own position vis-a-vis its British rival if only the German Empire had moved swiftly in the opening phases of the war to seize neutral Norway, whose ports would have exposed the northern flank of British naval defenses. Nonetheless, the German military buildup was decisively bipolar rather than tripolar; the navy was left underprioritized compared to the army and air force, whose combined operations were understood by German military and political planners as decisive in any land-based conflict. The German naval commander-in-chief, Erich Raeder, continuously attempted to gain Hitler's attention and support for a more decisive naval expansion program, such as when he held a presentation on his "Basic Thoughts about the Conduct of Naval Warfare" on 2 February 1937. The potentially exposed position of Northern Europe to a renewed British naval thrust worried German naval and economic planners, particularly with a view towards iron ore imports from Sweden. The "General Plenipotentiary for the Economy's" office calculated German iron ore imports in the year 1937 to have been ~20,620,000 tons, of which ~9,076,000 tons were produced in Sweden. Of this Swedish production, ~4,889,000 tons were transported from the iron ore mines in northern Sweden westwards across the Norwegian border and then shipped to Germany from the Norwegian seaport of Narvik. The iron ore transport ships thus travelled along the Norwegian coast, only protected by the diplomatic entanglements of Norwegian neutrality from a potential British attack in case of a war between Germany and the United Kingdom. British and German plans against Norway. With the outbreak of war in September 1939, the Western Allies began to calculate avenues of attack against Germany. In a draft paper by the British chiefs of staff on 4 September, the Chamberlain cabinet was warned that, although unlikely, a preemptive German occupation of Norway was a distinct possibility. In the minds of the British planners, the Germans could execute such a move to prevent any British intervention in Norway that could cut German iron ore supplies. One of the British ministers particularly enthusiastic about experimental military action was Winston Churchill. Not yet the prime minister, Churchill served under Chamberlain as the First Lord of the Admiralty, i.e. as the civilian minister for naval affairs. As early as 19 September 1939, Churchill suggested with some urgency that neutral Norway's territorial waters should be aggressively mined by the Royal Navy to force German ships out of Norway's neutrality zones, where they could more freely be attacked by the Royal Navy. This move was at this point rejected by the British cabinet, which feared the political ramifications that could be caused among neutral nations if Britain were to start mining neutral waters. Churchill later went on to win approval for a less internationally divisive scheme when he presented on 19 November a plan for a mine barrier between Norwegian waters and the Orkney islands, as had already happened once before during World War I. On 10 October 1939, German naval chief Raeder once again broached the topic of northern European security with Hitler; in a presentation to the dictator, Raeder stressed the critical dependency of Germany upon Swedish iron ore exports through Norwegian waters and also pitched the idea that Norwegian seaports might be useful as submarine bases for interdiction warfare against British naval imports. At this point in time, a quick victory over France seemed unlikely, which made Norwegian ports an appealing alternative to French ports in the support of German submarine warfare. Raeder was personally motivated by the rivalry inside the German armed forces; whereas army and air force received high degrees of attention, naval officers often felt neglected by the German political leadership. As a result, Raeder hoped to gain political influence and additional financial resources for his naval forces if Hitler were to decide on action against Norway. The importance of Northern Europe was drastically increased with the outbreak of the Winter War between the Soviet Union and Finland on 30 November 1939. Meanwhile, the Germans were aggressively courted by a potential collaborator in Norway. The Norwegian career soldier Vidkun Quisling, leader of a Nazi-inspired party in Norway, the "Nasjonal Samling", met with Raeder on 11 December, having done so on the suggestion to Raeder by Nazi chief ideologue Alfred Rosenberg which Rosenberg made to Raeder after another of the latter's attempts to convince Hitler of the necessity of preemptive German operations in Norway. Quisling, offering himself as the leader of a German-aligned government for the case of a German invasion of Norway, warned that a British preemptive attack against Norway might be imminent. Although unimpressed with Quisling as a person and potential leader of a German-allied Norway, Raeder was nonetheless alarmed at Quisling's suggestion that the Norwegian government might choose to not resist a British landing in Norway, thus effectively joining the Allies without a fight and denying Germany's iron ore supply. Quisling subsequently had two meetings with Hitler on 14 and 18 December 1939, again discussing the potential British threat against Norway. At about the same time, Churchill continued pushing the British cabinet towards naval action towards Norway. He reiterated his support for the mining of Norwegian waters on 16 December, and argued to the cabinet on 18 December that the United Kingdom could only gain advantages from the involvement of Norway and Sweden in the war. On 2 January 1940, Churchill pressed his case that British naval action in Norwegian waters might provoke a full-scale German invasion of the country, which would force Norway into the war on the Allied side and thus benefit Britain. Finally, on 6 January, the War Cabinet relented to Churchill's prodding and issued orders to senior naval commanders to assure British combat readiness for a Norwegian mission. However, the War Cabinet rejected on 12 January Churchill's own preferred plan of action towards Narvik, leading to lengthy discussions of a potential three-pronged attack on 19 January 1940, with "Operation Avonmouth" aimed at Narvik, "Operation Stafford" at Trondheim, Bergen and Stavanger, and "Operation Plymouth" at Trondheim, with the latter designed to bring sufficient forces to the Norwegian mainland to potentially involve Sweden in the war as well. The large associated resource allotment – 80,000 men, 10,000 vehicles, and a 60-day landing operation involving 12 passenger liners and 39 store ships, caused severe scepticism among the ranks of the War Cabinet, however. Just one day after the War Cabinet's discussion of the Avonmouth–Stafford–Plymouth trifecta, the German armed forces' high command (OKW) finished its own preliminary draft for an invasion of Norway, simply titled "Studie Nord", "Study North". Political pressure on the Allied leaders remained high after the unsanctimonious disaster on the Polish front and after the less than satisfying follow-up to the public outcry over the Soviet invasion of Finland, which was still ongoing due to the remarkable resilience of the Finnish defenders. On 5 February 1940, the Allied Supreme War Council decided during a Paris meeting to solve both the iron ore problem and the PR disaster with a joint strike against Norway and Sweden. British and French troops were to land on Norway's coasts, then to march into Sweden to seize the Swedish iron ore mines, and to eventually open supply routes to northern Finland so that the Western Allied could bring volunteer units to reinforce Helsinki's tenuous position. On 16 February 1940, the "Altmark Incident" accelerated German preparations for an attack against Norway. The German tanker "Altmark", which carried 299 British prisoners of war, was attacked and forced to surrender by "HMS Cossack" inside Norwegian waters. Hitler now fixated on the idea that Norwegian neutrality, which was obviously not respected by the Royal Navy, had become useless to Germany as a political assurance. He subsequently ordered the acceleration for the German operation,now dubbed "Weserübung", "Weser Exercise", on 19 February and agreed with OKW chief Wilhelm Keitel on army general Nikolaus von Falkenhorst as a viable candidate to lead the operation. Falkenhorst's experience as an operations officer in the Baltic theater of World War I was an important argument in his nomination. Falkenhorst was summoned to the Reich Chancellery, where he was informed by Hitler on 20 February about Weser Exercise. On Hitler's orders, he quickly worked out a preliminary operational plan based on a tourist guide he bought in Berlin, and presented it later the same afternoon to Hitler's approval. Falkenhorst took command of his staff, the subsequent XXI Army Corps, on 26 February, and presented a more elaborate plan to Hitler on the 29th. A critical innovation made by Falkenhorst was the occupation of northern Denmark, which had so far played only a minor role in the various considerations by German and British planners. Hitler approved this amendment, and promptly expanded it to include the entirety of Denmark. The dictator also definitively decreed that Vidkun Quisling was to play no role whatsoever in the invasion – even though Quisling would reappear later. Hitler, who guided the German war effort through several key directives over the course of the war, issued one such Leader Directive ("Führerweisung"), numbered 11, on 1 March 1940. The plan did not yet have a specific timetable, but placed the air force contingents under Falkenhorst's command. This would go on to outrage the commander-in-chief of the German air force ("Luftwaffe"), Hermann Göring, who during a conference on the 5th of March managed to change Hitler's mind, thus placing the air force back under his command. Falkenhorst was subsequently forced to visit Göring at his personal estate ("Karinhall") on the 7th to informally discuss the German air force's dispositions for Norway. The mutual distrust between the two sides peaked in March 1940; German naval high command warned of a British concentration of forces at Scapa Flow in preparation for an invasion of Norway on 8 March, which led to an all-out military emergency in Germany on 11 March. The imminence of this threat was however decreased when the Winter War ended with a negotiated settlement and a conditional Finnish defeat on 12 March. In spite of this strategic setback, Churchill continued to argue in the War Cabinet meeting of 14 March 1940 that the Royal Navy should undertake mining operations in Norwegian waters. Pressure on the British government was increased by the French ambassador, who delivered a frustrated letter from Prime Minister Daladier – later forced to resign due to the resentment of the French public – urging military action against the German iron ore supply routes through the maritime territories of Norway. Though under increasing pressure in the House of Commons, Neville Chamberlain nevertheless held his own and swayed the cabinet away from an invasion of Norway. Instead, the British undertook "Operation Royal Marine", hoping to cause damage to Germany by letting naval mines flow down the Rhine river from the French border into the German industrial heartland. On the German side, Raeder was anxious to finally get Weser Exercise on the road, arguing in a 26 March discussion with Hitler that a simultaneous operation against Norway and France was not feasible for the German armed forces. He pointed to the lunar cycle as a critical precondition for visibility during night operations and named new moon on 7 April as his preferred date for an invasion, and outright rejecting any date later than 15 April. On 28 March, the Supreme Allied War Council met and decided on a 'political horse trade', exchanging the British "Operation Royal Marine", the mining of the Rhine, for the French-desired "Operation Wilfred", the mining of Norwegian waters. This agreement was mainly designed to give the new French prime minister Paul Reynaud an important domestic political victory, and paid little attention to German schemes or reactions. While British naval chiefs began to inquire about naval ranges and French troops availabilities, the Germans prepared for final conferences to prepare the invasion. Hitler discussed Weser Exercise with all key commanders in a final conference on 1 April. On the following day, having received confirmations of operational readiness by all three branches of the armed forces, he placed the invasion date on 9 April 1940 (which was also dubbed "Weser Day") and the operational start at 04:15 in the morning (dubbed "Weser Hour"). Weser Exercise was also definitively separated into two sub-operations, with Weser Exercise North ("Weserübung Nord") being directed against Norway, whereas the forces of Weser Exercise South ("Weserübung Süd") were to attack Denmark. Although Chamberlain famously taunted in public that 'Hitler had missed the bus' on 5 April, the Royal Navy was now hurrying forces to undertake its mining operations in time. Though British minelaying began on 05:00 in the morning of 8 April, eleven German naval warship groups had already left their ports on the 7th, and were headed to their rendezvous points. Just after midnight on 8/9 April, the invasion troop ships departed Wilhelmshaven and Cuxhaven, en route for Norway and Denmark.

AI Art Generation Handbook/Censorship. In the beginning, AI Art Image Generations have safe guard rails to guards against the generations of the "unsavory" images but over the time, the safeguard is becoming more and more strict with examples such as below. "Note: All of this images were taken during the Bing Image Creator Great Purge in the late 2023/early 2024 , where more innocuous prompt token were blocked for Content Warning or Unsafe Image". " Some of the blocked prompts may or may not work since then" . DALL-E. Images with potential likeness to real people. Within the censorship context, as the images generation are getting good as it is approaching to realistic photo (from year to year), (See DALL-E2.5) the images generated by DALL-E maybe misused by persons with ulterior motives , therefore , the AI safety committees in various AI institutes put up stricter guardrails, especially for the names of famous persons/person of interest were added in . In case of Dall-E , even the generated images with human elements are saturated to the point that it looked cartoonish rather than realistic (i.e. In this example of Henry Kissinger). <br> <br> <br> <br> <br> <br> Images with political elements. In this another example, the prompts consists of political elements (especially if related to China at that time); where DALL-E blocked the possible combinations of Winnie the Pooh (with hidden connotations to the current China's premier) and the word Taiwan in the same prompt, triggering the alert for content warning and the blocking prompts from generating image. <br> <br> <br> <br> <br> <br> <br> <br> Images with elements of body diversity. In this example, during the "Bing" "Great Filter Purge", many of the body diversity (especially with potentially "offensive" tokens: fat , obese, skinny, dark skinned, etc ... ) are also believed to trigger the system alarm and blocking the prompt from generating the images. This is maybe misconstrue as the body shaming of such individuals or the inherent racism . <br> <br> <br> <br> <br> <br> <br> <br> Images with potential gore elements. In this example, the skeleton may be accidentally grouped in the gore categories and perhaps that is when then , the prompts consists of skeleton maybe blocked although the skeletons image category (See: Halloween celebrations) may seems benign compared to other type of gore images <br> <br> <br> <br> <br> <br> <br> <br> Images with religious significance. This is more sensitive topics in certain part of the world where certain tokens related with significant religious symbols are possibly unsafe to be generated due to its significant religious meaning. <br> <br> <br> <br> <br> <br> <br> <br> Images with sexual undertones. Although the prompt itself were not exactly requesting for the explicitly photos, DALL-E3 image AI models may have tendency to generate lewd types of imagery if similar keywords are presented in the prompts and/or the image filters maybe more restrictive in their DALL-E 3.5 If compared to SDXL image generations, most of the time will render closeup photo whilst showing the AI character wearing skimpy nightwear <br> <br> <br> <br> <br> <br> <br> Stable Diffusion. Unintentional Censorship. As per latest hoo-hah, release of both SD2.0 and latest SD3 Med also facing backlash over the prompt " codice_1" prompt which generate mutilated limbs At times, the censorship on the training dataset maybe too strict until it may causes unintentional censors on other similar subject such as the examples on the left. Cow udder is visually similar to human's female breast and the CLIP Vision may also pruned the dataset with visible cow udders unintentionally during the dataset pruning.

AI Art Generation Handbook/GUI/Fooocus. Fooocus is rather minimialised in the appearances among all other type of WebUI, but somehow it can still generate a high quality images. To use this, you just type the prompts or prompts in the bottom of the text field and click Generate button. You can also click the for Advance checkbox and apply different styles to the images, different image sizes and even the details and guidances.

Physics Explained Through a Video Game. This textbook explains algebra-based high school leveled mechanical physics through Bonk.io, a multiplayer physics browser game that runs on the . Materials from this video game are used to explain real-world physics concepts. Table of Contents. Note that the "Table of Contents" is subject to change and may not reflect recent structural changes. Unit 1: Kinematics. Topic 1.1 - Displacement1 Topic 1.2 - Calculating Displacement1 Topic 1.3 - Speed and Average Velocity1 Topic 1.4 - 1-D Uniformly Accelerated Motion Topic 1.5 - Motion in 2 Dimensions Unit 2: Forces and Newton’s Laws. Topic 2.1 - Systems and Center of Mass Topic 2.2 - Forces and Free Body Diagrams Topic 2.3 - Newton's Second Law Topic 2.4 - Newton's First and Third Laws2 Topic 2.5 - Resistive and Frictional Forces Topic 2.6 - Spring Forces Topic 2.7 - Tension and Circular Forces Unit 3: Work and Energy. Topic 3.1 - Introduction to Work3 Topic 3.2 - Introduction to Kinetic and Potential Energy Topic 3.3 - Conservation of Energy Topic 3.4 - Conservative and Nonconservative Forces Topic 3.5 - Power Unit 4: Linear Momentum and Collisions. Topic 4.1 - Introduction to Linear Momentum Topic 4.2 - Impulse (Change of Momentum) Topic 4.3 - Inelastic Collisions Topic 4.4 - Elastic Collisions Unit 5: Statics and Rotation. Topic 5.1 - Rotational Kinematics Topic 5.2 - Torque and Rotational Inertia Topic 5.3 - Translational and Rotational Equilibrium Topic 5.4 - Angular Momentum Topic 5.5 - Rotational Kinetic Energy Topic 5.6 - Rolling Objects Unit 6: Springs and Oscillations. Topic 6.1 - Introduction to Simple Harmonic Motion Topic 6.2 - Applying SHM Topic 6.3 - Physical Pendulums Unit 7: Gravitation. Topic 7.1 - Newton's Law of Universal Gravitation Topic 7.2 - Kepler's Laws Known existing issues: 1 "These topics were created before adopting an improved article structure. As such, these topics have fewer resources, such as not having end-of-section practice problems." 2 "For this topic, the Example 4 contains a mistake with the FBD of the koi fish and lily pad after collision. They should be treated as a single system for the problem. Also, editing likely needs to be done for clarity and accuracy with the remainder of the example." 3 For this section, there are relatively few in-depth examples, especially given the large scope of the topic. Also, the discussion of work as a scalar value probably can be simplified. Prior Knowledge. For this textbook, it is recommended that you have taken at least one year each of high-school leveled algebra and geometry.

The Complete Encyclopedia of Self-Help Techniques/Hypnotherapy. Self hypnosis: therapy steps into getting started /For self hypnosis therapy use a script./think write goals to achieve.//think write your reason to start a program for self hypnosis./// 1.enter your hypnosis therapy close your eyes of course get comfortable and just put your mind to sleep relax rid of any stress, anxiety or fear of past or future thought. Take deep slow breathing enter nose exit mouth. Keep your mind awake in deep thoughts. Now think how your vision of a perfect life without no negativity would be put yourself vision this and keep visioning thinking about that perfect life and how you got there in mechanical steps and how you are feeling like you are on top and financially free, sickness free.do this for about 15 minutes continuously over again. Don't overwhelmed yourself 2 significant changes are fine but thought as a whole collection is imperative that your already there as a whole mind and being perceptively. 2. Pray for forgiveness to any wrong you've done and pray for your god to guide you tell your god that your sacrifices are only to higher being on this life to learn and live eternal after this life. 3. Take slow and deep breathes enter nose exit mouth. Imagine all bad from your written list has been exited and your transformed to your perfect being you want. Tell yourself that you will be in hypnosis only for this only and nothing else after awake. 4. Now awake open your eyes, stand up and continue your physical day. Finish your daily self help hypnosis therapy written notes.

Proto-Turkic/Vocabulary table. To avoid confusion with modern sound forms, we have listed only the historical forms. Words in modern languages ​​came directly from these languages, to which they are descendants. Turkmen is included for lengths.

Biochemistry/The Cell. A cell is a small particle or organism which is found in the human body. These are some of the cells found in the human body. 1.Red blood cells. 2.White blood cells. IMPORTANCE OF RED BLOOD CELLS. 1.Helps in transportation of blood. 2.Helps in structure of the body. 3.Helps in body support. IMPORTANCE OF WHITE BLOOD CELLS. 1.Helps in prevention of diseases. 2.Helps your body to be strong. 3.Cleaning germs in the blood.

Orthopaedic Surgery/Patellar Tendinitis. Patellar Tendinitis (Patellar Tendinopathy), also known as Jumper's Knee, is a painful condition of the knee, which primarily occurs in sports requiring strenuous of jumping (as the name implies). The condition is caused by small tears in the patellar tendon which results in localized tenderness of the patellar tendon. The tears are typically caused by accumulated stress on the patellar or quadriceps tendon. The patellar tendon is a band of tissue that connects the patella (the kneecap; see fig 1) to the tibia (shin bone). The tendon is used when you extend your knee. When you contract your quadriceps muscles, they pull on the patellar tendon, which in turn pulls on the kneecap and extends the knee. Symptoms. The most common symptoms include: Causes and risk factors. The primary cause of patellar tendinitis is activities that place repetitive stress on the patellar tendon. These are activities like jumping, plyometrics, running, walking, bicycling. Treatment. Other treatments. TODO. Returning to activity. Volleyball. TODO Basketball. TODO Running. The Run Doctor, a running injury clinic, recommends the following checklist to be completed before considering returning to running: Not all runners experiencing patellar tendonitis will need to take time off from activity, however they may need to reduce their training load.

AI Art Generation Handbook/ControlNet/Scribble. I believe some of you are old enough , may remember this how to draw an owl meme Well , do you know that it is super easy to recreate this inside Stable Diffusion. Assuming you got latest Xinsir Union ControlNet models, you can do it inside both Auto1111 and SD.Next. SD.Next. We try to recreate above meme using the Scribble ControlNet. This model is designed to work with simple sketches or scribbles. It's perfect for turning basic shapes (like two circles ala snowman) into more detailed drawings of an owl as shown in the picture below. The result should be a detailed pencil sketch of an owl, where the two circles you drew have been interpreted as the owl's body and head, with the rest of the details (feathers, eyes, branch, etc.) filled in by the AI Model. The ControlNet is already integrated within SD.Next, therefore just go straight to the "Control" tab Using this prompt, we can try to recreate the images codice_1 Control Input. We upload this scribbles into the Input Image ( Look for : Drop Image Here ) on the left side Control Sampler. Double steps from 20 to 40 for more flushed out details Control Elements Parameters. Pre-Processor: Line Art Realistic ControlNet : Xinsir Scribble XL Recommended Parameters Values. Strength:1.6 Values above 1.0 can exaggerate ControlNet's influence, potentially leading to over-emphasis of the control image. Start: 0.35 A value of 0.0 means ControlNet starts influencing from the very beginning. End: 0.8 End Results. If done correctly, the AI model will try to adjust the diffusion to match as close as possible to the input images, provided that input images is closely resembled the Input image silhouette , also the input images of circles and rectangle will be not much visible

Astrodynamics/Perturbations. Perturbations in astrodynamics refers to the forces that act on an object that is not from the gravitational attraction of a single, more massive body. These forces can include gravity from other bodies, drag from an atmosphere(Such as the decay of satellite orbits because of collisions with gas molecules and the craft), or the off center attraction caused by an oblate or irregularly shaped body. Orbital resonance. Orbital resonance is when large bodies regularly perturb other bodies with gravity. This usually ends in instability in which the orbits of the two bodies are shifted until there is no more resonance. In some cases the resonant system can be self correcting, making the resonance stable. Orbital resonances are represented with either the ratio of orbits completed in a single time interval or the ratio of their orbital periods, with both representations giving ratios inverse to each other. Examples of this are the 2:3 resonance between Neptune and Pluto and the 1:2:4 resonance between Jupiter's moons of Ganymede, Europe and Io(Ratio of orbits in a single time interval used). Apsidal precession. Apsidal precession is the precession of the line of apsides of an object's orbit. It is related with the orbit's argument of periapsis. It is caused by perturbations from other bodies in the system, and by gravity perturbations explained by general relativity. Examples. An example of a major orbital perturbation is in April 1996, when Jupiter's gravitational influence decreased comet Hale-Bopp's orbital period from 4,206 years to 2,380 years.

Chess Opening Theory/1. e4/1...c5/2. Nf3/2...Nc6/3. d4/3...cxd4/4. Nxd4/4...Nf6/5. Nc3/5...e5/6. Ndb5/6...d6/7. Bg5/7...a6. The primary purpose of 7...a6 is to drive the knight on b5 away from its strong position. The knight on b5 threatens to jump to c7, which would be a fork on Black's king and rook, and puts pressure on d6. By pushing the a-pawn, Black gains space on the queenside, which can support further pawn advances like ...b5, potentially leading to active counterplay on that side of the board. Forcing the knight to retreat diminishes White's immediate threats and allows Black to better organize their pieces. The knight typically moves to a3, where it is less active compared to b5.

Cherokee/Verb Template. Cherokee verbs are very complex in their morphology. A Cherokee verb requires, at minimum, a pronominal prefix, a verb root, an aspect suffix, and a tense suffix. The aspect suffix will differ between verbs, so it is common for the verb root and and aspect suffix to be combined into a verb "stem." Every verb has five possible aspects, and, therefore, five possible stems. In addition to the required components of a Cherokee verb, there are optional components including the prepronominal prefixes, the reflexive prefix, the middle voice prefix, the incorporated noun, and the derivational suffixes. All of these components follow a predictable order, as shown in the following table. Components in the same column cannot co-occur within a Cherokee verb. Components in bold are the minimum required for a Cherokee verb.

Moving objects in retarded gravitational potentials of an expanding spherical shell/Table with computed results. Table with computed results. In the following table you will find different computed values based on the linear mass density of the black shell formula_1. The values had been numerically computed by a Java program. The Schwarzschild distance that is corresponding to the Hubble radius of the universe during the generation of the cosmic microwave background (CMB) at around 380,000 light-years (ly) is marked in bold. The same applies to a Schwarzschild distance of 858 million light-years, where the ratio between the mass of visible universe to the mass of the black shell is equal to one. This distance corresponds to the distance between the particle horizon and the visible horizon of the cosmic microwave background (CMB) according to the ΛCDM model. Note that to 380,000 years after the Big Bang, the slope of the Schwarzschild radius in relation to the slightly increasing linear mass density was very high. The Schwarzschild distance was grewing much faster than the mass of the black shell during this period.

Moving objects in retarded gravitational potentials of an expanding spherical shell/Java program. Java program. The following Java program can be used to numerically compute the Schwarzschild distances for different linear mass densities. It implements a numerical integration for the effective masses as well as the numerical solvation for the Schwarzschild distances. <syntaxhighlight lang="Java"> Source file: SchwarzschildDistance.java Program: Computation of the Schwarzschild Distance within the Hubble sphere Autor: Markus Bautsch Location: Berlin, Germany Licence: public domain Date: 18th July 2024 Version: 1.3 Programming language: Java This Java-Programm computes Schwarzschild distances for any black shell masses. The black shell is at the outer rim of the Hubble space. public class SchwarzschildDistance // Class constants final static double G = 6.67430e-11; // Gravitational constant in cube metres per kilogram and square second final static double c = 2.99792458e8; // Speed of light in metres per second final static double TropicalYear = 365.24219052 * 24 * 3600; // Tropical year in seconds final static double LightYear = c * TropicalYear; // Light-year in metres final static double R = 1.36e26; // Hubble length in metres final static double MUniverse = 2.97e53; // Mass of the visible universe in kilograms final static double lambdaMfirst = 3.36663e26; // start value for linear mass density of black shell final static double lambdaMlast = 3.36670e26; // startop value for linear mass density of black shell final static double deltaLambda = 0.00001e26; // step size for linear mass density of black shell // This method computes and returns the distance s of a mass m to a mass point dM in the black shell // distance is the closest distance of the mass m to to the mass point dM in the black shell in m // alpha is the angle beweteen the mass m and the mass point dM in the black shell as seen from the centre of the universe in rad private static double distanceToMassElementShell (double distance, double alpha) double cosAlpha = java.lang.Math.cos (alpha); double squareDistance = distance * distance; double argument = 2*R*R - 2*R*distance + squareDistance - 2*R*(R-distance)*cosAlpha; double s; if (argument <= squareDistance) // because of possible rounding errors s = distance; else s = java.lang.Math.sqrt (argument); return s; // This method computes and returns the angle of a half chord x in m as seen from the centre of the universe in rad private static double alpha (double x) double alpha = java.lang.Math.asin (x / R); return alpha; // This method computes and returns the integrand of the integral // distance is the closest distance of the mass m to to the mass point dM in the black shell in m // alpha is the angle beweteen the mass m and the mass point dM in the black shell as seen from the centre of the universe in rad private static double integrand (double distance, double alpha) double s = distanceToMassElementShell (distance, alpha); double e = R * (1 - java.lang.Math.cos (alpha)); // the auxiliary sagitta e double cosBeta = (distance - e) / s; double integrand = cosBeta / s / s; return integrand; // This method computes and returns the effective mass from 0 to alphaR for a given lambdaM and a given distance // lambdaM is the linear mass density of the black shell in kg/m // distance is the closest distance of the mass m to to the mass point dM in the black shell in m private static double integrateEffectiveMass (double lambdaM, double distance) final long steps = 1000000; // number of steps for numerical integration double squareDistance = distance * distance; double xMax = java.lang.Math.sqrt (2*R*distance - squareDistance); // maximum half chord for a given maximum angle alphaR double x = xMax; // x runs from xMax to 0 in steps of deltaX double alpha = alpha (x); // alpha runs from alphaR to 0 double integrand = integrand (distance, alpha); double integral = 0; long step = 0; while (step <= steps) x = xMax * (1 - (double) step / steps); double alphaNext = alpha (x); double integrandNext = integrand (distance, alphaNext); double deltaAlpha = alpha - alphaNext; double averageIntegrand = (integrandNext + integrand) / 2; double area = averageIntegrand * deltaAlpha; // numerical integration integral = integral + area; alpha = alphaNext; integrand = integrandNext; step++; double effectiveMass = R * lambdaM * squareDistance * (2*integral); return effectiveMass; // This method computes and returns the Schwarzschild distance d_S for a given effective mass mEff private static double schwarzschildDistance (double effectiveMass) double schwarzschildDistance = 2 * G * effectiveMass / c / c; return schwarzschildDistance; // This method computes and returns the effective mass mEff for a given Schwarzschild distance d_S private static double effectiveMass (double schwarzschildDistance) double effectiveMass = schwarzschildDistance * c * c / 2 / G; return effectiveMass; // This method solves and returns the Schwarzschild distance d_S for a given lambdaM // lambdaM is the linear mass density of the black shell in kg/m private static double solve (double lambdaM) final double limit = 5e-16; // for relative precision of determination of Schwarzschild distance double lowerDistance = 1e20; // first low guess for Schwarzschild distance double upperDistance = 1e25; // first high guess for Schwarzschild distance double delta; // delta shall become smaller than limit double distance; double schwarzschildDistance; double deltaDistance; do distance = (lowerDistance + upperDistance) / 2; double effectiveMass = integrateEffectiveMass (lambdaM, distance); schwarzschildDistance = schwarzschildDistance (effectiveMass); deltaDistance = (distance - schwarzschildDistance) / schwarzschildDistance; if (deltaDistance < 0) // Schwarzschild distance too large lowerDistance = (upperDistance + lowerDistance) / 2; else // Schwarzschild distance too small upperDistance = (upperDistance + lowerDistance) / 2; delta = (upperDistance - lowerDistance) / lowerDistance; } while (delta > limit); if (java.lang.Math.abs (deltaDistance) > 1e-6) java.lang.System.out.println ("unstable numerical result:"); java.lang.System.out.println ("Computed distance = " + distance); java.lang.System.out.println ("Schwarzschild distance = " + schwarzschildDistance); return schwarzschildDistance; // This method outputs the table header private static void printHeader () java.lang.System.out.print ("lambda_M in kg/m;"); java.lang.System.out.print (" M_S in kg ;"); java.lang.System.out.print (" M_S/M ;"); java.lang.System.out.print (" M_eff in kg ;"); java.lang.System.out.print (" d_S im m ;"); java.lang.System.out.print (" d_S/R ;"); java.lang.System.out.print (" d_S in ly ;"); java.lang.System.out.print (" (R-d)/R "); java.lang.System.out.println (); // This method outputs a table result line // lambdaM is the linear mass density of the black shell in kg/m // distance is the closest distance of the mass m to to the mass point dM in the black shell in m // mEff is the effective mass of a sphere in the black shell private static void printResults (double lambdaM, double distance, double effectiveMass) double mShell = lambdaM * R * 2 * java.lang.Math.PI; // mass of black shell double mRatio = mShell / MUniverse; double dRatio = distance / R; double dLightYears = distance / LightYear; double deltaRatio = (R - distance) / R; java.lang.System.out.printf ("%15.7e", lambdaM); java.lang.System.out.print (" ;"); java.lang.System.out.printf ("%15.6e", mShell); java.lang.System.out.print (" ;"); java.lang.System.out.printf ("%15.6f", mRatio); java.lang.System.out.print (" ;"); java.lang.System.out.printf ("%15.2e", effectiveMass); java.lang.System.out.print (" ;"); java.lang.System.out.printf ("%15.2e", distance); java.lang.System.out.print (" ;"); java.lang.System.out.printf ("%15.2e", dRatio); java.lang.System.out.print (" ;"); java.lang.System.out.printf ("%15.0f", dLightYears); java.lang.System.out.print (" ;"); java.lang.System.out.printf ("%15.9f", + deltaRatio); java.lang.System.out.println (); // This method has a loop for computing a sequence of effective masses and Schwarzschild distances for different lambdaM values // lambdaM is iterated from lambdaMfirst to lambdaMlast in steps of deltaLambda private static void computeScharzschildRadii (double lambdaMfirst, double lambdaMlast, double deltaLambda) double lambdaM = lambdaMfirst; // linear mass density of black shell at Hubble radius in kg/m lambdaMlast = lambdaMlast * 1.00000001; // for the reason of numerical rounding errors printHeader (); do double schwarzschildDistance = solve (lambdaM); double effectiveMass = effectiveMass (schwarzschildDistance); printResults (lambdaM, schwarzschildDistance, effectiveMass); lambdaM = lambdaM + deltaLambda; } while (lambdaM <= lambdaMlast); // Main program public static void main (java.lang.String [] arguments) computeScharzschildRadii (lambdaMfirst, lambdaMlast, deltaLambda); </syntaxhighlight >

Machine Learning/Bank Fraud Detection. There are many applications for machine learning and binary classification, and one of these is the detection of illegitimate transactions on ATMs at banks. The recall is defined as number of true positives divided by the total number of true positives plus false negatives.

Cherokee/Vocabulary. Body Parts Clothing Verbs

Cherokee/Body Parts. Description, Citation Please help us elicit the missing information in this chart! BMA only tells us so much!

Czech/Nouns/Case/Accusative. =Accusative Case in Czech (4th)= The accusative case is primarily used to indicate the direct object of a transitive verb. It is also used after certain prepositions and in time expressions. Uses. Used as the direct object of transitive verbs. Note: the object in Czech may be translated as the subject in some languages. These verbs use the accusative case for the person experiencing the feeling, while the cause of the feeling is in the nominative case. After certain prepositions. Some common prepositions that take the accusative case include: Also, v, před, pod, nad, o, mezi In time expressions. = References =

Czech/Nouns/Case/Genitive. =Genitive Case in Czech (2nd)= The genitive case in Czech is used to express possession, after certain prepositions, with numbers, and in various other constructions. It often corresponds to the English "of" or possessive "'s". Declension. Some common words have irregular genitive plurals, for example Uses. To express possession. With proper nouns. When expressing possession with proper nouns, the same principle applies. The possessor's name is put into the genitive case: Note that there are two ways to express possession: Using the genitive case of the name (Cyrila) after the noun; and using the possessive adjective form (Cyrilův, Cyrilova, Cyrilovo), which agrees in gender and number with the possessed object. Both forms are correct and commonly used. The possessive adjective form is more common in spoken language, while the genitive form might be perceived as slightly more formal. Examples with other proper nouns: After certain prepositions. Some common prepositions that take the genitive case include: Note that some prepositions can take different cases depending on their meaning or usage. The examples provided here specifically demonstrate their use with the genitive case. Partitive Expressions and Quantifiers. With numbers above 5 (genitive plural). Note that after numbers 5 and above, the genitive plural is always used, regardless of whether the noun is countable or uncountable. In "Pět litrů vody." The word "litrů" is genitive plural, while "vody" is genitive singular. After some units With mass nouns, the genitive singular is used: With count nouns, the genitive plural is used: Kolik (how much/many). "Kolik" always requires the genitive case: Všechno, každý, žádný. These words meaning "every" and "no" also require the genitive case in certain constructions: After certain verbs. Some verbs always take the genitive case: Negative sentences. In negative sentences, the direct object often changes to the genitive case: Note that in modern colloquial Czech, the accusative is often used instead of the genitive in negative sentences, but the genitive is still considered more correct in formal language. In dates. The genitive case is used in expressing dates in Czech. Days of the month. When expressing a specific date, the day is in the genitive case: Note that the ordinal number (first, second, third, etc.) is in the genitive case, as is the month. Months. The names of months are always in the genitive case when used in dates: Years. When expressing a full date including the year, the year is typically in the genitive case as well: However, in informal speech, the nominative form of the year is often used. Examples of full dates. Note that in Czech, unlike in English, prepositions are typically not used before dates. The genitive case itself indicates the temporal relationship.

Famous Theorems of Mathematics/π is transcendental. Introduction. The proof of the transcendence (inalgebraicy) of formula_1 is in many ways similar to that of formula_2. However, we must introduce a number of very important issues before proceeding to the full, detailed proof. Basic prior knowledge about complex numbers is required.

Famous Theorems of Mathematics/π is transcendental/Monomial ordering. Definition 1. Let formula_1 be an N-tuple. Let us define: For functions formula_3 let us define: This abbreviated notation will be of great use to us in the following pages and in the proof. Definition 2. Let formula_5 be a field. Then formula_6 is the polynomial space in variables formula_7 with coefficients in formula_5. A monomial is a polynomial of the form formula_9, such that formula_10 and formula_11. Definition 3. Let formula_12, and let formula_13 be an exponent vector. Let us define: Properties. Monomial multiplication maintains exponent vector addition: Definition 4. Let formula_17 be monomials. We say that formula_18 is of lower order than formula_19 (and denote it by formula_20) if there exists an index formula_21 such that In other words, the vectors formula_23 have a lexicographic ordering. In a polynomial formula_24, the monomial of maximal order is called the leading monomial, and is denoted by formula_25. Lemma. Let formula_27 be polynomials. Then formula_28. Proof. Let formula_29 be monomials, with formula_30. 1. Let us assume that formula_31. We will show that formula_32 for all formula_33.<br> By definition, there exists an index formula_21 such that 2. Let us assume also that formula_36. We will show that formula_37.<br> By definition, there exist indexes formula_38 such that respectively hence: formula_41 Definition 5. Let formula_42 be a polynomial. Let us define: Meaning, the set of all monic monomials of degree formula_44 which are of lower order than formula_25.

Famous Theorems of Mathematics/π is transcendental/Symmetric polynomials. Definition 1. A permutation is a bijective function from a set to itself. Let formula_1 be a finite set. The function formula_2 is called a permutation if and only if it is one-to-one and onto. Meaning, for all formula_3 there exists a unique formula_4 such that formula_5. The set of all permutations of the elements of formula_6 is denoted by formula_7. Example. For formula_8 there are formula_9 different permutations: In general, if formula_11 then formula_12. Definition 2. Let formula_13 be a polynomial. Let us define: Properties. Let formula_15 be polynomials. Then we have: Definition 3. Let formula_31 be a polynomial. Then it is called symmetric if for all permutations formula_33.

Famous Theorems of Mathematics/π is transcendental/Elementary symmetric polynomials. Let there be a polynomial of degree formula_1 By the Fundamental Theorem of Algebra, it has formula_3 complex roots (with multiplicity). Then we can write: As we know, Vieta's formulae link between the coefficients and roots of a polynomial: As we can see, these sums are symmetric polynomial, and are called elementary symmetric polynomials. Definition. The elementary symmetric polynomials in variables formula_6, are defined as such:

Canadian Refugee Procedure/IRPR ss. 230-234 - Stay of Removal Orders. IRPR ss. 230-234. The text of the relevant sections of the "Immigration and Refugee Protection Regulations" read: IRPR s. 231: Stay of removal — judicial review. History of IRPR s. 231(1) and 231(2). In 2009, prior to the amendments to the IRPA that are referred to as "refugee reform", the above provisions read as follows: IRPR s. 231(1): Subject to subsections (2) to (4), automatic stays are provided for judicial reviews of RAD decisions, but not RPD decisions. IRPR s. 231(1) provides that a removal order is stayed if the subject of the order makes an application for leave for judicial review with respect to a decision of the Refugee Appeal Division. Prior to the implementation of refugee reform, section 231 of the IRPR provided an automatic stay of removal when failed refugee claimants filed an application for leave and judicial review of a refugee claim rejected by the RPD, subject to certain exceptions and limitations (e.g. serious criminals and claims with no credible basis). On December 15, 2012, amendments to section 231 of the IRPR provided an automatic stay of removal to claimants with a claim rejected by the RAD, rather than by the RPD, when a refugee claimant seeks judicial review of his/her rejected claim. As a result of the amendment to section 231 of the IRPR, an automatic stay of removal is no longer available to a claimant who has applied for leave and judicial review of claims rejected by the RPD; rather, the automatic stay of removal is available to a claimant who has applied for leave and judicial review of claims rejected by the RAD, subject to certain exemptions and limitations.

Famous Theorems of Mathematics/π is transcendental/Fundamental theorem of symmetric polynomials. Let formula_1 be a field, and let formula_2 be a symmetric polynomial.<br> Then formula_3 can be expressed uniquely as a polynomial formula_4, such that: Proof. First, we shall descibe an algorithm for finding the desired polynomial formula_5. Let us define initial conditions formula_10 and formula_11. In order to prove the algorithm we need two lemmas. Lemma 1: The leading monomial in formula_3 satisfies formula_21. Proof: Let us assume there exists an index formula_22 such that formula_23. Then there exists a permutation formula_24 such that But the polynomial formula_26 contains the monomial formula_27, which is of higher order than formula_28. A contradiction. formula_29 Lemma 2: The leading monomial in the expansion of formula_30 is formula_31. Proof: We have The last equality holds if and only if formula_29 We shall now prove the theorem: 1. Let formula_35 be a symmetric polynomial in variables formula_36.<br> The proof is by strong induction on formula_37 (see definition). If formula_38 then formula_3 is a constant polynomial, and it is easy to show the algorithm holds. Let us assume the algorithm holds for all symmetric polynomials formula_3 with formula_41, for some formula_42.<br> We will show that the algorithm holds also for a symmetric polynomial formula_43 with formula_44, such that formula_45. By lemma 2, we get: The function formula_47 is a polynomial, since formula_21.<br> In addition, by the properties of symmetric polynomials formula_49 is a symmetric polynomial in variables formula_36, therefore so is formula_51.<br> The polynomials formula_52 both contain formula_53, hence it is cancelled in their subtraction.<br> If formula_54 then formula_55.<br> If formula_56 then formula_57, meaning formula_58.<br> Thus, the inductive assumption holds for formula_51, and therefore the algorithm yields a polynomial formula_60 such that 2. The properties of the theorem hold: formula_69 Important results. Theorem. Let formula_70 be a field, and let formula_71 be a polynomial of degree formula_72 with roots formula_73.<br> Let formula_74 be a symmetric polynomial. Then formula_75. Proof. By Vieta's formulae, we get By the fundamental theorm above, formula_5 can be expressed as a polynomial formula_29 Theorem. Let formula_70 be a field, and let formula_71 be a polynomial of degree formula_72 with roots formula_73.<br> Let formula_84, and let formula_85 be the sums of every formula_22 of the roots formula_87 (namely formula_88).<br> Then there exists a monic polynomial formula_89 of degree formula_90 with roots formula_85. Proof. We will show that By Vieta's formulae, its coefficients are all symmetric polynomials in formula_85. Let formula_94 be a symmetric polynomial, and let formula_95 be the sums of every formula_22 of the variables formula_36.<br> Then formula_5 can be expressed as a polynomial It is easy to see that by applying a permutation on formula_36, we also apply a permutation on formula_95.<br> Hence formula_102 is a symmetric polynomial, and by the previous theorem we get formula_29

Famous Theorems of Mathematics/π is transcendental/Proof. The mathematical constant formula_1 is a transcendental number (or inalgebraic). In other words, it is not a root of any polynomial with rational coefficients. Proof. Let us assume that formula_2 is algebraic, so there exists a polynomial such that formula_4. Part 1. Lemma: If formula_2 is algebraic, then formula_6 is algebraic. Proof: We get hence formula_6 is a root of the polynomial formula_10 Therefore, there exists a polynomial formula_11 of degree formula_12 with roots formula_13, such that formula_14. By Euler's identity we have formula_15. Therefore: The exponents are symmetric polynomial in formula_13, and among them are formula_18 non-zero sums. That is: As we previously learned, for all formula_20 there exists a monic polynomial formula_21 of degree formula_22 such that its roots are the sums of every formula_23 of the roots formula_13. Therefore: After reduction we get that: Multiplying by the least common multiple formula_27 of the rational coefficients, we get a polynomial of the form Part 2. Let formula_29 be a polynomial of degree formula_30. Let us define formula_31. Taking its derivative yields: Let us define formula_33. Taking its derivative yields: By fundamental theorem of calculus, we get: Now let: Summing all the terms yields: Part 3. Lemma: Let formula_29 be a polynomial with a root formula_39 of multiplicity formula_40. Then formula_41 for all formula_42. Proof: By strong induction. Let us write formula_43, with formula_44 a polynomial such that formula_45. For formula_46 we get: Assume that for all formula_48 the claim holds for all formula_42.<br> We shall prove that for formula_50 the claim holds for all formula_51: The is of multiplicity formula_53, with formula_54 a polynomial such that formula_55.<br> Hence their product satisfies the induction hypothesis. formula_10 Part 4. Let us now define: and formula_58 is a prime number such that formula_59. We get: hence for all formula_61, the function formula_62 is a polynomial with integer coefficients all divisible by formula_58. By parts 1 and 3, we get: The is an integer not divisible by formula_58.<br> The is an integer divisible by formula_58.<br> The is the most important:<br> By Vieta's formulae we get and the sums are symmetric polynomials in formula_68. Therefore, these can be expressed as polynomials In addition, we get: Hence, the blue part is an integer divisible by formula_58. Conclusion: formula_73 is an integer not divisible by formula_58, and particularly formula_75. Part 5. By part 2, we get: By the triangle inequality for integrals, we get: On the other hand, we get hence for sufficiently large formula_58 we get formula_80. A contradiction. formula_10 Conclusion: formula_6 is transcendental, hence formula_2 is transcendental. formula_84

Physics Explained Through a Video Game/Displacement. Introduction. Displacement is the change of an object's position between two times. It is a vector quantity, meaning that it has both a direction and a magnitude. Direction of Displacement. Displacement's direction is dependent on our choice of what is the positive direction and the negative direction are. This choice can vary between situations. However, we tend to consider the Cartesian grid convention to figure out whether a displacement is positive or negative. To the right is an image of a 2D-Cartesian grid. We generally use the convention that an upwards motion (+Y direction) has a positive vertical displacement and a rightwards motion (+X direction) has a positive horizontal displacement. With this in mind, if an object is found to have a motion in a certain direction, we can decide if it has a negative or positive displacement, as listed in the table below. Displacement with the Cartesian Grid Convention. As a general example, consider a handful of balloons above that are rising relative to the ground in Balloon Fight by MuadDib. These balloons are above the ground and becoming farther from it, traveling in an upwards motion. Therefore, we can use the Cartesian grid convention to show that the balloons are having a positive vertical displacement relative to the ground. Exceptions to the Cartesian Grid Convention. The car in Cruisin' 4A Bruisin' by Raspy 667 is facing and moving leftwards. To note, it's typical for cars to move in the way in which they are facing. Therefore, if we were to track the motion of this car on the bridge above, it may be more practical to assume that when the car moves forewords (in the leftward direction in this case) that it is moving in the positive direction. As such, we may want to define the leftward motion of the car to instead result in a positive displacement, overruling the Cartesian grid convention. Magnitude of Displacement. The magnitude of displacement is dependent on the shortest distance that exists between the starting and ending point during the observation of an object. Thus, the path taken between the start and end points doesn't effect the displacement. Instead, only the location of the start and end points do. As an example, suppose you are playing Battlegrounds 2 by HeyListen (on the right) and your goal is to cross to the other green hill. You can either jump up onto the ladder to cross or dodge cannon balls to get across. Assuming that you have the same start and end points, you will have the same displacement in either case. As such, although there are paths of different lengths between Point A and Point B in the image, the displacement magnitude remains the same.

Physics Explained Through a Video Game/Calculating Displacement. Topic 1.2 - Calculating Displacement. We can calculate for the exact displacement of an object in a particular direction by considering the final and initial positions of an object in that direction. Then, we subtract the initial position value from the final position value to get the change in position, which is displacement. This allows for us to quantitatively look at the change in an object's position, which will later make it possible to make clearer predictions about objects in motion. Formula for displacement in the X (horizontal) direction: formula_1 Formula for displacement in the Y (vertical) direction: formula_2 If we have a displacement that has both a vertical and a horizontal component, we can add these displacement vectors together to get the net displacement vector.Question 1: Suppose that there's an in-game situation where the player Hello (colored white)" wants to knock Jensen" "(colored grey)" off the board. At a particular time in the game, "Hello" needs to have a formula_3 of formula_4 and a formula_5 of formula_6 in order to reach "Jensen". How far does "Hello" need to travel in order to reach "Jensen", assuming that "Jensen" is at the same spot when "Hello" arrives? Solution:In order to calculate how far "Hello" needs to travel, we need to find formula_7. Using the image on the right, we can see that if we were to draw formula_8 from where formula_9 ends, we form a right angle between these two displacement components. Also, we know that because the net displacement is the shortest path between two points, formula_7 must be a line directly connecting "Hello" to "Jensen". Thus, we can form a right angle triangle that is outlined by formula_7 and its components. Because of the Pythagorean Theorem where formula_12 where a and b are components and c is formula_7, we can show that: formula_14 formula_15 formula_16 formula_17 Note that any displacement can either be positive or negative, depending on what is convenient for the situation. In this case, because we are looking for the magnitude of distance between Hello and Jensen, we can allow for formula_7 to be positive such that: formula_19 (using 1 significant figure).

Themes in Literature/Isolation and Community/Beautiful Dreamer. Introduction. "Beautiful Dreamer", directed by David Gaddie, is a science fiction short film adapted from Ken Liu's short story, "Memories of My Mother." The film chronicles Amy's life through intermittent visits from her mother. Diagnosed with a terminal illness, Amy’s mother undergoes an experimental treatment based on Einstein’s Theory of Relativity. This treatment allows her to extend her life expectancy from two years to approximately eighty, enabling her to witness her daughter’s growth. However, this temporal manipulation means that Amy's mother is only able to see Amy at certain points in her life, often disappearing for years in between these visits. The short film depicts the moments in which they are reunited, as Amy grows from childhood to adulthood to old age. Analysis. It can be said that opposing themes attract one another. Sadness amongst a happy story is more powerful than a tragic story alone. Gaddie utilizes this idea to depict the bond between Amy and her mother. We see this first in the setting of the film. Despite the plot taking place over decades, we see each scene depicted among a background of winter. Not only is the film set in winter, the lens of the film is altered with a grey/blue-tinted filter. The full scheme of the colors that are used are dulled slightly, making the mother the only character to stand out, clad in bright red. The cold, grey, wintry landscape provides an apt juxtaposition to the warmth radiated by the mother and the love between mother and daughter. In fact, Gaddie begins the film with a montage of mothers and babies within nature. He shows animals, vulnerable and innocent, being protected from the harsh elements of winter by their mothers. Gaddie then cuts to Amy and her mother. Immediately, the mother stands out in a bright red coat, red stockings, and high heels. As Amy’s mother leaves for the first time, she passes a young Amy to her father, who is clad fully in grey and black, a stark contrast to the brightness and warmth of Amy’s mother. Another prominent visual motif of the film is Gaddie’s use of doors. In the scene where Amy is depicted as a teenager, we see Amy, her father, and her mother entering, exiting, opening, slamming, and knocking on doors. In almost every frame at the beginning of the scene, doors are being opened and closed, and tensions are high. Because of this, the setting continues changing. We go between the hallway, to Amy’s room, to the father’s room. In doing this, Gaddie represents the feeling of being drawn into the minds of these different characters. It feels as though Gaddie is pulling the audience back and forth into different areas of the home, the camera constantly moving and changing. Each time the doors are opened and closed, the camera cuts and the audience is left wondering where the focus will shift to next. While the characters are being shut out, the audience is forced to be present, going exactly where Gaddie thinks they need to be, in whichever character’s feelings are most prominent in the moment. Think of the phrase, “Where one door closes, another opens”. Gaddie uses this in a stressful, almost over stimulating way, leaving the audience wondering which door will ultimately be left open or closed. A door itself is an invitation in, or an abrupt goodbye. A door is an opening to opportunity, or a security blanket for privacy. What better way to exhibit our characters’ inner turmoil of solitude versus acceptance than using the door as a symbol? Ultimately, the door remains open for the mother to continuously enter into Amy’s life, and Amy continuously lets her in. Amy continues to let her mother in at various points in her life, and their interactions provide a reprieve from life’s struggles. However, the overarching turmoil that Amy faces is still prominent in each scene. While Liu’s short story touches on these life struggles through narrative and Amy’s reminiscing, Gaddie instead focuses them on dialogue. For the purposes of a film adaptation, this makes more sense when writing the script. However, in the context of the story, it actually feels more palpable that Amy is saying these feelings out loud to her mother. Their relationship feels more real, as opposed to Amy feeling these things more on her own, with her mother roaming in the background. In one scene, we see the mother wipe a tear from Amy’s eyes. It is vulnerable and a subtle but impactful representation of a mother’s love. From the confusion of teenagehood, to her divorce as an adult, the death of her father, and her isolation from her own children in old age, we always see her interactions with her mother as a constant. Despite the character’s individual turmoil, their bond is a reminder of unwavering connection. The film ends with Amy at age 80. She and her mother have their final interaction, a full circle moment in which the mother’s initial goal is achieved, and Amy has been able to spend a lifetime with her mother. The film is a beautiful adaptation of love persevering in spite of the elements, an unbreakable bond between these two women. Characters. Amy (age 3) - our introduction to Amy’s character; the time at which her mother decides to leave Amy (age 10) - the first stage in which the mother comes back to visit Amy; we see childlike innocence; solely an appreciation at seeing her mother again Amy (age 17) - the first change in tone, with stronger themes of isolation among community; Amy is resentful of her mother; the father is most prevalent in this scene Amy (age ~26) - a pivotal moment - the only scene where Gaddie departs from Liu’s short story. This stage of Amy’s life is not included in the original story. Amy and the mother are the same age; their roles go from being the same (as they have reached the same point in life) to reversed (as Amy must now recognize her mother as an equal, and care for her in a time of need) Amy (age 38) - Amy is fully in adulthood; we see her with her own children Amy (age 80) - Amy is at the end of her life; we see an elderly Amy reconnecting with her mother, who has remained physically unchanged; true love and closure as the film comes to an end Mother - Amy’s mother who has been diagnosed with a terminal illness; uses futuristic technology to extend her lifetime from 2 years to 80 years to remain with her daughter. The mother is often absent, but returns at different stages of Amy’s life Father - Amy’s father who is the main parent throughout her lifetime

Themes in Literature/Isolation and Community/Preface. This volume of "Themes in Literature" explores the complex relationship between isolation and community through the lens of literature. The individual lessons collected here were written by college students. With a few exceptions, these lessons examine single literary texts that are readily available on the web. The texts presented explore how societal forces and individual choices intersect to shape our understanding of belonging, displacement, and the search for meaning in a world characterized by both profound connection and stark isolation. The selected works include a diverse collection of short stories examining the complexities of human connection and isolation. Anton Chekhov's "The Bet" delves into the psychological effects of solitude on a man imprisoned for fifteen years. This theme of isolation continues in Ray Bradbury's "Night Call, Collect," a science fiction story about an aging astronaut haunted by phone calls from Earth, and Julio Cortázar’s "The Southern Thruway," which chronicles a group of travelers trapped in a massive traffic jam. Ken Liu's "The Paper Menagerie" explores how cultural differences can lead to isolation between a Chinese immigrant mother and her American-born son. Oscar Wilde's parables, "The Happy Prince" and "The Selfish Giant," use allegorical figures to consider the consequences of selfishness and the transformative power of compassion. Finally, the collection examines "Dream House as Choose Your Own Adventure," a chapter from Carmen Maria Machado's memoir "In The Dream House", recounting her experiences in an abusive relationship. Two novels by Octavia Butler, "Kindred" and "Dawn", explore the power dynamics in relationships shaped by historical trauma and the potential for connection and transformation despite seemingly insurmountable obstacles. "Kindred" follows Dana, a young African American woman who travels back in time to a pre-Civil War plantation, forcing her to confront the horrors of slavery and her own family history. "Dawn", the first book in the "Xenogenesis" series, imagines a future where humanity has been rescued from the brink of extinction by the Oankali, an alien species with a unique approach to genetic engineering and community building. The volume also analyzes a selection of poems that use vivid imagery and unconventional syntax to examine feelings of loneliness and the search for connection. Three poems by Emily Dickinson—"Much Madness is divinest Sense," "They shut me up in Prose," and "The Soul selects her own Society"—use the metaphor of confinement to explore themes of social conformity, self-expression, and the search for genuine connection. Two poems by Langston Hughes, "Let America Be America Again" and "Mother to Son," consider the challenges faced by marginalized communities in their pursuit of the American dream, using the extended metaphor of a difficult journey to convey resilience and hope. Two short films, "Belonging" by Pierre Ieong and "Beautiful Dreamer" by David Gaddie, offer meditations on the search for connection across cultures and time. "Belonging" tells the story of George, a young Syrian refugee who finds solace and community in a diverse international school in Paris. "Beautiful Dreamer", a science fiction film adapted from Ken Liu's short story "Memories of My Mother," explores the complexities of a mother-daughter relationship impacted by an experimental treatment that allows the mother to live for decades but only visit her daughter at certain points in her life. Last but not least, the lesson "The Deluded Self" examines the complexities of self-delusion and its impact on perception and relationships by contrasting two texts, “"Dream House" as Choose Your Own Adventure” by Carmen Maria Machado and the play "No Exit" by Jean-Paul Sartre. Machado story uses a choose-your-own-adventure format to recount experiences in an abusive relationship, emphasizing the difficulty of escaping self-deception as choices ultimately lead to the same outcome. Sartre's play explores the concept of bad faith and the façades people construct to avoid facing their true selves.

Physics Explained Through a Video Game/Speed and Average Velocity. Topic 1.3 - Speed and Average Velocity. When using the concept of displacement, we can also describe the motion of an object over a period of time. The rate at which the displacement of an object is called its velocity. We can define the average velocity as the "average" rate at which the displacement is changing over a period of time. Since displacement is a vector, velocity is also a vector. Because velocity is a vector, it describes a object by (1) identifying the direction of motion and (2) finding the average rate at which an object is moving over some distance relative to time. By using the described definition for average velocity, we can represent average velocity by using the formula in terms of displacement and time as shown below. formula_1 To better understand velocity, consider the video on the right. In the map "DumbBell Brown" by GayfishDeluxe players are moving back and forth to fight against one another. Because velocity is a vector, it varies depending in which direction an object is traveling in. For instance, in the clip, Hot Wheel "(reddish-orange color)" fires an arrow in both the leftward and rightward directions. By the end of this topic, we will explore the idea that even though the arrow may have been shot just as fast in both cases by "Hot Wheel", the arrows will have had different velocities immediately at launch. Block on Ice Activity. To apply how average velocity is calculated, an interactive demo is provided below. For this activity, we will be observing the motion of a block sliding across ice to gather information about its average velocity. Task: "Note:" "Task Solution:" Here's a table and graph of the collected positions of the block and their associated times. The graph allows for us to see that the position is linearly increasing with respect to time. We can include a curve fit to showcase this relationship. Also, because formula_7, formula_8 is our vertical change and formula_9 is our horizontal change on the graph. As such, formula_4 is the slope of the graph on the right. Since the graph is linear, this means that the slope (formula_4) will be the same regardless of what our final and initial (Time, Position) data points were. This explains why the task described above did not specify which two specific data points to collect; formula_4 should be similar regardless of the chosen values. As an example of two selected data points, consider the (Time, Position) data points of (0.71, 0.0) and (3.28, 4.0). Since formula_13 occurs after formula_14, this means that the data point (3.28, 4.0) are the final values and (0.71, 0.0) are the initial values. We can substitute these data points into the equation for formula_15 and simplify. formula_16 formula_17 formula_18 To clarify, because our measured values for calculating formula_15, distance (m) and time (s), respectively have 2 and 3 significant figures, it's convention that our final answer should only have 2 significant figures. Also, note that answers may vary slightly depending on which exact data points were collected while observing the video. The Difference Between Speed and Velocity. On the right is a video showcasing circles moving in different directions but having the same speed. Unlike the average velocity of a system, the speed of the system is only dependent on the magnitude of the rate at which an object is moving. Thus, there's a few notable properties of speed: The total distance traveled refers to the length of the path in which an object has taken. This is separate from the displacement, which only considers the start and end points of the path. Using the videos below, monkey butler "(orange color)" is able to jump over the barrier and fill in the Capture Zone in a single jump. In contrast, Other Guy "(green color)" tries several times in order to cross over the barrier until he is able to do so. Although both "monkey butler" and "Other Guy" have the same displacement between their spawning location and the Capture Zone, "Other Guy" moved a greater total distance because of him running back and forth. Since speed is always a non-negative value, this means that average speed is not necessarily equal to the average velocity. We can see this through an example of a clip played in Football Mode as shown below. From this video, we are able to take a record of whenever the football passes over one of the grid lines and record its time. This allows for us to begin getting an idea of the horizontal motion of the football during the clip. Through the previous graph of the position of the football relative to time, we can consider each two blue data points from the X-Value Passed vs. Time graph. From this, on the Average Velocity vs. Time graph, we can plot red data points representing the average velocity of the football between each pair of blue data points. To note the differences between the average velocity and the average speed in this situation, because the average speed of the object can never be negative, whenever formula_23 Also, if the football was to change direction while between the grid lines, the average velocity cannot equal the average speed. As previously mentioned, average speed is dependent on the total distance of the path taken. This is dissimilar to using displacement to calculate average velocity. Therefore, if the football loops back and forth while traveling, the total distance traveled will be greater than the displacement of the football during the same time interval. An example of this is between 0.90 and 1.20 seconds in the Football video above. Because the football bounces off the wall and back onto the 9.0 meter line, there's no displacement during this time interval. However, the ball travels a non-zero total distance towards the wall and back. Thus, although formula_24, the average speed over this time interval must be greater than 0 m/s.

Hej, Jonathan! (Germanisch). Hej, Jonathan! is a course that teaches Germanisch, a zonal international auxiliary language created by Olivier Simon, the creator of Sambahsa. The course is a translation of "Salute, Jonathan!", originally authored by Dave MacLeod and wrriten in Occidental. Först kap'itel (kap'itel 1). Den förste Mai (1 Mai). En mann sto in en stad. Den mann schryv en dagbuk. Den mann sie en stad. Sto den mann in en stad? Ja, hi sto in en stad. Sto den mann in...en mann? Nee, hi sto nit in en mann. Hi sto in en stad. Sto den mann in en trein? Nee, hi sto nit in en trein. Hi sto in en stad. Den mann sto in en stad. Schryv den mann en dagbuk? Ja, hi schryv en dagbuk. Schryv dat dagbuk den mann? Nee, dat dagbuk schryv nit den mann. En dagbuk schryv nit. En mann schryv. Den mann schryv dat dagbuk. Sie den stad den mann? Nee, den stad sie nit den mann. En stad sie nit. En mann sie. Sie den mann den stad? Ja, den mann sie den stad. Waar sto den mann? Hi sto in en stad. Wat schryv den mann? Hi schryv en dagbuk. Wat sie den mann? Hi sie en stad. Hi sto in en stad, oh hi schryv en dagbuk, oh hi sie en stad. "Sto" den mann in en dagbuk? Nee, hi "schryv" en dagbuk. Sie den mann en mann? Nee, hi sie nit en mann; hi sie en stad. Den mann is grot. Den mann is gud, oh den mann es kluk. Hi denk. Hi denk over den stad. Hi denk: "Wat is den stad? Is den stad gud? Is den stad grot?" Schryv den mann en stad? Nee, hi schryv nit en stad; en stad is grot. Hi schryv en dagbuk; en dagbuk is nit grot. Denk dat dagbuk over den mann? Nee, en dagbuk denk nit. En mann denk. Hi denk over dat dagbuk, oh denk over den stad. Oh hi schryv in dat dagbuk. Hi schryv over den stad. Hi schryv: "Den stad is gud, oh den stad is grot." Hi denk: "Den stad is gud"; hi denk dat den stad is gud. Hi denk: "Den stad is grot"; hi denk dat den stad is grot. Hi denk over den stad, oh hi denk over dat dagbuk. Wat is dat dagbuk? Dat dagbuk is waar den mann schryv; hi schryv in dat dagbuk. In dat dagbuk, den mann schryv over den stad. In dat dagbuk, den mann schryv nit over den trein; den mann sto in den stad, nit in den trein. Hi denk nit over den trein; hi denk over den stad, den stad München. Wat is den stad? Et is München. Waar is München? Et is waar den mann sto. Waar is den mann? Hi is in München. Ja, München is en groten stad, oh en guden stad. Den mann denk dat München is en guden stad, oh hi denk dat et is en groten stad. Den mann es kluk. Hi sej: "Hej, München!" Hi is en guden mann! Den mann sto oh denk: "Waar is den trein?" Hi sie...hi sie den trein! Hi denk: "Den trein!" Nu denk hi nit over dat dagbuk oh denk hi nit over den stad; hi denk over den trein! Twodet kap'itel (kap'itel 2). Den twode Mai (= 2 Mai). Nu is den mann in den trein. Hi is nit in München; hi is in en trein. Hi rys. Hi denk: "Nu rys ik fram München tu Wien. Et is en guden ryse. Ik lik ryse." Hi denk over München. Hi denk: "Nu är ik in den trein, mä gester was ik in München. Oh nu schryv ik en dagbuk in den trein, mä gester schriv ik en dagbuk in München. Oh nu denk ik in den trein, mä gester daht ik in München. Gester daht ik over München in München, oh nu denk ik over Wien in den trein. Nu är ik in den trein, nit in Wien. Mä ik denk oh schryv over Wien." Denk den mann in München nu? Nee, hi denk nit in München nu. Hi denk in den trein. Gester hi daht in München. Hi sej: "Hej, trein!" Den mann is in den trein, oh hi rys tu en stad. Den stad is nit München; München is gesters stad. Den stad is Wien; Wien is disdags stad. Den mann denk over München oh Wien. Hi denk: "München was gesters stad, oh München was gud. Nu is et disdag, oh ik är in en trein; den trein is gud. Will Wien wese gud?" Den mann denk over München: München was gesters stad. Hi denk in den trein: hi is in den trein disdag. Oh hi denk over Wien: Wien will wese tu'morgens stad. Oh hi denk: "München was grot. Den trein is grot. Will Wien wese grot?" Oh hi denk: "In München ik schriv in en dagbuk. In den trein schryv ik in en dagbuk. Will ik schryve in en dagbuk in Wien? Ja, tu'morgen will ik schryve in en dagbuk in Wien. Ik lik dagbük." Den mann denk mang (hi denk mang = hi denk oh denk oh denk), oh hi schryv mang. Ja, hi is en kluken mann. Kluke männ schryv mang, oh denk mang. Hi is Jonathan; Jonathan is en kluken mann. Hi schryv: "Ik är Jonathan. Ik är in en trein. Gester stode ik in München; tu'morgen will ik sto." Hi denk, oh schryv: "Den trein...et is gud, mä ald. Et is nit neu; et is ald. Är de trein in München ald? Ja, Münchens trein är ald. Mä Münchens trein är gud, oh ik lik Münchens trein. Gester likte ik den trein in München, oh disdag lik ik den trein nu, oh tu'morgen will ik like den trein in Wien. Ik lik trein!" Jonathan schryv: "München is en guden stad oh en alden stad, oh Wien is en guden stad oh en alden stad. München oh Wien är nit ald, mä är gud. München oh Wien är alde, mä gude städ. De städ är nit neu, mä gud. Ik lik städ!" Jonathan denk, dat den förste rysedag is gud. Hi sej: "Disdag was en guden twoden rysedag. Ik lik ryse!"

Physics Explained Through a Video Game/Systems and Center of Mass. Topic 2.1 - Systems and Center of Mass. Before looking discussing forces later in this unit, we need to consider a few definitions first. Introduction. The center of mass (sometimes abbreviated as CoM) of an object refers to the average position of all of the parts that form the object. In classical mechanics, considering this can be useful in variety of circumstances as it oftentimes enables us to view an awkwardly-shaped object as a singular point. Later on in this article series, this will be particularly helpful in simplifying many problems and concepts. In some cases, such as in "Ferris Wheel "by antidonaldtrump / Armin van buren (on the right), it may be useful to locate the center of mass of a system of objects. To note, a system in physics is just some collection of objects or parts of an object that we're interested in observing together. Example 1: Ferris Wheel. As an example of how to calculate for a system's center of mass, we need to first consider the objects with mass that compose it. In the case of the Ferris wheel, for simplicity, we're using the assumption that only the six carts of the Ferris wheel have non-negligible mass. Thus, the rest of the Ferris wheel (such as the base or the rim) won't be considered as contributing to its center of mass. With this out of the way, we can define that there are six objects that compose the system. For each of these objects, we need to know their own center of masses. Using the diagram on the right hand side, which represents the locations of each of the six cart's center of mass, assume that: In order to calculate for where the Ferris wheel's center of mass is, we need to use the definition for the center of mass in a particular direction, as provided below. Essentially, this allows for us to separately calculate the center of mass of a system in the horizontal and vertical directions (assuming a 2-Dimensional situation). From this, we can describe where the center of mass (in a specific direction) by knowing the relative mass and coordinate of each cart. "With the formula above, formula_3 represents the mass of each object in the system. Also, formula_4 represents that respective object's center of mass in the (X) direction." To reflect on the information that we've gathered so far, refer to the table below. In this, we're assigning each cart and its center of mass location as a part of the formula above. With the information labeled above, we can begin calculating for the center of mass in the formula_5 and formula_6 directions. We can start with calculating the formula_5 direction's center of mass, formula_8. To do this, we need to employ the formula specified above for the center of mass in a particular direction, formula_9. Using the information from the table above and then substituting the known variables, we will find that: formula_10. <br> This can be algebraically simplified such that:<br> formula_11 formula_12 Practice Question: Using the provided table and the process shown for solving for the center of mass in a specific direction, solve for formula_13. Using this information, where is the center of mass of the Ferris wheel system, (formula_14, formula_13)? Using the table on the right, we can consider the columns dedicated for the assigned mass (formula_16) and the Y-coordinate (formula_17) of the parts of the system (the Ferris wheel). Also, since the center of mass in the Y-direction is being solved for, the related equation can have the particular direction be specified as formula_6 such that: formula_19 Similar to how formula_8 was calculated, we can then use substitution of variables and algebraically simplify to find formula_21. This derivation results in finding formula_21, as shown below. formula_23 With knowing both the values for the formula_14 and formula_13, we can now specify the center of mass of the Ferris wheel's as being located at formula_26. For more context, we can again consider the reference image of the Ferris wheel system. By using the surrounding labeled coordinates, we could plot where the center of mass as being just below the center shaft (where the large grey rods in the image meet together). See the Ferris wheel image below with a superimposed bold yellow star indicating the approximated center of mass's location. Example 2: Blocky Snakes. Section I: Content Primer. In the previous example about a Ferris wheel, a specific moment was considered for calculating for the system's center of mass at that instant. However, it required a considerable amount of effort despite the system only having a few considered parts. This raises a couple of questions: In this example, we'll address these questions, providing more of an insight towards the intuition and applications behind a system's center of mass. Consider the map "Strange Snake" by "11vanya11". In this, a snake made out of seven blocks circles around a rectangular perimeter, as seen on the right. Suppose that all of these seven blocks with a uniform density* each with the same mass. Because the snake's block has a uniform density, the removed portions will have the same mass. Also, because they are on opposite sides, they won't change where the snake block's center of mass is. This is because the center of mass in other words is the average point of where an object's mass is. To note, with many basic flat geometric shapes, such as a circle, square, equilateral triangle, etc., if it has a uniform density, then the center of mass is where the geometric center is. Section II: Helpful Techniques for Finding the CoM. When considering the information of the snake's blocks (and the sidenote's information above), in many cases, we can simplify the process of finding the center of an object's mass. To explain this concept, let's consider a still image of "Block Snake" by "O_o O_o O_o," a similar map as shown on the right. Suppose that each block of the snake has a mass of formula_27 and is formula_28 in length and width. We now know that if (1) a system or system part creates a simple geometric shape and (2) that system or system part has a uniform density, then it has a center of mass where the geometric center is. When looking at the image on the right, it is clear that the snake as a whole does not create a simple geometric shape. However, there are parts of the snake that each create a rectangular shape. Because the entire snake has a uniform density, each of these snake parts (shown below) has its own center of mass at its geometric center.Through using the method shown above, if we were to use the formulas for formula_8 and formula_21 (provided a coordinate plane), the computation is much simpler. Instead of having to consider the formula_31 coordinates of 16 different blocks, we only have to consider four distinct groups to find the entire snake's center of mass at that moment. Continue this example: If we were aware of the coordinate position of each of center of masses or otherwise were able solve for them, we would be able to figure out where the snake's entire center of mass is. Recall that the snake's blocks each have a length and width of formula_32. If we allow for the green segment's center of mass to be located at the formula_31 coordinate formula_34, we can manually figure out where the other center of masses are located as shown in the diagram on the right. From this point, we could use the formulas for formula_8 and formula_21, as introduced in Example 1. formula_9 formula_38 formula_39 formula_40 formula_19 formula_42 formula_43 With both formula_44 and formula_45, we can define that the center of the snake's mass is approximately at formula_46, as labeled by the image now with a superimposed star on the right. To clarify, it is okay for the center of an object's mass to be not on the object itself. For instance, consider a doughnut that (1) has a hole in the middle of it and (2) has uniform density. Although there isn't mass from the doughnut itself inside of the hole, the doughnut mass surrounding the hole averages out such that it's center of mass is in the hole. As such, as also seen by the example just shown with the snake, the center of an object's (or a system's) mass does not necessarily have to be inside of the object itself. Section III: Velocity of the CoM. As seen in the previous sections, we've idealized finding the center of mass of a system by object considering still frames. However, oftentimes, the center of mass is undergoing motion, making it necessary to account for this in some situations. To explore this concept, consider the map Climb by Fantao. Assume that: What is the velocity of the center of mass of the system of visible platforms? Approaching this problem may be intuitive. To explain, if an entire system (such as a car, or in this case: a set of green platforms) is moving at the same velocity, then it makes sense for the center of mass of that system to be moving at that velocity. This would be regardless of where that center of mass is positioned. More formally, we can also solve this problem by considering the formula for the velocity of the center of mass: formula_50<br> formula_51<br> formula_52 Alongside this, because the platforms aren't moving in the horizontal direction, there is a formula_53 that equals formula_54. We can add these vectors together similar to how displacement was calculated for in Topic 1.2. As such, we can specify that formula_55. Thus, formula_56 <br> formula_56 formula_58. Question 1: Icebergs. Consider the map "Breaking Ice" by "Semi_Cow124", as pictured on the right.<br>Part (a): Calculate the location of the center of mass of parts of the three icebergs that are above water. Assume that 90% of each of the iceberg's total mass is below water. It's been discovered that the center of mass for the three iceberg's portions that are underwater is at formula_59.<br>Part (b):(i): How much mass do three icebergs combined have underwater? <br>(ii): Calculate the location of the center of mass of the three entire icebergs.Suppose a new in-game situation (above) where players are actively jumping on the icebergs. Because of their impacts, the largest iceberg in the center of the map begins to fracture and fall apart. At a certain moment, 70% of the largest iceberg that is above water has its vertical velocity recorded, as diagrammed on the right. Assume that the remainder of the iceberg remains stationary. <br>Part (c):How quickly is the center of mass of the center iceberg's portion above water falling vertically? "Consider discussing your solutions on this article's , where you find help from others."

Physics Explained Through a Video Game/Motion in 2 Dimensions. Topic 1.5 - Motion in 2 Dimensions. There are situations where we may want to track the motion of an object that is moving in more than one dimension. Oftentimes, this is illustrated through projectile motion. Projectile motion is when an object travels through the air while only being accelerated by gravitational force (also known as the force due to gravity). For now, all we need to know about forces in general is that they make objects accelerate in a certain direction. With gravitational force specifically, it results in all objects with mass to have an attractive force. This makes objects with mass to accelerate towards each other. As a real life example, if one was to drop a banana peel on Earth, the banana peel will accelerate downwards until striking the ground. The Earth also would accelerate upwards towards the banana peel at the same time. However, the displacement of Earth because of the banana would be extremely small because the Earth has far, far more mass than the banana peel. As such, for almost all purposes, we can allow for the Earth's movement because of the banana's mass to be negligible. <br><br> "For more of an introduction into gravitational attraction and other related information, check out this video by VSauce (from 2:40 to 7:27)." Consider the map "Banana jungle" by "O_dot" (below). We can see bananas accelerating towards the ground (in the downward direction) after they're released by the monkey. Until the bananas are either in contact with the ground or the jungle, the bananas are in free fall. This means that the bananas are only accelerating because of the force of gravity from the Earth. As such, the bananas have their vertical velocity become more negative while falling. However, their horizontal velocity stays the same. This property can be seen by tracking the motion of the first banana thrown by the monkey (at about 2 seconds into the video). In this, the banana in free fall is accelerating at a constant rate downwards but has no acceleration horizontally. As such, note the following: Air Resistance. In real life, air resistance affects all objects on Earth while they are in projectile motion in Earth's atmosphere. This is because objects collide directly with the air when moving differently than it. As such, the air can apply a drag force on the object. This force slows the object down and shortening its flight path. The drag force exerted on an object is dependent on a variety of factors, including the exact shape of the flying object and its velocity magnitude and direction. <br><br> In this content series, unless otherwise specified, we will assume that the drag force (air resistance) is negligible for all experiments to make calculations more manageable. <br> As explained above, air can apply a significant drag force on objects in some situations. For instance, in Emergency Landing by Raspy 667 (on the right), the visualized plane is falling at a constant vertical speed rather than accelerating towards the ocean. This specific example will be further explored in Unit 2: Forces and Newton's Laws. However, for now, it is important to discuss this from a kinematics perspective. Gravitational force is making the plane accelerate downwards; however, an equally strong drag force is making the plane accelerate upwards. As such, because both of these forces are happening at the same time in the video, the plane is neither accelerating upwards or downwards. Instead, the plane has an acceleration of 0 m/s2 until it lands. Because of this, the plane has a constant velocity (in the downwards direction) until it impacts the ocean. Airborne in a Cavern. "For this section, a physical situation will be broken down and fully explained. It is meant to provide an in-depth example for the following Practice Problems for this topic. For better understanding, it is recommended that you go through this section carefully beforehand." With a stronger understanding of what makes objects in projectile motion accelerate, we'll now look into the trajectory of Clare "(lime green color)" in a modified version of Gemspark Cave by G3nius (down below). Suppose "Clare" is launched upwards and towards the right in projectile motion. Assume that any vectors describing her motion have the components in (X,Y) notation in the Cartesian grid convention from Topic 1.1 - Displacement. It is given that "Clare" has an initial velocity vector of (10., 20.) (m/s). Also, assume that air resistance (drag force) is negligible and that the magnitude of "Clare's" acceleration due to gravity is 9.8 m/s2. Example 1: Calculating Angles Calculate the angle (in degrees) at which "Clare" is initially launched at relative to the +X direction. When "Clare" is initially launched, we're already given the initial velocity components. With this, we can calculate directly for the angle of her launch (with respect to the +X direction) by constructing a right angle triangle from the known initial vertical and horizontal velocity vectors. More specifically, since the initial velocity is (10., 20.) m/s and is in (X,Y) notation, formula_1has a magnitude of 10 m/s and is facing rightwards. Also, formula_2would have a magnitude of 20. m/s and be pointed upwards. We can sketch these vector components onto "Clare" as shown on the right. With this, we can consider the right angle triangle constructed by the formula_3vector and its components. We can solve for the direction of a vector through considering its components and performing trigonometry. With this, we can make a derivation for formula_4 by noting that formula_5. From this, we can substitute in the known values for the sides of the drawn triangle. For reference, the side lengths represent the magnitudes of the components of formula_6. Therefore, formula_7. Then, we can compose both sides of the equation by the inverse of formula_8, formula_9. This allows for us to rewrite the equation as formula_10 As such, we are able to calculate that at the instant when "Clare" is initially launched, she is traveling about 63 degrees above the +X direction. Example 2: Considering Acceleration and Velocity Together Consider the horizontal and vertical velocity of "Clare" while in the air. Because only the force of gravity is acting on "Clare" during then, in what direction is she accelerating towards? Also, how does this affect the horizontal and vertical velocities of "Clare" while in projectile motion? With the understanding that there is only the force of gravity on "Clare", the player is accelerating in the downwards direction while in projectile motion. Because of this, "Clare" only has a changing vertical velocity while her horizontal velocity stays constant. To delve into this further, we know that the player's acceleration due to gravity is 9.8 m/s2 in the downward direction. This is the only acceleration that is occurring on "Clare". As such, the gravitation acceleration is acting fully in the vertical direction and not in the horizontal direction. Therefore, "Clare" has a vertical velocity that is decreasing by 9.8 m/s every second. This can be seen by the equationformula_11from Topic 1.4 - 1-D Uniform Accelerated Motion. To get an intuition of "Clare's" changing velocity, check out Ohio State University's Bonk.io Physics Coding Project. With this interactive demo (pictured on the right), we can find that if only gravitational force is acting on a player while airborne: "What if the vertical acceleration was different?"In this example, we directly modify the overall vertical "acceleration" of the player while keeping the horizontal acceleration as 0. In the clip's first half, the down arrow key is held (making the player have a more negative acceleration). As such, for when the player is airborne, the velocity vector will quickly move downwards. This is in contrast to the clip's second half where the up arrow key is held (making the player have a less negative acceleration). In this test, the velocity vector noticeably moves downwards more slowly. Thus, we can see that by having a smaller acceleration magnitude in some direction, the velocity in that direction will change more slowly. Example 3: Solving Multi-Step Kinematics Problems Suppose "Clare" impacted the cave wall 3.5 seconds after launching with the previously specified conditions. What is "Clare's" net displacement magnitude during this time interval? Although it appears that with only a time interval, the initial velocity vector, and the acceleration due to gravity, that calculating a precise value for the displacement of "Clare" is daunting, this problem becomes much simpler if we break it down. Firstly, because the acceleration of "Clare" being vertical, the horizontal velocity stays the same for the entirety of the time interval. Because velocity is the rate at which displacement is occurring, if the velocity is constant in some direction x, formula_12. Because formula_13and formula_14, formula_15. Considering the vertical velocity, refer back to the general kinematic equation table given constant acceleration. To note, we already know formula_16, formula_17, and formula_18. With this, we're looking for the vertical displacement formula_19, which is another way of saying formula_20. With the table, we can use the equationformula_21to solve this problem given the variables that we either know or need to solve for. To clarify, because the direction in the formula is merely a placeholder, we can use the equation with the direction being vertical. Thus, we are solving for vertical displacement withformula_22. formula_22 formula_24 formula_25 formula_26 <br> With the two displacement components formula_27 and formula_28, we can solve for the net displacement through creating a right angle with the components and using the Pythagorean Theorem (as shown to the right) in order to solve for formula_29. formula_30 formula_31 formula_32 formula_33 Thus, "Clare" has a net displacement of 36. meters in the cave during the 3.5 second time interval between her launching and her collision with the cave wall. "**If needed, check out Topic 1.2 - Calculating Displacement for additional help with the derivation." Practice Problems. "Note:" The (table of kinematic equations given constant acceleration) may be helpful for some question parts. Question 1: Suppose a snake is sliding across boulders in the map "Catalina" by "monkey butler". The snake falls off a large boulder with a horizontal velocity of -0.3 m/s and enters free fall motion until hitting the grassy ground 1.0 m below. While in free fall, the snake is accelerating downwards at a constant magnitude of 9.8 m/s2. The snake experiences negligible air resistance. Part a) (i) For when the snake is free falling, which acceleration vector would model the overall acceleration of the snake? Pick one of the four choices from the image on the right. (ii) Using the Cartesian Grid Convention, would the acceleration vector of the snake be in the positive or negative direction? Part b) If we were to solve for the amount of time that would elapse during the snake's fall, which formula should we use with the given information? Part c) Evaluate the amount of time that elapses during the snake's fall. Part d) Using your response from part c), determine the horizontal displacement of the snake during its fall. Question 1 Solutions: By considering the initial prompt, we can collect the information that is already known about the snake's free fall. Through considering the prompt carefully, we're now able to quickly answer Part a): With Part a) (i), because acceleration due to gravity is the only type of acceleration acting on the snake while it is in free fall motion, the snake's acceleration vector can be represented by an arrow pointing downwards. Thus, Choice D in the diagram shown again on the right would be the correct answer. To answer Part a) (ii), since the snake's acceleration is pointed downwards and we're using the Cartesian grid convention, the acceleration of the snake would be in the negative direction. In order to answer Parts b) and c), we need to find a way to directly calculate for the time it takes for the snake to fall. By considering the (kinematic equation table), we can decide which equation would be the best to use in this situation. One way to do this is by ruling out the equations that we can't use. We can accomplish this by asking ourselves one question: If we subsistence the known variables into some equation, would be be able to directly solve for a singular unknown value? To cut to the chase, we can't easily introduce more unknown values and be able to solve for the unknown variable that we're asked to look for. By using this process of elimination, we can decide against using Equations 1, 2, and 4. This is because all of them would introduce the unknown variable formula_38. Because we don't know what the final vertical velocity of the snake is, choosing one of these equations would either be introducing additional work or be leading us to a dead-end. As such, let's consider Equation 3 because of a lack of an alternative: formula_22. We know the vertical displacement of the snake, which would be equal to formula_40 . Thus, by definition, we can rewrite the equation as: formula_41. Additionally we know that the initial vertical velocity, formula_42, is equal to 0. If we substitute this into our equation, then the term formula_43would also equal 0. Thus, we can remove this term from the equation. To clarify, even if we were considering a different situation where formula_44, we could still eventually arrive with directly solving for the time elapsed. However, this would involve dealing with simplifying a quadratic equation, such as through using the quadratic formula. After this, we would also likely need to eliminate an extraneous solution. For this particular problem with the snake, the solving process will be simplified because of the term formula_43 being removable. As such, we can write Equation 3 in the form formula_46 for the situation with the snake. Because the vertical acceleration (which isformula_47) and the vertical displacement (formula_48) are both known, we can solve directly for the time elapsed. Therefore, Equation 3 can be used to solve for the time elapsed, answering Part b). Because of the derivations made above, we can use the rewritten form of Equation 3, formula_46, and use algebra to directly solve for formula_50as shown below: formula_46 formula_52 formula_53 formula_54 formula_55 formula_56 Thus, the snake falls for 0.45 sec between the time at which it falls off the boulder and when it impacts the grassy ground, answering Part c). Finally, in order to approach Part d), we need to be aware that the horizontal velocity of the snake is constant. This is because there are no other forces acting on the snake apart from gravitational force. As such, we can consider our initial horizontal velocity, formula_34. If our horizontal velocity is constant, then it will remain the same at all times while the snake is falling. Therefore, formula_58. As introduced in Topic 1.3 - Speed and Average Velocity, we know that formula_59. Because formula_60 would be equal to formula_61during the snake's fall, we can substitute formula_60into the expression and simplify, as shown below: formula_59 formula_64 formula_65 formula_66 formula_67 Thus, the snake has a horizontal displacement of -0.14 m during the fall, answering Part d). Question 2: Suppose a great white shark swallows a krill in the map "SHARK!!" by "O_dot". The krill accelerates rapidly past the shark's jaws before disappearing in the shark's stomach 0.500 seconds later. The krill's acceleration vector relative to the shark has a constant magnitude of 19.5 m/s2 and is pointed below the -X direction by 5.00 degrees. To note, the krill is assumed to initially be at rest when entering the shark's jaws. Part a) (i) Consider if the krill's position was recorded by a dot every 0.100 seconds between when it entered the shark's jaws formula_68to its arrival at the shark's stomach formula_69. Which of the three choices below would best represent the krill's changing position during this time interval? <br> (ii) Referencing the diagram to the right, calculate the magnitude of the krill's acceleration in the horizontal direction. <br><br> <br> <br> Part b) Calculate the horizontal displacement of the krill between 0.000 and 0.500 sec. Part c) Suppose a researcher is solving for the krill's overall speed when it arrives in the shark's stomach formula_70. What should the researcher use as the acceleration value when solving for the krill's overall speed? Also, write a justification for your choice. Part d)* In a new situation, suppose the researcher later spots the shark swallowing a second krill. Both krills travel into the shark with the same acceleration and travel the exact same path. However, the second krill travels into the shark with an initial velocity magnitude of 2.0 m/sec. The initial velocity is directed 5.0 degrees below the -X axis. How much time elapses for the second krill to reach the shark's stomach from the shark's jaws? Question 2 Solutions: Just as we did in Question 1, considering the initial prompt allows for us to quickly get an idea of what exactly is going on. By reading the prompt: With this, we know that the krill is starting from rest and is accelerating such that it is quickly moving leftwards and downwards. Thus, with Part a) (i), we need to find the choice where the average velocity of the krill is increasing as time passes. Because the dots displayed on Choices A, B, and C are meant to be position records of the krill at 0.000, 0.100, 0.200, ... , and 0.500 seconds. Therefore, 0.100 seconds passes between each recorded dot. Since formula_75such that formula_76, if we know that the average velocity between each time interval is increasing, then the displacement between each subsequent dot must be greater. In other words, the krill is becoming faster because it is accelerating from rest. As such, the dots should be spaced further apart as the change in time between the dots are the same but the krill's velocity is increasing. When considering our choices, only Choice B would satisfy the situation as the dots are becoming further. To rule out the other choices, Choice A incorrectly has the dots all evenly spaced, implying that the krill has a zero acceleration and that it doesn't start from rest. Also, Choice C incorrectly shows the dots becoming closer to one another as time passes. This would falsely imply that the krill does not start from rest and that the krill's acceleration would not be in the left-down direction. Thus, Choice B (displayed on the right) is the answer for Part a) (i). For the following section, Part a) (ii), we're presented with a computational problem with the support of a diagram (reproduced on the right) and need to calculate for the magnitude of the krill's horizontal acceleration. To note, we're already given the krill's net acceleration, formula_77and its angle relative to the -X axis, formula_73. However, we need to figure out the horizontal component of already known net acceleration. Using the provided diagram, we can see that the components construct a right angle triangle in which formula_79extends in the horizontal direction. Using trigonometry, because formula_80in which formula_81and formula_82, we can rewrite this equation as formula_83and then solve for formula_79. formula_83 formula_86 formula_87 formula_88 Thus, the krill has a horizontal acceleration magnitude of 19.4 m/s2 when it is being swallowed by the shark, answering Part a) (ii). Note that if we weren't solving for the magnitude of the krill's horizontal acceleration, then this value would have been negative as the krill by the Cartesian grid convention. This is because the krill is accelerating to the left. When beginning to approach Part b), in order to approach solving for the horizontal displacement, we need to consider what we currently know about the krill's horizontal movement. For one, we're already given that the krill starts from rest. Thus, we know that formula_89. Also, we had just solved for the krill's horizontal acceleration in Part a) (ii). Therefore, we also know that because of the Cartesian grid convention, formula_90. Finally, we already know the time elapsed for the krill, formula_71. As seen in the solution for Question 1: Part b), we can use the kinematic equations table and eliminate Equations 1, 2, and 4 because of the final velocity being unknown. This is with the understanding that by introducing more unknown values, we can't solve directly for any individual unknown equation, unless we were simultaneously using multiple equations. After this, we can simplify for the change in displacement in a very similar fashion to Question 1: Part b) by using Equation 3. Note that with the current shark and krill problem, we're instead evaluating for the horizontal displacement, not the change in time. formula_92 formula_93 formula_94 formula_95 formula_96 Thus, the krill traveled 2.44 m horizontally on its way from the shark's jaws to the shark's stomach, answering Part b). The next section of the question, Part (c) is conceptual in nature and requires a justification. Justifying a conceptual answer doesn't mean that we need to quantitatively calculate the situation. Instead, we need to make an argument of which acceleration value should we use when solving for the krill's overall speed at formula_97. When considering the overall speed of the krill, this is just another way of saying the magnitude of the overall velocity. We're aware that the krill is moving for the full time interval between 0.000 and 0.500 seconds 5.00 degrees below the -X direction. This is because the krill is starting from rest, motionless, and then only accelerates in this direction. As such, there's both a horizontal and a vertical component for the velocity vector. Another way of looking at this problem is by recalling Choice B from Part a) (i). In this, the krill's path is marked by a straight line of dots. Because of the krill's path, we understand that if we were to calculate for the krill's final velocity, while already knowing the initial velocity and some acceleration value, we can use Equation 1 from the kinematic equations table in which formula_98 where x is some direction. If we were to use this equation to solve for the overall final velocity, then we would need to substitute in the overall acceleration, formula_99, into this equation. This is because formula_99shares the same direction angle as the velocity. Therefore, one possible response to Part c) would be: PRIMARY SAMPLE RESPONSE Choice: The krill's acceleration vector, as given before Part a. Reason: Suppose we choose to use the equation formula_101to solve for the overall final velocity. This would be solved in the direction angle of 5 degrees below the -X direction. Because the vectors of velocity during the time interval and the overall acceleration share the same direction, we could use the overall acceleration to solve for the final overall velocity. To note, selecting "The horizontal component of the acceleration vector, as calculated in Part a (ii)" or "Neither" as your choice for Part c) does not necessarily mean that you're incorrect. There are alternate solutions to this problem, two of which are provided below. The previous derivation and the "primary sample response" above, however, may showcase a simpler approach. ALTERNATE SAMPLE RESPONSE #1 Choice: The horizontal component of the acceleration vector, as calculated in Part a (ii). Reason: We can use the equationformula_98to solve for the overall final velocity in the horizontal (X) direction. Then, because we know the direction angle of the overall velocity vector at the final time, we can use trigonometry to solve for the krill's final overall speed. ALTERNATE SAMPLE RESPONSE #2 Choice: Neither. Reason: We can calculate first for the acceleration of the krill in the vertical (Y) direction with trigonometry. This is because we are given the direction angle at which the krill's acceleration is occurring. After this, we can use the equationformula_103to solve for the overall final velocity in the Y direction. Then, because we know the direction angle of the overall velocity vector at the final time, we can use trigonometry to solve for the krill's final overall speed. The final section, Part d), was particularly challenging. As such, the derivations have been attempted to be explained as thoroughly as possible. Two methods are presented below. Note that there may be versions of these approaches that arrive at the same solution. ANALYTICAL METHOD for Part d): First, we need to interpret what has changed in the modified situation with the second krill. From the prompt for Part d), we understand that although the path and acceleration of the second krill is the same, it will have a different initial velocity of 2.00 m/s. This new velocity is directed 5.00 degrees below the -X direction. Because of this, we should expect for the second krill to travel faster to the shark's stomach. The second krill, with a greater initial velocity and the same acceleration, will be traveling faster than the first krill at all times between its entrance into the shark's jaws until reaching the shark's stomach. Therefore, our answer for Part d) should be less than 0.500 seconds, the given time elapsed for the first krill. Since we have a known initial velocity,formula_104, a net acceleration formula_72, and a horizontal displacement (found in Part b) formula_106; we can start by trying to see if there's any equations can we can use through the kinematic equations table. With only these variables, we would be forced to introduce another unknown in order to solve an equation for the time elapsed. To avoid this, we can consider if we can solve for another new variable beforehand. This would make it possible for us to then solve directly for the time elapsed. One possibility would be to consider Equation 1: formula_107. Although we don't yet know the final velocity, we could attempt solve for it with Equation 4: formula_108. Once again, we run into the situation where we're introducing a new, unknown variable. This time, however, we have already solved for the horizontal displacement in part b). Also, because we are told that the paths of the two krills are the same, we can use trigonometry to solve for the net displacement of the second krill. Since the velocity and acceleration vectors are pointed 5.00 degrees under the -X direction for the entirety of the elapsed time for both krills, we can write the equation formula_109. Then, we can solve directly for formula_110such that: formula_109 formula_112 formula_113 formula_114 Then, we can again look back at Equation 4 where formula_108and solve directly for the final velocity of the second krill. formula_108 formula_117 formula_118 formula_119 Finally, we now have the variables necessary to solve directly for the time elapsed with Equation 1: formula_107as shown below. formula_121 formula_122 formula_123 formula_124 formula_125 Thus, 0.408 seconds elapsed for the second krill between when it entered the shark's jaws to it arriving at the shark's stomach. This answers Part d) through an analytical method. GRAPHICAL METHOD for Part d): Note: This solution method involves using this custom Desmos graph. Have the graph open alongside the solution presented below. Another way to solve Part d) involves a partial graphical approach. However, this method also begins by analytically considering the variables that we currently have. These include a net initial velocity,formula_104, a net acceleration formula_72, and a horizontal displacement (found in Part b) formula_106. After collecting these terms, we refer to the kinematic equations table and consider Equation 3: formula_129. Although we are introducing an additional unknown variable (net displacement) formula_130into our problem by considering Equation 3, we can directly solve for this value beforehand with trigonometry. This is because we already know the horizontal displacement and the krill's angle of displacement. Using the same approach as the Analytical Solution, we can find that the net displacement of the second krill with the horizontal displacement. As such, with holding onto significant figures, formula_131. For the derivation of this, please refer to the Analytical solution. Now, we can directly solve for the time elapsed by using Equation 3 as shown below: formula_129 formula_133[Definition of displacement.] formula_134[Setting the quadratic equation equal to 0.] By allowing for the quadratic equation above to equal 0, we can figure out for which values of formula_50the equation is true. To explain, formula_50 is the only remaining unknown variable in the equation above. Thus, we can evaluate if a certain value of formula_50is the answer by checking if it satisfies the equation. We could do their either by using the quadratic formula or by using a graphing calculator to look for the equation's intersection points with the X-Axis (where the equation would equal 0). However, because a graphing calculator was provided for this question, the rest of the solution will focus on graphically solving the problem. To solve this with the graphing calculator, we need to first decide which function should be used to model the equation that we have derived. On the custom graph, we have four possible graph selections. However, only Function Choice 1: formula_138contains the same variables from the derived equation. To clarify, with the other functions, they: Following this, we scroll down and enter in our values for formula_139,formula_140, and formula_99. Note that formula_142can be skipped as it is not in our selected function. Using the information gathered so far, the entered values should be: formula_139:formula_144 formula_140: formula_146 formula_147: formula_148 Note: To prevent confusion, all of these values must share being either all positive or all negative. This is because the values above all are vectors in the same direction. Thus, it's our choice depending on whether we want to follow the Cartesian grid convention. For this solution, we all for the values to all be positive. However, both ways result in the same answer. With this, we are able to create a parabolic curve that intersects the X-axis at formula_149 and formula_150. As explained on the Desmos graph, only the positive solution is correct. We can confirm this logically as the krill arrived at the shark's stomach after passing through the shark's jaws. Thus, a positive change in time must of occurred. Therefore, with three significant figures, the second krill traveled from the shark's jaws to its stomach in formula_151 , also answering Part d).*

Pokémon Scarlet and Violet/Victory Road. "Victory Road" follows the classic "Pokémon" formula of beating eight Gyms and then taking on the League. Gyms. There are eight Gym Leaders in the region. They can be challenged in any order, but the recommended order is as follows: Katy, Brassius, Iono, Kofu, Larry, Ryme, Tulip, Grusha. Cortondo Gym. Gym Leader Katy. Katy specializes in Bug-type Pokémon. Artazon. Gym Leader Brassius. Brassius specializes in Grass-type Pokémon. Levincia Gym. Gym Leader Iono. Iono specializes in Electric-type Pokémon. Cascarrafa Gym. Gym Leader Kofu. Kofu specializes in Water-type Pokémon. Medali Gym. Gym Leader Larry. As a Gym Leader, Larry specialized in Normal-type Pokémon. Montenevera Gym. Gym Leader Ryme. Ryme specializes in Ghost-type Pokémon. Her battle is a double battle. Alfornada Gym. Gym Leader Tulip. Tulip specializes in Psychic-type Pokémon. Glaseado Gym. Gym Leader Grusha. Grusha specializes in Ice-type Pokémon. League. Like other Pokémon Leagues, the Paldea League features an Elite Four. Unlike others, rather than only having one Champion, anyone who completes the League Assessment is forever considered a Champion, with there being one Top Champion. The Elite Four are fought in a fixed order. Rika. Rika specializes in Ground-type Pokémon. Poppy. Poppy specializes in Steel-type Pokémon. Larry. As an Elite Four member, Larry specializes in Flying-type Pokémon. Hassel. Hassel specializes in Dragon-type Pokémon. Nemona. Nemona is fought several times throughout the "Victory Road" storyline.

SearXNG. SearXNG is a free and open-source metasearch engine software that anyone can host themselves. It was forked from Searx in 2021. This is a guide on how to install it for yourself and host your own instance.

SearXNG/What you'll need. Creating a SearXNG instance is not too difficult. You will need the following items: Once you have all of these, proceed to Installation.

SearXNG/Installation. There are multiple ways to install SearXNG. Docker. This is the recommended method, as it is the easiest. First, create a directory for the instance and enter it: mkdir my-instance cd my-instance/ Next, specify a port for the container to run on. 8080 is used for this example: export PORT=8080 Then, pull the container from the repository: docker pull searxng/searxng Now, run the container. Replace "my-instance" with whatever you want to call your instance: docker run --rm \ -d -p ${PORT}:8080 \ -v "${PWD}/searxng:/etc/searxng" \ -e "BASE_URL=http://localhost:$PORT/" \ -e "INSTANCE_NAME=my-instance" \ searxng/searxng If your web browser is running on the same computer that you installed SearXNG on, go to "localhost:8080" in your web browser. If not, enter the IP address of the computer SearXNG was installed on, and specify port 8080 (or whichever port you chose). If you installed it correctly, you should get the SearXNG homepage. Installation script. The installation script is also fairly easy to do, though depending on your server it may have more issues than Docker. First, clone the repository and enter its directory: git clone https://github.com/searxng/searxng.git searxng cd searxng Then, run the script as root. Replace "sudo" with whichever privilege elevation program you use: sudo -H ./utils/searxng.sh install all Manual step-by-step installation. This method is intended for advanced users and is not necessary in most cases. First, install the necessary packages: On Debian-based systems: sudo apt install python3-dev python3-babel python3-venv uwsgi uwsgi-plugin-python3 git build-essential libxslt-dev zlib1g-dev libffi-dev libssl-dev On Arch-based systems: sudo pacman -S python python-pip python-lxml python-babel uwsgi uwsgi-plugin-python git base-devel libxml2 On Fedora/RHEL systems: sudo dnf install python python-pip python-lxml python-babel python3-devel uwsgi uwsgi-plugin-python3 git @development-tools libxml2 openssl Next, create the searxng user: sudo useradd --shell /bin/bash --system --home-dir "/usr/local/searxng" --comment 'Privacy-respecting metasearch engine' searxng Create the searxng user's home directory, and give the directory the correct permissions: sudo mkdir /usr/local/searxng sudo chown -R searxng:searxng /usr/local/searxng Next, change to the searxng user: su - searxng Alternatively, use the interactive mode of sudo: sudo -u searxng -i As the searxng user, create virtualenv: python3 -m venv "/usr/local/searxng/searx-pyenv" echo ". /usr/local/searxng/searx-pyenv/bin/activate" » "/usr/local/searxng/.profile" Exit the session and start a new one. Then, use pip to install the necessary dependencies: command -v python && python --version pip install -U pip pip install -U setuptools pip install -U wheel pip install -U pyyaml cd /usr/local/searxng/searxng-src pip install -e . Next, create a file called "settings.yml" and copy this configuration into it. Modify it to your preferences. After that, copy the configuration file to the correct directory: sudo mkdir -p /etc/searxng sudo cp /usr/local/searxng/searxng-src/utils/templates/etc/searxng/settings.yml /etc/searxng/settings.yml Enable debugging and start the web app: sudo sed -i -e "s/debug : False/debug : True/g" /etc/searxng/settings.yml sudo -u searxng -i cd /usr/local/searxng/searxng-src export SEARXNG_SETTINGS_PATH="/etc/searxng/settings.yml" python searx/webapp.py If you ever want to disable debugging, use the following command: sudo sed -i -e "s/debug : True/debug : False/g" /etc/searxng/settings.yml If you did this correctly, SearXNG should be running on port 8888.

SearXNG/Hosting a public instance. There are several public instances of SearXNG, and you can host your own. DNS configuration. Create DNS A records (and AAAA records if your server uses IPv6) with your domain name registrar pointing to the server that you are hosting your instance on. Setting up the HTTP server. There are several HTTP servers available. The most popular ones all work with SearXNG. Caddy. This is the recommended server, as it is the easiest to configure and automatically obtains a TLS certificate from Let's Encrypt. Docker container. If you installed the SearXNG docker container, add the following to your Caddyfile, and reload Caddy. Replace "example.com" with your fully-qualified domain name: example.com { reverse_proxy localhost:8080 Obtaining a TLS certificate. A TLS certificate is necessary in order to use HTTPS, which is preferred by most modern web browsers. For this tutorial, we will use Let's Encrypt for a certificate, as Let's Encrypt provides certificates for free. Caddy. Caddy automatically obtains a TLS certificate, so you do not need to do anything if you used Caddy. Other web servers. Visit the Certbot website and follow the instructions there. Adding your instance to the list. An official list of public instances is maintained at searx.space. You can add your own by going to the Github repository and creating a new issue to add your instance. Ensure you meet the requirements.

Physics Explained Through a Video Game/Newton's Second Law. Topic 2.3 - Newton's Second Law. In order to apply the concepts of kinematics to a greater amount of physical situations, we need to consider how forces can be used to explain the movement of objects. Example 1: Rocket Propulsion. As illustrated with a rocket*, objects accelerate in the direction of the net force applied onto them. In this case, the only non-negligible force acting on the rocket is its propulsion, which is pushing the rocket in the right-upwards direction. Thus, the rocket's propulsion force is its net force, formula_1. With this, we can see that the rocket is accelerating from rest in the right-upwards direction. In other words, the rocket is speeding up in the right-upwards direction because of the propulsion force. The behavior of the rocket's acceleration can be generalized by noting the definition of Newton's Second Law below: With this definition, we can create the equation formula_2. This allows for us to link together the concept of forces being applied onto objects and the object's acceleration. Example 2: Skateboard Riding. Consider a modified version of "Skate Duel "by "TiM MELL0". In this, "monkey butler" (left) and "Hello4409" (right) both try to move their skateboards by applying a force that alternates between the left and right directions on a skateboard. To note, both players have a horizontally applied force of the same magnitude. As shown by the video above, "Hello4409's" skateboard moves back and forth much more slowly than "monkey butler's" skateboard. This is because the skateboards have different masses. More specifically, "Hello4409's" skateboard has more mass. Thus, when "Hello4409" tries to apply a force onto his skateboard, it accelerates more slowly. This can be expressed by the diagram on the right, illustrating that given some known net force, formula_3, if the object has more mass, then the net acceleration of that object, formula_4, will be less. In the case of the skateboard example, "monkey butler" and "Hello4409's" skateboards are undergoing the same forces when the skateboard is at the bottom of the ramp. More precisely, both player's skateboards have gravitational and contact forces that cancel out. As such, the net force for both players is the applied force. Although the applied forces are equal in magnitude, we can consider Newton's Second law where formula_5. Because "monkey butler's" skateboard (colored in blue) has less mass, it will have a greater magnitude of net acceleration. Also, because "monkey butler" is able to give his skateboard a greater acceleration, his blue skateboard is able to travel further up the ramp. This concept will be further explored later on in Unit 3. Example 3: Car Chase. For a further look into the applications of Newton's Second Law, consider a modified version of "Car Chase "by "O_o O_o O_o" below. In this map, a police car is pursuing a red vehicle where both vehicles gradually slow down and come to a stop. Task: <br>Assume that <br> • Air resistance is negligible. <br> • The police car's gravitational force and contact force are found to be negligibly different in magnitude. <br> • The police car's applied force from engine thrust and the car's friction force from the rough ground are acting only horizontally. Considering the video overall, which of the forces mentioned below has a greater magnitude while the police car is slowing down? Then, create a free body diagram (FBD) of a police car. Answer: The police car's friction force (from the rough ground). To explain, we already know that the police car is slowing down. Because the police car is initially moving towards the right and then enters rest relative to the ground, the police car has a net acceleration, formula_6 in the leftward direction. Using the Cartesian Coordinate Convention, the net acceleration is pointed in the negative horizontal direction. Because of Newton's Second Law, the direction of the net force is the same as the net acceleration. Thus, the police car has a net force that is pointing leftwards. With the vertical forces, including the gravitational and contact forces are equal in magnitude, they fully cancel each other out. Also, we're aware that the friction force is acting towards the left and the applied force is acting the right. With this information, we can begin creating a free body diagram (FBD) of the situation, as shown on the right. Since only the friction and applied forces are in the horizontal direction, for the net force to be in the negative (leftward) direction, the frictional force must have a greater magnitude. This is reflected by the FBD where the vector of the frictional force is drawn longer than the applied force. What if the frictional force magnitude wasn't greater than the applied force?If we were to have assumed that one of the other choices were correct, then the physical situation involving the police car would have been different. Example 4: Propeller Planes. To expand the discussion of Newton's Second Law to quantitative situations, let's consider the map "Maneuver Planes" by "TiM MELL0". In this, players have the ability to influence a small airplane's motion by applying a force onto its surface. Suppose that the player Sam (red color) is flying an airplane, initially applying a downwards force while moving horizontally towards the right. When "Sam" enters the dotted line track, he stops applying a force onto the airplane. Soon afterwards, the direction of the plane's motion approaches a steady, gradual descent. Assume that: Task: Calculate the the acceleration of the airplane in the horizontal direction while it is traveling on the dotted line track. Answer: formula_17 [2 sigfigs] <br> Note: The horizontal acceleration is being considered, not the overall acceleration. In order to solve this problem, we need to consider Newton's Second Law. However, we haven't been explicitly given formula_18. Instead, we have to use the listed information above to figure out formula_18. We can find that formula_20 and formula_21 directly oppose and cancel each other out. As such, we can ignore these forces while finding formula_18. (They're also acting only in the vertical direction, making them not have an impact on the horizontal acceleration. However, we'll get to that shortly.) As such, only formula_23 remains uncancelled. Therefore, formula_24. With this information, we can derive the overall acceleration of the plane, as shown below: formula_25 [Definition of Newton's Second Law] formula_26 [Solving for overall acceleration vector.] formula_27 [Substitution of known variables; formula_28]. formula_29 [Division; formula_30; 2 sigfigs]. However, we're not done just yet. This is because the task specifically asked for us to consider the horizontal acceleration of the plane, not the overall acceleration. As such, we need to use trigonometry to get the final answer. We can graph that the overall acceleration, formula_31, has a magnitude of formula_32 and has a direction angle that is formula_11 below the +X Axis. Concerning the direction angle of formula_31, we know this because of formula_35's specified direction angle. Also, objects accelerate in the direction of its net force. Using the acceleration vector formula_31 and its components, we can construct a right angle triangle which a known angle formula_37 which equals formula_38. Because formula_39 is adjacent side to formula_37 and formula_31 is the hypotenuse of the triangle, we can solve directly for formula_39 as shown below: formula_43[Trigonometric definition of the formula_44 function.] formula_45 [Solving for the adjacent side length.] formula_46 [Substitution of variables.] formula_47 [Computation (using degree mode).] formula_48 [2 sigfigs] The plane is horizontally accelerating at a rate of formula_49. To note, formula_39 with 2 significant figures is equal to formula_31. This is because formula_37 is a small angle where formula_53. Question 1: Frog Shrine. Consider the map "Frog shrine" by "O_dot". In this map, players are engaging in direct combat with one another. One of the players accidentally touches a trap sensor (magenta colored), resulting in a decorative frog to suddenly launch its tongue. Shortly afterwards, Spy Coder X (blue colored) is snatched and pulled away to their demise. Assume that: Use the information provided above to answer the following: <br>Part (a): Determine the direction in which the net force on the frog's tongue, formula_35, is acting towards. Part (b):Using Newton's Second Law, calculate the acceleration of frog's tongue. Part (c):Draw a free body diagram of all of the forces acting on "Spy Coder X" after landing on the frog's tongue (as imaged on the right). Note that some of the forces mentioned in the image may not be needed. Part (d):(i) Let the moment in which the frog's tongue be initially released be formula_61. At what time formula_62 does the frog's tongue return to its mouth? (ii) At what time formula_62 is the frog's tongue fully extended? (iii) Calculate how much time elapses between "Spy Coder X" sticking to the frog's tongue and them entering the frog's mouth. "Consider discussing your solutions on this article's , where you find help from others."

Physics Explained Through a Video Game/Newton's First and Third Laws. Topic 2.4 - Newton's First and Third Laws. This section will explore the remainder of Newton's Laws, providing a broader understanding of the general properties of forces and their interactions with objects. Example 1: A Swimming Koi Fish. Consider the map "fish pond" by "O_dot". In this, a player is able to maneuver a koi fish at the bottom of the map. As such, the fish is able to spin and move gently across the body of water. We can see that when the fish is moving in a straight path, it appears to move at a constant velocity. To examine this further, consider the free body diagram (below). It includes the forces that are acting on the fish while it is moving in a straight line. Using the free body diagram, we can find that the fish's formula_1 is formula_2. This can be done through vector addition, as visualized on the right. By using the definition of Newton's Second Law from Topic 2.3, we can solve directly for the net acceleration of the fish, formula_3, as shown below. formula_4 [Definition of Newton's Second Law] formula_5 [Algebra.] formula_6 [Substitution. We can assume that the fish does have mass in it.] formula_7 [Algebra, formula_8 ] We can see that because formula_1is formula_2, the fish has a formula_3 of formula_12. To note, the fish would have zero net acceleration regardless of how much mass is contained within it. Therefore, we can generalize this concept further below. Because formula_5, if there is a zero formula_1 and an object has any positive amount of mass, then formula_15"." Essentially, this is the definition of Newton's First Law, which is really just a special case of Newton's Second Law. As a formal definition, it states that: <br>Thus, from Newton's First Law, we can find that if an object has mass then if it isn't accelerating, then it must have a zero net force. To note, this property can help us significantly with solving for the magnitude of a specific force provided that we know the other forces, such as seen in Example 3. Example 2: Sliding on Sandstone. Consider the map Sandstone Cave by Sirestyx in which players slip around on a smooth inclined surface. To note, in Topic 2.2, we discussed that for an inclined plane system with an object resting on it: In the case of "Sandstone Cave", there isn't an obvious flat plane that the players are slipping on. Instead, the surface appears curved. However, we can still use most of the techniques from Topic 2.2 to investigate the interactions between the players and the cave wall. The only difference is that we need to note that the normal force that would be acting on the player will change as the slope of the surface changes.More specifically, we previously had found that we could describe the normal force of an object on an inclined plane by the formula formula_18 where formula_19 is the angle of elevation, in degrees. Because formula_20 decreases from formula_21 to formula_22 as formula_19 increases from formula_24 to formula_25 (graphed on the right) where: formula_26 formula_27 formula_28 As such, for where the surface angle is somewhere between being horizontal (formula_24) and vertical (formula_25), the steeper the slope is, the lesser the normal force will be. We can see this in action for when a player is rolling along the surface of the sandstone cave. For instance, in the video above, after striking "monkey butler", the player "ez01" slides up along some of the right-hand wall before falling back down. Several instances of ez01's position and their associated free body diagram can be considered as diagrammed on the left below. From the left diagram, we can find that formula_31appears to share the same length as the Y-component of formula_32. We can prove this using trigonometry on the right diagram, as shown above. With knowing that formula_33 in magnitude, one way that we can describe the magnitude of the normal force is that it can be defined as: In other words, the normal force from the sandstone wall "pushes" back on the player just as much as the player pushes into it. To note, the horizontal component of the player's gravitational force, formula_36 is always perpendicular to the sandstone wall's slope (see left diagram). Thus, it does not "push" the player into the wall. Rather, only the formula_34 does. We can generalize this behavior through Newton's Third Law, whose definition is provided below: From this, we can verify that there must be a formula_31 of an equal magnitude but opposite direction of formula_34. To rationalize this, suppose Newton's Third Law wasn't the case. If we were to calculate the net force on the player in the Y-axis (which is perpendicular to the wall's slope), formula_40. In this situation, the player, "ex01" would have been either accelerating downwards through the sandstone or had been floating into the air. However, if we consider the video clip, "ex01" slides gently with the sandstone after colliding with "monkey butler", never losing contact with the surface of the wall. Thus, Newton's Third Law holds. Example 3: Sinking Container Ship. Continuing to build up a foundation of Newton's Third Law, suppose a container ship strikes a small rock island and subsequently begins to sink in the map "Sinking Container Ship" by "O_o O_o O_o." Before then, the shipping containers are stacked vertically on top of each other. For this exercise, assume that we're considering the period of time before the ship hits the rock and that: Exercise 1: Consider the reference image and the other information provided to draw a free body diagram for the yellow shipping container. Then, calculate the normal force acting on it. Answer 1: To approach this problem, we first need to consider that there needs to be a gravitational force acting on the yellow shipping container that is acting directly downwards. Because the gravitational force is acting directly downwards, all of this force is pushing the yellow container into the orange container beneath it. Therefore, by Newton's Third Law, there must exist a force of an equal magnitude but opposite direction of the gravitational force. Using the definitions of the types of forces from Topic 2.2, this would be a normal force. To calculate the magnitude of the normal force, we can compute formula_43. Because formula_44 such that formula_45, formula_46. Then, we can use algebra to find the normal force on the yellow container, as shown below: <br> formula_47 <br> formula_48 [Substitution of variables.]<br> formula_49 [Algebra; Definition of a Newton (N)] <br> formula_50 [2 sigfigs; Using Scientific Notation] The yellow shipping container has a normal force acting on it of formula_51 that is pointing in the upward direction. Exercise 2: Draw a free body diagram for the orange shipping container that is underneath the yellow shipping container from Example 1. Calculate the normal force acting on this orange shipping container. Answer 2: For this problem, we can consider that in Exercise 1 that there is a gravitational force that is being exerted downwards on the yellow container (formula_52). Because it is exerting downwards (into the orange container), this means that the orange container is being pushed downwards by it. In addition, the orange container is being pushed downwards by its own gravitational force ((formula_53). Because of this, the orange container is pushing into the ship with both the gravitational force of itself and of the yellow container. Since the ship's deck is flat and directly under the containers, the ship is being pushed by the entirety of the downwards forces acting on the orange container. Because of Newton's Third Law, there must be an equal magnitude and oppositely directed force that opposes the previously mentioned downwards forces on the orange container. This will be our normal force for the orange container, formula_54. Since formula_54 is equal in magnitude to formula_56 and formula_56 combined, in terms of magnitude: <br> formula_58<br> From here, we can solve directly for the magnitude of the normal force acting on the orange container: <br> formula_59 [Substituting the orange box's gravitational force from Exercise 1, using as many sigfigs as possible.] <br> formula_60 [Because the orange box has the same mass, it will have the same magnitude of gravitational force.] <br> formula_61 [Addition.] <br> formula_62 [2 sigfigs; scientific notation.] <br> The orange container has a normal force with a magnitude of formula_63 that is pointing in the upward direction. Example 4: A Swimming Koi Fish Continued (Challenging). Swapping back to the map "fish pond" by "O_dot", we will gain a further understanding of Newton's Third Law. Suppose the fish continues to swim in the pond. Then, the fish begins to interact with a lily pad in front of it. For this example, we'll be looking into the force interactions between the fish and a lily pad. Assume that (for a extended period of time): Why did we not account for formula_67 's equality? To note, the drag force of water on an object is dependent on a wide variety of factors, including the velocity of the object, the surface area of the object, the shape of the object, and other factors. For simplicity and convenience, we're assuming that formula_69 and formula_70 in Example 2 to not be necessarily tied to formula_70 in Example 1. Put as simply as possible, if formula_72, then the lily pad and fish would briefly accelerate together after collision. This is because the water either wouldn't be tugging on them as much as beforehand (or would be tugging on them more). However, the lily pad and fish thus would have a changing speed. Therefore, their respective drag forces will change over time. This would result in the lily pad and fish system to approach a constant velocity, regardless of the specific drag forces. Exercise: Consider the two free body diagrams of the fish and the lily pad before their interaction and the other information provided to draw two free body diagrams for the fish and the lily pad while they are interacting with each other. Answer: By Newton's First Law, because fish and the lily pad both have a constant velocity, the system has a formula_1 of zero. However, it is clear that the fish is pushing the lily pad forward by the formula_74 force. This is because this is the only force on the fish (that we know currently) that is directed towards the lily pad. Therefore, there must also exist some formula_75 force of an equal magnitude and opposite direction by Newton's Third Law. Another factor to consider is that once the fish begins to interact with the lily pad, the lily pad as previously mentioned begins to move. This would result in a drag force on the lily pad, which can be called formula_76. Since a drag force opposes the direction of movement, the drag force vector would start at the lily pad's center of mass and point backwards. In other words, formula_76 is pushing on the fish. Because of this, we can again apply Newton's Third Law, including for there to be an equal magnitude, opposite direction force of formula_76 from the fish to the lily pad. We can call this vector formula_79. With this, we can draw free body diagrams for the fish and the lily pad during their interaction, as diagrammed below. Question 1: A Humble Boat. Consider a modified version of "Don't Drown" by "Hexigon" where a small sailboat is on a deck. The boat is being pushed into the water by a player applying a leftward force on it. Assume that: Part (a):(i) Define which forces are acting on the player in the vertical direction when he and the sailboat are moving at a constant velocity. <br> (ii) Also, define which forces are acting on the player in the horizontal direction at this time. <br> (iii) Construct a free body diagram of the player at this time. <br> Part (b):(i) Using the options on the right and the information provided above, decide where the sailboat's center of mass is located.<br> (ii) Construct a free body diagram of the sailboat when the player and sailboat are moving at a constant velocity.<br> Part (c): Suppose that soon after the player begins pushing on the boat, the player-boat system has a maximum horizontal acceleration magnitude of formula_82. This acceleration occurs in the leftward direction.<br> (i) Calculate the magnitude of the force due to kinetic friction on the player-boat system when it reaches its maximum acceleration. <br> (ii) Calculate the magnitude of the force due to kinetic friction on the player-boat system when it is traveling at a constant leftward velocity. <br> "Consider discussing your solutions on this article's . On there, you can find help from others."

Chinese (Mandarin)/Lesson 12. = Lesson 12: Mandarin is so interesting/ 第十二课：汉语真有趣！ = 繁体字/Traditional Characters. （仲文在走廊遇到了他的好友思源） 思源：嗨！仲文！你吃早飯了嗎？ 仲文：早上好，思源！我吃過早飯了。 思源：根據天氣預報，這一天都會有雨。我沒辦法到操場踢毽子了，真無聊啊。 仲文：確實很無聊，不如咱們來玩斷句遊戲吧！ 思源：好啊，你先來！ 仲文：這幾天天天天氣不好，請根據語意斷開它吧！ 思源：我看看。嗯，答案應該是“這幾天/天天/天氣不好“對吧！ 【1】 仲文：答案正確，換你來考我吧。 思源：這句怎麼樣？小明明明明白白的卻裝作不明白。 仲文：答案是“小明/明明/明明白白/的/卻裝作不明白“對吧？ 【2】 思源：答案正確，你真厲害！ 仲文：過獎了，你也很厲害！ 思源：漢語真有趣啊！我們兩個的句中都有反復出現的同一個字，可它們卻有不同的意思，還成功構成了一個完整的句子。 仲文：確實。漢語真有趣！ （上課鈴響起） 思源：上課了，我先走了。回見！ 仲文：回見！ 简体字/Simplified Characters. （仲文在走廊遇到了他的好友思源） 思源：嗨！仲文！你吃早饭了吗？ 仲文：早上好，思源！我吃过早饭了。 思源：根据天气预报，这一天都会有雨。我没办法到操场踢毽子了，真无聊啊。 仲文：确实很无聊，不如咱们来玩断句游戏吧！ 思源：好啊，你先来！ 仲文：这几天天天天气不好，请语意断开它吧！ 思源：我看看。嗯，答案应该是“这几天/天天/天气不好”对吧！【1】 仲文：答案正确，换你来考我吧。 思源：这句怎么样？小明明明明明白白的却装作不明白。 仲文：答案是“小明/明明/明明白白/的/却装作不明白”对吧？【2】 思源：答案正确，你真厉害！ 仲文：过奖了，你也很厉害！ 思源：汉语真有趣啊！我们两个的句中都有反复出现的同一个字，可它们却有不同的意思，还成功构成了一个完整的句子。 仲文：确实。汉语真有趣！ （上课铃响起） 思源：上课了，我先走了。回见！ 仲文：回见！ 汉语拼音 / Chinese Pinyin. (Zhòng wén zài zǒuláng yù dàole tā de hǎoyǒu sīyuán) sīyuán: Hāi! Zhòng wén! Nǐ chī zǎofànle ma? Zhòng wén: Zǎoshang hǎo, sīyuán! Wǒ chīguò zǎofànle. Sīyuán: Gēnjù tiānqì yùbào, zhè yītiān dūhuì yǒu yǔ. Wǒ méi bànfǎ dào cāochǎng tī jiànzile, zhēn wúliáo a. Zhòng wén: Quèshí hěn wúliáo, bùrú zánmen lái wán duànjù yóuxì ba! Sīyuán: Hǎo ya, nǐ xiān lái! Zhòng wén: Zhè jǐ tiāntiāntiān tiānqì bù hǎo, qǐng yǔyì duàn kāi tā ba! Sīyuán: Wǒ kàn kàn.ng , dá'àn yīnggāi shì “zhè jǐ tiān/tiān tiān/tiānqì bù hǎo” duì ba![1] Zhòng wén: Dá'àn zhèngquè, huàn nǐ lái kǎo wǒ ba. Sīyuán: Zhè jù zěnme yàng? Xiǎomíngmíngmíngmíngmíngbáibái de què zhuāng zuò bù míngbái. Zhòng wén: Dá'àn shì “xiǎo míng/míng míng/míng míngbái bái/de/què zhuāng zuò bù míng bái” duì ba?[2] Sīyuán: Dá'àn zhèngquè, nǐ zhēn lìhài! Zhòng wén: Guòjiǎngle, nǐ yě hěn lìhài! Sīyuán: Hànyǔ zhēn yǒuqù a! Wǒmen liǎng gè de jù zhōng dōu yǒu fǎnfù chūxiàn de tóng yīgè zì, kě tāmen què yǒu bùtóng de yìsi, hái chénggōng gòuchéngle yīgè wánzhěng de jùzi. Zhòng wén: Quèshí. Hànyǔ zhēn yǒuqù! sīyuán: Shàngkèle, wǒ xiān zǒule. Zàijiàn! Zhòng wén: Zàijiàn!

Cherokee/Verbs. Ahh ... the mighty Cherokee verb! For now refer to the "Verbs (Raw)" tab in this sheet: https://docs.google.com/spreadsheets/d/16I5CmBxTY2gb1n9HzMMqW1RuwvyBNrB7rJ5R6tgZUxw/edit?usp=sharing On Memorization. It "is" useful to memorize bare roots of verbs. There will be a set of flashcards for the purpose of helping you memorize bare roots. Later the flashcards will include more important information that should be memorized for each verb. Start small, though.

Hindi/Honorifics. Hindi speakers use three different pronouns for "you" ( tu, tum, and aap), and which one people use depend on the level of intimacy or context. "Tu" is the most intimate, only used to refer to another person whom one is most familiar with, like a husband, as well as to children. However, when used in inappropriate situations, "tu" can sound offensive and insulting even. "Tum" is more respectful. "Aap" implies the highest level of respect and Hindi speakers use "aap" when talking to their parents, elders, authorities, or superiors. Each of these "you"-pronouns has their own be-verbs. "Hai" comes with "tu," "ho" with "tum," and "hain" with "aap." Note that in English, in the context of the second person, all of these be-verbs are translatable to "are" as in "you are." But "hai" in other situations can be translated to "is" and "hain" to "are."

Hindi/Grammar. Also known as Hindi Vyākaraṇa, the Hindi grammar is the set of rules that define the proper usage of the language. Hindi Grammar is mainly divided into three parts-

Physics Explained Through a Video Game/Forces and Free Body Diagrams. Topic 2.2 - Forces and Free Body Diagrams. Unlike most other sections of this course, this topic is reserved for reading and as a reference. As such, there are fewer practice materials for this topic. Consider the later topics in Unit 2 to broaden and apply the skills discussed below. Dynamics and Forces. So far during much of this course, we have focused on kinematics, a way of describing the motion of objects, such as through their position, velocity, and acceleration. However, we also need to discuss what causes for objects to move in the first place. In this unit, we will be largely focusing on dynamics. This content area focuses on linking together the existence of forces and how they impact the motion of objects. A force can be described as a push or a pull that is acting on an object. It is a vector quantity, meaning that it has both a magnitude and direction. We can use the symbol formula_1 to denote the presence of a force. Also, forces have their units in Newtons (formula_2). As will be explored closely in the next topic, a Newton is equivalent to formula_3. In other words, a force's magnitude is related to the mass of the object (described in formula_4 multiplied by the acceleration of that object (described in formula_5). This can give us an intuition that forces—through their ability to push or pull—influence objects in accelerating towards a particular direction. Characterizing Forces. There are a wide variety of forces that exist, ranging from frictional forces, gravitational forces, tension forces, and many more. To characterize these forces, we distinguish between forces that are external and internal. An external force is a force that is originating from outside of an object or system and is applying a push or pull on it. For instance, assume that we're considering a rhinoceros as a system of interest. The gravitational force exerted by the Earth that pulls the rhinoceros down towards the ground is an example of an external force on the creature. In contrast, there also exists numerous forces "inside" of the rhinoceros as well, such as skeletal and muscular forces that enable it to be able to eat, walk, and breathe. These can be characterized as internal forces. Often, we take a particular interest towards the external forces that are acting on a system. This is because if we combine add all of the external forces vectors together**, we can derive the net external force, denoted by formula_6. This vector is important as the direction angle of formula_6 determines the direction in which an object is accelerating. With this property and others that will be discussed later in this unit, we can garner information concerning the kinematic behavior of an object directly through considering the forces that are acting on it. "** This requires the use of vector addition. Although this concept will later discussed in Topic 2.3 - Newton's Second Law, consider reading the introduction of vector addition by FHSST Physics, another Wikibooks project." The Difference Between Mass and Weight. In Topic 2.1 - Systems and Center of Mass, we introduced the concept of mass to describe the distribution of it within an object. However, there's an important distinction to make between mass and weight. To note, these two terms are often used interchangeably in everyday speech. The amount of mass within an object is the measure of how much matter is within an object. In other words, it describes the amount of "stuff" that is present in an object. In contrast, an object's weight is a measure of the gravitational force that is acting on an object. Mass is an intrinsic property. That means that any object with mass will have the same amount of mass regardless where it is located. Unlike mass, an object's weight varies depending on where the object is because the gravitational force that is exerted on an object varies. We can define an object's weight with the formula: formula_8. To note, using the previously discussed material in this unit: Thus, we could express the formula above as formula_14. One way of considering how mass and weight differ is by considering a water bottle that has a mass of formula_15. Then, we place this water bottle on the Earth. Because the Earth has an acceleration due to gravity of approximately formula_16, we can calculate for the weight of the water bottle on Earth, formula_17: formula_18 formula_19 Likewise if placed on the Moon, the water bottle would still have a mass of formula_15. However, the moon has an acceleration due to gravity of approximately formula_21. With this information, we can solve for formula_22, as shown below: formula_23 formula_24 Therefore, the weight of an object varies depending on where the object is. Unlike mass, weight is not a intrinsic attribute of an object; it is only a description of the force due to gravity acting on it. Free Body Diagrams. To reemphasize, gravitational force is not the only type of force that can exist on an object. In fact, there's oftentimes several significant forces acting on an object in a given physical situation. As such, there's an importance towards being able to find a way to illustrate these forces easily and clearly. One such method is called a free body diagram (FBD). It is a visualization of all of the external forces that are acting on an object. Through creating a free body diagram, it can become easier to find the net external force, formula_6 (as introduced earlier in this topic) as well as gaining an understanding for what is happening in a physical situation. What's included in a free body diagram: What's optional in a free body diagram: What's not included in a free body diagram: Reference Table of Important Forces. For the rest of the course, we will often be solving problems involving forces or otherwise applying an understanding of them to physical situations. As such, here is a reference table for the types of forces that will be discussed. To note, many of the listed forces have dedicated sections considering them, providing further information.

Isotonic regression. Linear order isotonic regression can be thought off as approximating given series of 1-dimensional observations with non-decreasing function. It is similar to smoothing splines, with the difference that we use monotonicity, rather than smoothness, to remove noise from the data. General isotonic regression is the approximation of a given series of values with values that satisfy a given partial order. Problem. Given values formula_1 and their positive weights formula_2, approximate them as closely as possible by formula_3, subject to a set of constraints of kind formula_4 : If all weights are equal to 1, the problem is called "unweighted isotonic regression", otherwise it is called "weighted isotonic regression". The graph formula_14 must to be acyclic. For example, hte constraints formula_15 and formula_16 cannot exist simultaneously. Example: Minimize formula_17 subject to Linear order. Of a particular interest is linear order isotonic regression. For linear order isotonic regression, there is a simple linear algorithm, called #Pool Adjacent Violators Algorithm (PAVA). If all weights are equal to 1, the problem is called "unweighted linear ordering isotonic regression". Linear order isotonic regression as a model. Sometimes, linear ordering isotonic regression is applied to a set of observations formula_32, where formula_33 is the explanatory variable, y is the dependent variable. The observations are sorted by their formula_33's, then the isotonic regression is applied to the formula_35's with the additional constraint formula_36 for all formula_8. Other variants. Non-Euclidean metrics. Sometimes other metrics are used instead of the Euclidean metric, for example formula_38 metrics: or unweighted formula_40 metrics: Points on a grid. Sometimes, values are placed on a 2d or higher dimensional grid. The fit value must increase along each dimension, e.g. Algorithms. Pool Adjacent Violators Algorithm. The Pool Adjacent Violators Algorithm (PAVA) is a linear-time (and linear-memory) algorithm for linear order isotonic regression. Preliminary considerations. The algorithm is based on the following theorem: Theorem: For an optimal solution, if formula_44, then formula_45 Proof: Suppose the opposite, i.e. formula_46. Then for sufficiently small formula_47, we can set which decreases the sum formula_50 without violating the constraints. Therefore, our original solution was not optimal. Contradiction. Since formula_45, we can combine both points formula_52 and formula_53 to a new point formula_54. However, after combining two points formula_52 and formula_56 to the new point formula_57, this new point can violate the constraint formula_58. In this case it should be combined with the formula_59-th point. If the combined point violates its constraint, it should be combined with the previous point, and so on. Algorithm. Input: Output: The algorithm: Here formula_60 defines to which old points each new point corresponds. Arbitrary case algorithms. In the arbitrary case, this can be solved as a quadratic problem. The best algorithm takes formula_61 time, see: Implementations. R. isoreg. The function isoreg performs unweighted linear ordering isotonic regression. It does not require any packages. For many simple cases, it is enough. Example of usage: x=sort(rnorm(10000)) y=x+rnorm(10000) y.iso=isoreg(y)$yf plot(x,y,cex=0.2) lines(x,y.iso,lwd=2,col=2) The isoreg function also implements linear ordering isotonic regression as a model: x=rnorm(10000) y=x+rnorm(10000) y.iso=isoreg(x,y)$yf plot(x,y,cex=0.2) lines(sort(x),y.iso,lwd=2,col=2) Iso. The package Iso contains three functions: Example of usage: install.packages("Iso") # should be done only once library("Iso") # should be done once per session x=sort(rnorm(10000)) y=x+rnorm(10000) y.iso=pava(y) plot(x,y,cex=0.2); lines(x,y.iso,lwd=2,col=2) isotone. This is the most advanced package. It contains two functions: Example of usage: install.packages("isotone") # should be done only once library("isotone") # should be done once per session x=sort(rnorm(10000)) y=x+rnorm(10000) y.iso=gpava(y)$x plot(x,y,cex=0.2); lines(x,y.iso,lwd=2,col=2) Comparison of speed. Since all three libraries somehow implement the PAVA algorithm, we can compare their speed. As we can see on the graph below, for unweighted linear order isotonic regression (LOIR) with formula_62, isoreg should be used. For weighted LOIR and unweighted LOIR with formula_63, pava should be used. As for gpava, it should be used only for non-Euclidean metrics. Also, the implementations of weighted simple ordering isotonic regression on R are far from perfect. Java. Weka, a free software collection of machine learning algorithms for data mining tasks written in the University of Waikato, contains, among others, an isotonic regression classifier. The classifier is deeply ingrained into the Weka's hierarchy of classes and cannot be used without Weka. Python. While scikit-learn implements the isotonic regression, Andrew Tulloch made the Cython implementation of the algorithm for linear ordering isotnic regression, which is 14-5000 times faster, depending on the data size. See [Speeding up isotonic regression in scikit-learn by 5,000x https://tullo.ch/articles/speeding-up-isotonic-regression/]. If you just want the code, click [here https://gist.github.com/ajtulloch/9447845#file-_isotonic-pyx]. Usage. Calibrating the output for categorical probabilities. For more information, see "Predicting Good Probabilities With Supervised Learning", by Alexandru Niculescu-Mizil and Rich Caruana. Most of the supervised learning algorithms, for example Random Forests, boosted trees, SVM etc. are good in predicting the most probable category, but not the probability. Some of them tend to overestimate high probabilities and underestimate low probabilities. (One notable exception is neural networks, which themselves produce a well calibrated prediction.) In order to obtain the correct probability, unweighted linear order isotonic regression can be used: where To find f, the model's predictions on the validation set are matched with the output variable, and the isotonic regression is applied to the pairs. An alternative approach is to pass formula_66 through a sigmoid function: This is called Platt calibration. For small data sets, Platt calibration is better than isotonic regression. Starting at 200-5000 observations, the isotonic regression slightly outperforms Platt calibration. Note also that this kind of isotonic regression is simpler and faster than the logistic regression required by Platt calibration. Calibration of recommendation models. The problem. The following example is for recommendation engines, whose general purpose is to predict the grade given by user formula_70 to item formula_8. We have a model M that predicts the residuals r of another model, trained on the training set. For cases with small number of users or items, the model is overfitted, and the predictions need to be relaxed: where Solution. Given the set of triplets formula_85 from validation database, we have: Therefore, for formula_87-th observation we should set weight formula_88 and value formula_89 . The triplets formula_90 then sorted with respect to formula_60, and the isotonic regression applied to them. When a function directly predicts the rating or acceptance rate, rather than predicting the residual of some other model, we can define formula_92 as the difference between this model and some simpler model (i. e. average rating for the item plus average rating correction for the user). This model is called basic model. Analysis and calibrating input variables. Often, we know that an output variable depends monotonically on an input variable, but otherwise we have very little prior knowledge about the dependence. In this case, isotonic regression can provide important clues about the dependence. Non-metric multidimensional scaling. A multidimensional scaling (MDS) algorithm, given the matrix of item–item similarities, places items into N-dimensional space such that similar items would be closer to each other than dissimilar items. This is especially useful for visualization purposes. There are several types of MDS, depending on the relationship between similarity and distance, as well as the definition of distance. For non-metric MDS, the distance can be any monotone function of similarity. The function is typically found using isotonic regression, see Non-metric multidimensional scaling in wikipedia. Non-metric multidimensional scaling is implemented as in the R's MASS library. Advantages, disadvantages, and post-processing. Advantages: Disadvantages: These drawbacks can be improved by smoothing the result of the isotonic regression. In this way, we can get the best of both worlds (smoothness and monotonicity).

Chess Opening Theory/1. e4/1...e5/2. Nf3/2...Bc5/3. Nxe5/3...f6. 3...f6? This move, seemingly attacking the knight, allows White to win with 4. Qh5+! because it allows 4...g6 5. Nxg6! hxg6 6. Qxh8 when White is up material. Other legal moves instead of 4...g6 would lead to checkmate: 5. Qf7+! Kd6 6. Nc4+ Kc6 7. Qd5+ Kb5 8. Ne5+ Ka4 9. Qc4+ Bb4 10. Qb5# 1. e4 e5 2. Nf3 Bc5 3. Nxe5 f6?

Physics Explained Through a Video Game/Introduction to Work and Energy. <br> Topic 3.1 - Introduction to Work and Energy. Goals: "Topic 3.1.1 - Work". Example 1: Rolling City Boulder. <br>To further describe the behavior of a physical object, we can consider work. Work is a type of energy within a system, which we'll discuss shortly. It involves considering the magnitude of a selected force on an object and the displacement that the object has had. In its simplest form, work is the product of a force's magnitude and an object displacement, given that (1) the force is constant and (2) the force and displacement vectors are pointing in the same direction. In this special case, formula_1. We could use the convention that work can be represented by the symbol formula_2 such that formula_3. As an example of this, consider a rolling circular boulder in a modified version of "lack of comfort" by "swiped3." The boulder accelerates towards the right while on a flat surface. This is because wind is pushing on the boulder, providing it a constant formula_4 that is pointed towards the right. Additionally, during the video, there is a displacement vector towards the right. Exercise: Suppose that formula_4 has a magnitude of formula_6 and the displacement magnitude is formula_7. What would be the net work done on the boulder? Answer: In this case, we are told that the force is constant. Also, we can see in the diagram that the formula_4 and formula_9 vectors are pointing in the same direction (towards the right). Thus, we can use the formula formula_3, as shown below to find the work done on the boulder by the net force, formula_4. formula_3 formula_13 [We're considering the work done by the net force on the ball. As such, we're calculating for the net work on the ball.] formula_14 [Substitution.] formula_15 [2 sigfigs] The boulder has a net work of formula_16 while it is pictured rolling on the rooftop. Example 2: Propeller Planes. As previously seen in Topic 2.3 - Newton's Second Law, let's again consider the map "Maneuver Planes" by "TiM MELL0" where the player Sam (red) descends a plane towards the ground. We'll use this example to generalize the concept of calculating the work for a constant force on an object. To note, we can also calculate for the work done by a force on an object even when the displacement vector isn't parallel with the force vector. With this, we can use the general formula for the work done on an object, given a constant force: With the equation above, note two things: If dot products are unfamiliar to you, we can also write the general equation for work done by constant force alternatively as formula_17. In this, Exercise: Suppose that there's a gravitational force of formula_23 "(directed downwards)" on the plane as it is descending. Using two points on the plane's path, labeled as "START" and "END," there is found to be a distance of formula_24 between them. Additionally, the displacement vector of the plane between these two points is directed formula_25 below the +X Axis. What is the work done by the gravitational force between the two labeled points? Answer: To solve this problem, we can first list the information mentioned in the prompt. With this, we can find the variables needed for substitution into the equation formula_29. More specifically, the gravitational force vector's magnitude is our formula_20 variable. In addition, the displacement vector represents the formula_9 variable. Therefore, we can partially substitute in our variables into the equation such that: formula_29 [Formula] formula_33 [We're considering the gravitational force acting on the plane.] formula_34 [Substitution of the gravitational force and displacement vector magnitudes.] To consider the value of formula_22, we can graph formula_36 and formula_9 as diagrammed to the right. Using the information given and the diagram, we can find that formula_38. We can substitute this value into our equation for formula_39 and continue simplifying. formula_40 [Substitution of formula_22] formula_42 [Using degree mode; Algebra; 2 sigfigs] While the plane is descending, the gravitational force does formula_43 of work on the plane. Example 3: Assembly Line. We can also calculate the work done by a force even when a force is changing. Suppose that in the map "Factory of Copy Maps" by antidonaldtrump / Armin van Buren that colored boxes are being pushed along an assembly line. On the top row of the assembly line, the blocks have a net force that is pushing them forward along the line, accelerating them in the leftward direction. Unlike previous examples, the force that is being exerted on the object is varying. This means that the equation formula_29, which assumes that the force is constant, cannot be directly used. Instead, we can consider calculating for the work on the colored boxes graphically. Walkthrough: Suppose that on the top row of the assembly line where: In this example, we're going to be calculating graphically for the work done on the yellow box by the net force. To do this, we can create a function of the force on the yellow box with respect to its displacement as pictured on the right. To go further, we need to consider some theory regarding work. The work done by a force is the "accumulation" of that force over a distance. In other words, we can graph a function of force with respect to distance. Since work is the accumulation of force with respect to distance, the area under the above-mentioned function is the work. Using this definition, we can create a graph of the yellow box's net force, formula_49 with respect to formula_50. Because the force exerted on the yellow box is "dependent" on where the box is positioned, formula_49 is the independent variable (formula_52) and formula_50 is the dependent variable (formula_54). As mentioned above, because work is the accumulation of force with respect to distance, if we were to calculate the area between the green curve and the formula_55 axis, this would be work done on the yellow box between formula_56 and formula_7. This can be done through breaking apart the pictured graph into two rectangles as also pictured. From here, we can easily calculate the area of each of these rectangles, giving us the amount of work done. With this, we can combine the two rectangle areas to find the work done: Thus, when the yellow box was being pushed along the top portion of the assembly line, it had a work of formula_58 done by the net force acting on it. "Section 3.1.2 - Kinetic Energy". Example 1: Arrows in the Forest. In this example, there will be an introduction of kinetic energy, what it means conceptually, and how to calculate for it in a physical situation. [Content] Example 2: Flying Squirrel. [Intro] Suppose the in-game situation in the map "Flying Squirrel" by Raspy 667. In this map, a flying squirrel is gently moving through the air back and forth. What would be the kinetic energy of the squirrel at each of the labeled points from the video [cont] "Consider discussing your solutions on this article's . On there, you can find help from others."

Chess Opening Theory/1. e4/1...c6/2. d4/2...d5/3. e5/3...c5/4. c3. =Caro-Kann - Botvinnik-Carls Defense - 4. c3= White’s most popular response to the Botvinnik-Carls Defense according to the Lichess database, accounting for more than half of white’s moves in the preceding position. While the move is neither the engine’s top choice, nor that of most grandmasters, it still leaves white with a small advantage. The objective of the move is to strengthen the center and enable a recapture with the pawn in the event of 4... cxd4. It is best for black to put more pressure on the d4 pawn before taking the pawn, which can best be done with the move 4... Nc6.

Chess Opening Theory/1. e4/1...c6/2. d4/2...d5/3. e5/3...c5/4. c3/4...cxd4. =Caro-Kann - Botvinnik-Carls Defense - 4. c3 cxd4= Black's second most popular response to the move 4. c3 in the Botvinnik-Carls Defense. It is not the best move, as it is better for black to have added more pressure to the d4 pawn first before taking.

Chess Opening Theory/1. e4/1...c6/2. d4/2...d5/3. e5/3...c5/4. c3/4...Nc6. =Caro-Kann - Botvinnik-Carls Defense - 4. c3 Nc6= This is the best continuation for black after the move 4. c3 in response to the Botvinnik-Carls Defense. It adds pressure on the d4 pawn. The move 5. Nf3, defending the pawn, should be met with 5... cxd4 6. cxd4 Bg4. This sequence allows for sustained pressure on the d4 pawn, which can be increased via the g8-e7-f5 manoeuvre of the black knight.

Antenna Television/Market Listings/Oregon. Klamath Falls. Counties. Oregon California Medford. Counties. Oregon California Portland. Counties. Oregon Washington

Synesthetic Explorations. Synesthetic Explorations : An Autoethnographic Research into Music, Color, and Creativity Author: Arnaud Quercy Introduction. I am Arnaud Quercy, an artist dedicated to exploring the intersections of visual arts, music, and creativity. My practice encompasses painting, sculpture, musical composition, and performance, with a particular focus on how synesthesia—a phenomenon where one sensory experience involuntarily triggers another—shapes and informs my artistic work. While I do not hold an academic title, my approach to this research is guided by a commitment to rigor and meticulousness. This work does not seek to position itself within traditional scholarly frameworks but rather aims to offer a deeply personal and experiential contribution to the broader understanding of synesthesia. The insights presented here are grounded in my own experiences, yet I strive to maintain a level of analysis that is both reflective and critically engaged. This study is positioned as an autoethnographic investigation into the creative potential of synesthesia, specifically focusing on how it can be systematically applied and analyzed across the chromatic scale and the color wheel. Although synesthesia is a phenomenon that many individuals experience, the uniqueness of this research lies in its structured application to artistic practice. My intention is to bridge the gap between personal sensory experiences and their broader implications for creative expression. As an autoethnographic work, I acknowledge the inherent subjectivity and potential bias in this research. However, I approach this study with a genuine intent to contribute valuable insights to the field, offering a unique perspective that may serve as a useful resource for researchers, artists, and others interested in synesthesia and its applications. The significance of this research extends beyond mere documentation of synesthetic experiences. It proposes a framework for understanding how synesthesia can be harnessed as a tool for artistic creation, providing a new lens through which to view the relationship between sound, color, and emotion. This work aims to fill a critical gap in both the academic study of synesthesia and its practical applications within the arts. "Synesthetic Explorations" is a living document, open to ongoing refinement and contributions from others. It aspires to deepen our understanding of synesthesia not only as a neurological phenomenon but also as a powerful creative force with the potential to transform artistic practices.

Physics Explained Through a Video Game/Introduction to Work. <br> Topic 3.1 - Introduction to Work. Goals: Example 1: Rolling City Boulder. <br>To further describe the behavior of a physical object, we can consider work. Work is a type of energy within a system, which we'll discuss shortly. It involves considering the magnitude of a selected force on an object and the displacement that the object has had. In its simplest form, work is the product of a force's magnitude and an object displacement, given that (1) the force is constant and (2) the force and displacement vectors are pointing in the same direction. In this special case, formula_1. We could use the convention that work can be represented by the symbol formula_2 such that formula_3. As an example of this, consider a rolling circular boulder in a modified version of "lack of comfort" by "swiped3." The boulder accelerates towards the right while on a flat surface. This is because wind is pushing on the boulder, providing it a constant formula_4 that is pointed towards the right. Additionally, during the video, there is a displacement vector towards the right. Exercise: Suppose that formula_4 has a magnitude of formula_6 and the displacement magnitude is formula_7. What would be the net work done on the boulder? Answer: In this case, we are told that the force is constant. Also, we can see in the diagram that the formula_4 and formula_9 vectors are pointing in the same direction (towards the right). Thus, we can use the formula formula_3, as shown below to find the work done on the boulder by the net force, formula_4. formula_3 formula_13 [We're considering the work done by the net force on the ball. As such, we're calculating for the net work on the ball.] formula_14 [Substitution.] formula_15 [2 sigfigs] The boulder has a net work of formula_16 while it is pictured rolling on the rooftop. Example 2: Propeller Planes. As previously seen in Topic 2.3 - Newton's Second Law, let's again consider the map "Maneuver Planes" by "TiM MELL0" where the player Sam (red) descends a plane towards the ground. We'll use this example to generalize the concept of calculating the work for a constant force on an object. To note, we can also calculate for the work done by a force on an object even when the displacement vector isn't parallel with the force vector. With this, we can use the general formula for the work done on an object, given a constant force: With the equation above, note two things: We can also write the general equation for work done by constant force alternatively as formula_17. In this, Exercise: Suppose that there's a gravitational force of formula_23 "(directed downwards)" on the plane as it is descending. Using two points on the plane's path, labeled as "START" and "END," there is found to be a distance of formula_24 between them. Additionally, the displacement vector of the plane between these two points is directed formula_25 below the +X Axis. What is the work done by the gravitational force between the two labeled points? Answer: To solve this problem, we can first list the information mentioned in the prompt. With this, we can find the variables needed for substitution into the equation formula_29. More specifically, the gravitational force vector's magnitude is our formula_20 variable. In addition, the displacement vector represents the formula_9 variable. Therefore, we can partially substitute in our variables into the equation such that: formula_29 [Formula] formula_33 [We're considering the gravitational force acting on the plane.] formula_34 [Substitution of the gravitational force and displacement vector magnitudes.] To consider the value of formula_22, we can graph formula_36 and formula_9 as diagrammed to the right. Using the information given and the diagram, we can find that formula_38. We can substitute this value into our equation for formula_39 and continue simplifying. formula_40 [Substitution of formula_22] formula_42 [Using degree mode; Algebra; 2 sigfigs] While the plane is descending, the gravitational force does formula_43 of work on the plane. Example 3: Assembly Line. We can also calculate the work done by a force even when a force is changing. Suppose that in the map "Factory of Copy Maps" by antidonaldtrump / Armin van Buren that colored boxes are being pushed along an assembly line. On the top row of the assembly line, the blocks have a net force that is pushing them forward along the line, accelerating them in the leftward direction. Unlike previous examples, the force that is being exerted on the object is varying. This means that the equation formula_29, which assumes that the force is constant, cannot be directly used. Instead, we can consider calculating for the work on the colored boxes graphically. Walkthrough: Suppose that on the top row of the assembly line where: In this example, we're going to be calculating graphically for the work done on the yellow box by the net force. To do this, we can create a function of the force on the yellow box with respect to its displacement as pictured on the right. To go further, we need to consider some theory regarding work. The work done by a force is the "accumulation" of that force over a distance. In other words, we can graph a function of force with respect to distance. Since work is the accumulation of force with respect to distance, the area under the above-mentioned function is the work. Using this definition, we can create a graph of the yellow box's net force, formula_49 with respect to formula_50. Because the force exerted on the yellow box is "dependent" on where the box is positioned, formula_49 is the independent variable (formula_52) and formula_50 is the dependent variable (formula_54). As mentioned above, because work is the accumulation of force with respect to distance, if we were to calculate the area between the green curve and the formula_55 axis, this would be work done on the yellow box between formula_56 and formula_7. This can be done through breaking apart the pictured graph into two rectangles as also pictured. From here, we can easily calculate the area of each of these rectangles, giving us the amount of work done. With this, we can combine the two rectangle areas to find the work done: Thus, when the yellow box was being pushed along the top portion of the assembly line, it had a work of formula_58 done by the net force acting on it. Question 1: Colorful Canyon. Consider the map "Valley" by "monkey butler". In this map inspired by the Aravaipa Canyon Wilderness in Arizona, U.S., a player is falling off the side of a rock face and falls onto a boulder resting in the river, as diagrammed on the right. In the diagram, as labeled by a solid green path, the player briefly slides across a portion of the rock face. Then, labeled by a solid yellow path, the player slides off the rock face, entering free fall motion until striking a boulder in the river. Dotted green and yellow lines are provided, indicating the magnitude of displacement in both the horizontal and vertical directions for both parts of the diagrammed path. Assume that: Use the information provided above to answer the following question parts.Part (a):"Consider when the player is sliding across the rock face portion under the green path." (i) Calculate the magnitude of the gravitational force acting on the player. (ii) Calculate the normal force that is acting on the player. (iii) Draw a completed free body diagram. Use the unlabeled reference image on the right. Part (b):"Continue to consider when the player is sliding across the rock face portion under the green path." (i) Determine the net force magnitude and direction of the player. (ii) Calculate the work done by the net force on the player when traveling along the green path. Part (c):"Now consider when the player is traveling in free fall." (i) Determine the net force magnitude and direction of the player. (ii) Calculate the overall displacement of the player while on the yellow portion of the path. (iii) Calculate the work done by the net force on the player when in free fall.Part (d):"Now consider the player over the entire labeled path." (i) Create a graph of the net force acting on the player with respect to the total displacement of the player. (ii) Calculate the work done by the net force on the player when traveling along the green path. "Consider discussing your solutions on this article's . On there, you can find help from others."

AI Art Generation Handbook/FLUX. FLUX 1.0 is one of the alternate AI Art model coming from Black Forest Labs (which is coincidentally formed by former Stable Diffusion engineers) in August 2024. In August 2024, Flux 1.0 is fully supported by ComfyUI. In September 2024, lllyasviel's Forge with the community support have integrated support for FLUX 1.0 into Forge WebUI

AI Art Generation Handbook/FLUX/Workflow JSON. Copy the whole code below the break line and paste it to Notepad (for Windows) . 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Canadian Refugee Procedure/117-121.1 - Human Smuggling and Trafficking. IRPA Sections 117-121.1. Sections 117-121.1 of the "Immigration and Refugee Protection Act" read: Human smuggling is an offence in Canada under section 117 of the IRPA, but Canadian law recognizes a defence to section 117 of the "IRPA" for those providing humanitarian aid to asylum seekers. Human smuggling is an offence in Canada under section 117 of the "IRPA" and carries a maximum term of imprisonment of at least 10 years. However, Canadian law recognizes a defence to section 117 of the "IRPA" for those providing humanitarian aid to asylum seekers. Specifically, Canadian law recognizes defences of mutual aid, family member aid, and humanitarian aid. By way of example, the Canadian humanitarian aid defence has four elements:

Chess Opening Theory/1. e4/1...e5/2. Nf3/2...Nc6/3. c3/3...Nf6/4.d4. Ponziani Opening: Jaenisch Counterattack. 4.d4. With 4.d4, White ignores the attack on the e-pawn and counterattacks Black's central pawn on e5. Black now has a choice - they can take the e-pawn, take the d-pawn, strike in the center with d5, or defend their own pawn with d6. This is the most common line in the Ponziani opening at both beginner and master level.