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Latest 25 Comments  0  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. 
5  Click here to jump to the problem!  webtechcoupons
20181105 07:03:03  Aw, I thought this was a significant pleasant post. Godaddy Renewal coupons In idea, I should set up composing like this also – investing energy and real exertion to make an extraordinary article… yet precisely what do I say… I hesitate a lot through no methods seem to go did. 
6  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/ 
7  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 
10  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). 
11  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. 
12  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. 
13  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. 
14  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. 
15  Click here to jump to the problem!  ryanmes
20181024 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. 
18  Click here to jump to the problem!  kronotsky
20181023 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. 
20  Click here to jump to the problem!  kronotsky
20181023 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 nontrivial 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). 
21  Click here to jump to the problem!  kronotsky
20181023 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). 
23  Click here to jump to the problem!  kronotsky
20181023 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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