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 2 Click here to jump to the problem! TimToolMan2018-04-12 16:27:04 We know that $\\beta=v/c=pc/E$ (using relativistic momentum and energy). Since $v=c$ we have $1=pc/E$ which implies $E=pc$. And, from the relation $E^2 = (pc)^2+(mc^2)^2$, we can say that $E=pc$ implies $m=0$.Click here to jump to the problem!
 7 Click here to jump to the problem! whereami2018-04-09 03:39:50 My memory goes that selection rule regards only delta l and delta ml. I didn\'t even believe selection rule works on those aspects. \r\nTurns out that delta s =0 is one of the selection rules. And there are none regarding n values. My thought against A is that n could be 0. But of course that is not forbidden by the selection rules. Click here to jump to the problem!
 9 Click here to jump to the problem! whereami2018-04-09 02:01:42 special relativity doesn\'t really change how things are, i think. It only supplies you with different coordinates of space and time. In this case no matter what frame you choose, the doors will be opened/closed at the moment when it is perceived to be needed to be opened/closed. \r\nIf you don\'t believe this you can calculate for both frames. \r\nAnswer is E bytheway. Click here to jump to the problem!
 11 Click here to jump to the problem! enterprise2018-04-07 14:27:53 There is another way to work this out. Any two-body problem in mechanics can be reduced to one body problem using the concept of reduced mass. So we should expect that the masses should appear as reduced mass in the solution. The only choice in which this happens is D which have K(1/M_1+1/M_2) =K/reduced mass. So , this is the solution.Click here to jump to the problem!
 12 Click here to jump to the problem! whereami2018-04-07 03:27:19 I need help. \r\n1. electron has degrees of freedom than atom because they are in potential well. I don\'t understand this logic. Atoms in metals are pretty much fixed while conduction electrons surely seems to have much more freedom than these atoms. \r\n2. thermal equilibrium. if they are in thermal equilibrium, how can electrons have much more kinetic energy than a few KTs ? \r\n3. I don\'t even know what is a fermi gas...I need to read.\r\nI think a good explanation must be much better than what I see now. What I see now is basically favoring certain logic while ignoring other logic without clear explanations.Click here to jump to the problem!
 13 Click here to jump to the problem! whereami2018-04-06 14:20:34 look at any regular polygon. draw the lines that connect each vertex to the center of the polygon. the lines themselves can form another regular polygon. This means their vector sum is zero. so is E. Click here to jump to the problem!
 14 Click here to jump to the problem! whereami2018-04-06 04:41:31 At x=0. Particle m has total energy = E. when it returns from either direction to this point x=0, its energy will be E. x<0 can depict a scenario in which it hit a spring at equilibrium.x>0 can depict another scenario in which it jumps into the sky and eventually comes back. Click here to jump to the problem!
 15 Click here to jump to the problem! whereami2018-04-06 03:32:09 i think the best way to solve this problem is some sense or knowledge of how the solution of coupled harmonic oscillators look like. Look at m1 and m2. They are on the opposite sides of a spring. This means: you will never see m1 +m2 as a factor in the solution. If you are familiar with some particular solutions of the coupled harmonic oscillator system, you will know this. This mathematical expression is simply un-physical. The only way you see m1+m2 is when you put m2 and m1 together on the same side of a spring. Click here to jump to the problem!
 16 Click here to jump to the problem! whereami2018-04-06 02:01:22 i see some people are saying average momentum can\'t be zero due to uncertainty principle. But that has nothing to do with principle. average momentum is not delta p. Option I is wrong because average momentum can be not zero. there is no reason its momentum should be fixed to zero. Click here to jump to the problem!
 17 Click here to jump to the problem! whereami2018-04-06 01:23:24 i think the answers are all wrong, or maybe I am making a mistake. \r\nThe problem states that orbital angular momentum is zero. That\'s not even spin. And everybody is saying that spin is zero. This really bugs me. \r\nI don\'t think L has anything to do with S. \r\nCan somebody help me ? Click here to jump to the problem!
 22 Click here to jump to the problem! TimToolMan2018-04-04 16:21:50 Another way to think about it:\\\\r\\\\nTake $E^2=(pc)^2+(mc^2)^2$ if $m=0$ then $E=pc$ and then $c=E/p$ which means $c=\\gamma mc^2 / \\gamma mv$ cancelling like terms gives $v=c$ when $m=0$Click here to jump to the problem!
 23 Click here to jump to the problem! TimToolMan2018-04-03 00:15:02 Simple Unit argument.\\\\r\\\\nEnergy is in units of $kg(m^2/ s^2)$\\\\r\\\\nSo we can get the units of k to be $kg/(m^2 s^2)$\\\\r\\\\nForce must be in units of $kg (m/s^2)$\\\\r\\\\nKeeping k\\\\\\\'s units in mind, there are only 2 solutions with these units of force: B & E\\\\r\\\\nNo gravity in this question, so it must be B Click here to jump to the problem!