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djcluv
2019-02-28 09:13:42
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2019-02-28 09:12:02
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djcluv
2019-02-28 09:10:37
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2019-02-28 09:09:40
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2019-02-26 10:06:47
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2019-02-26 10:04:26
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andymark
2019-02-26 10:01:05
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2019-02-26 09:59:12
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andymark
2019-02-26 09:57:41
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johnbhura1
2019-01-18 11:29:34
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tdl17
2019-01-05 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.
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tdl17
2019-01-03 21:16:15
Since the particles only took one half-life, which is 2.5\\\\\\\\times10^{-8} seconds, to travel 15m, it must be traveling close to the speed of light. \\\\r\\\\n\\\\r\\\\nAssume the speed of the particles is some fraction of the speed of light and let this fraction be a, we now have v = ac. \\\\r\\\\n\\\\r\\\\nUsing v=\\\\\\\\frac{L\\\\\\\\sqrt{1-\\\\\\\\frac{v^2}{c^2}}}{t_0}, we can replace v with ac and from there, you can solve for a and that should get you to a = \\\\\\\\frac{2}{\\\\\\\\sqrt{5}}c, which is choice C.
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djcluv
2018-12-14 00:44:47
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djcluv
2018-12-14 00:41:42
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2018-11-05 07:04:26
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2018-11-05 07:03:03
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raytyler
2018-11-01 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/
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raytyler
2018-11-01 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
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petyr
2018-10-30 06:41:46
dp=v*dm=v*rho*dVolume=v*rho*A*dx=v*rho*A*v*dt\r\n\r\ntherefore dp/dt=F=rho*(v^2)*A.\r\nshowbox
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TheBridge
2018-10-27 02:12:16
But 1 + i - i is not equal to 0. The eigenvalues are actually 1, -\\\\\\\\frac{1}{2} \\\\\\\\pm \\\\\\\\frac{\\\\\\\\sqrt{3}}{2}i.
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TypT
2018-10-26 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 4^2 = 16 (E), to give a little more confidence in (C).
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The Bridge
2018-10-26 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 g, 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 g\\approx10 m/s2 in the table of information at the top of the test anyway, so the left side of the equation is completely unneeded.
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kronotsky
2018-10-26 03:27:43
The relative differences in wavelengths is given by the factor (n-1), so this is what you are equating to 2e-4. 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.
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Ryry013
2018-10-25 12:05:41
My half-BS 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.
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kronotsky
2018-10-25 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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