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Latest 25 Comments  5  Click here to jump to the problem!  andymark
20190226 10:04:26  Hi ! Nice post .Points are really good . Great article .Over all nice content , Thanks for posting.\r\nGame Lover  https://games.lol/puzzle/ 
9  Click here to jump to the problem!  johnbhura1
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10  Click here to jump to the problem!  tdl17
20190105 22:45:55  You can also use dimensional analysis. Whatever that goes into the arc tangent should be purely numerical without any units, and the only answer where this happens is choice A. Vq cancels with mv^2 because they both have units of Joule. L cancels with d because they both have units of length. 
15  Click here to jump to the problem!  webtechcoupons
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16  Click here to jump to the problem!  raytyler
20181101 03:14:57  \\lambda_1 = \\frac{1}{2} + i\\frac{sqrt{3}}{2}\r\n\r\n\\lambda_2 = \\frac{1}{2}  i\\frac{sqrt{3}}{2}\r\n\r\n\\lambda_3 = 1\r\n192.168.0.1 http://19216801help.com/ 
17  Click here to jump to the problem!  raytyler
20181101 03:13:02  1 + i  i is not equal to 0. The eigenvalues are actually 1, \\\\\\\\\\\\\\\\frac{1}{2} \\\\\\\\\\\\\\\\pm \\\\\\\\\\\\\\\\frac{\\\\\\\\\\\\\\\\sqrt{3}}{2}i.\r\n192.168.0.1 login 
20  Click here to jump to the problem!  TypT
20181026 22:10:37  Quick, dumb way I did it without having to figure out what exactly the question was asking:\r\n\r\nSpace time intervals have a Pythagorean relationship, the two numbers we are given are 3 and 5, to complete the Pythagorean triple we need a 4 so we get (C). It even has the trap answer of (E), to give a little more confidence in (C). 
21  Click here to jump to the problem!  The Bridge
20181026 19:15:17  That won\'t work. The formula on the left is for centripetal acceleration of someone standing on the Equator due to the rotation of the Earth, and works out to about .88 m/s2. This is in fact many times less than , the acceleration due to gravity of something on the Earth\'s surface, which is the right side. If you knew the orbital period of an object at the Earth\'s surface off the top of your head (84 min), then this method would work, but it gives you 10 m/s2 in the table of information at the top of the test anyway, so the left side of the equation is completely unneeded. 
22  Click here to jump to the problem!  kronotsky
20181026 03:27:43  The relative differences in wavelengths is given by the factor (n1), so this is what you are equating to 2e4. Other than that, the situation happens to be exactly analogous, since increasing the index of refraction essentially increases the effective optical path length, which is the quantity used to derive the thin film equation. 
23  Click here to jump to the problem!  Ryry013
20181025 12:05:41  My halfBS solution: I tried to figure out which of the answers could reasonably come from 1.5. I thought 1.5 = 3/2, saw 2/3, and then picked it. A guessing strategy for if you know zero Physics. 
24  Click here to jump to the problem!  kronotsky
20181025 02:57:36  I\'m sorry, but you are misinformed about Gauss\'s law. Classically, Maxwell\'s equations hold at any instant in a given frame. Gauss\'s law simply follows from the divergence theorem and the first equation. It may even be more appropriate to say that the first equation is Gauss\'s law. Gauss\'s law certainly applies for radiated fields (and causal propagation fields traveling at the speed of light), though radiation typically breaks the electrostatic symmetry. But in this case, the symmetry argument you use to apply the integral form of Gauss\'s law has exactly the same content as the symmetry argument you use to exclude radiation. 




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