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djcluv
2018-12-14 00:44:47
thanks for this blog!! multiplayer racing games
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djcluv
2018-12-14 00:41:42
thanks for this blog!! multiplayer racing games\r\n
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2018-11-05 07:04:26
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webtechcoupons
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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ryanmes
2018-10-24 21:32:38
The easiest way to look at so many problems on this exam is limits. If d increases, then you need a smaller velocity to stay on the track. There is only one answer where this is the case (when d is in the denominator). And this is the answer.
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BruceLyu
2018-10-24 07:35:28
This problem involves two plane waves (de-Broglie waves) in two regions. Initially, it is e^{ikx} and after entering the region with potential energy V, the kinetic energy becomes E-V_0 and the corresponding wave is e^{ik\'x}. For a plane wave, the energy is E=\\hbar^2 k^2/2m so \r\n(\\frac{k}{k\'})^2 = \\frac{E}{E-V_0} = (\\frac{\\lambda\'}{\\lambda})^2\r\nand solve it we get the final answer E.
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ryanmes
2018-10-24 01:19:02
I did the same thing... it works
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kronotsky
2018-10-23 04:29:06
This is not correct. The cube still has a symmetry, and we can apply Gauss\'s law in the same way as with a sphere as long as our integrals are over surfaces which respect that symmetry.
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kronotsky
2018-10-23 04:23:24
In classical regimes, equipartition of energy.
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kronotsky
2018-10-23 04:11:55
I really wish that every QFT/particle physics class spent like one or two whole lectures on the cross section right at the beginning.\r\n\r\nAs to the \"particularly brutal\" part, the essential reason why the formula is exponential is that the probability of multiple collisions along an initial trajectory becomes non-trivial in the limit of higher densities. By the problem statement, we are in the limit of low *single* collision probability, so the probability for scattering is proportional to the thickness of the sample. It is also inversely proportional to the density. This is the essential information you need to know to solve the problem (without dimensional analysis, which you should definitely use).
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kronotsky
2018-10-23 03:51:23
Filled subshells are the singlet component of the direct product of the component orbitals, so they have zero orbital angular momentum. So does 4s. The rotation operator(s) is an exponent of the orbital angular momentum operator, so it is equal to 1 (i.e. rotational symmetry).
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kronotsky
2018-10-23 03:44:30
Right hand rule trick: if you put your thumb in the direction of I, and curl your fingers up into a fist, the fingers point along B. We want q \\vec{v} \\times \\vec{B} = \\vec{I}. Cyclic permutations and setting q = -1 gives \\vec{I} \\times \\vec{B} = \\vec{v}.
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kronotsky
2018-10-23 03:10:02
You should buy it for many other reasons, but I\'ve found that \"Modern Classical Physics\" from Thorne and Blandford has the best philosophical explanation of thermodynamics and statistical mechanics I\'ve encountered so far. The treatment really is limited to a basic course, however, so it\'s not what you should get if you want lots of detail (there are only a few chapters on thermal physics!).
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ryanm
2018-10-23 02:56:31
This is the solution that I used and it worked. It is simple because it is easy to take the derivative.
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kronotsky
2018-10-23 02:49:28
Who cares if energy is conserved? The energy lost by the nail is due to the forces on it.
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kronotsky
2018-10-23 02:46:41
This solution is essentially the same as the one given, because in both cases the assumption is made that the nail experiences a constant force. This is the correct approach for the time average of the force, but we still need to know the time dependence. Similarly, conservation of energy is the right approach to find the distance average of the force.
 
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