Latest comments


Latest 25 Comments  0  Click here to jump to the problem!  deneb
20181015 23:29:30  My reasoning is the breaking happens continuously. It\'s not quantized like an electron jumping energy levels. You don\'t go from 10 m/s to 5 m/s without slowing down to 9, 8, etc. 
2  Click here to jump to the problem!  UNKNOWNUMBER
20181014 22:55:11  This is how I got the correct answer, can someone confirm if it\'s viable method or if I just got lucky? \r\n\r\nWe\'re given a constant magnetic field, this means that according at Faraday\'s law, the curl of the electric field = 0. Since the electric field is what propels the charge, it must move in a straight path because there is no curl; there is no rotational aspect so it can\'t be any of the other answers. \r\n 
3  Click here to jump to the problem!  deneb
20181013 23:23:14  If you know enough about orbitals to eliminate 4 answers, it would be way quicker to just figure out the correct configuration for 11 electrons and pick C 
4  Click here to jump to the problem!  deneb
20181013 21:57:38  Damnit I thought this was rolling down an incline. So I added a downward acceleration due to gravity to the inward pointing centripetal acceleration and picked A... 
5  Click here to jump to the problem!  deneb
20181013 21:26:52  I\'m confused what the answer is describing. Is the s=1 referring to the entire helium atom? This particular helium atom has two electrons which each have s=1/2, so together s=1? And then since it\'s in the S orbital, l=0, then j=1? 
6  Click here to jump to the problem!  Hassan0013
20181012 15:08:27  I think the better way is to just use the \\\\\r\n\r\n \\\\\r\n\r\nand again combine both 2 waves to make the vector for the electric field.\r\n\r\nnow if we calculate the vector\'s magnitude we easily find the solution for choice (A)\r\n\r\nthis is recombination after portions : \\\\\r\n\r\n \\\\\r\n\r\n\r\nnow it needs just to calculate the magnitude of E to get the answer...\r\n 
7  Click here to jump to the problem!  Hassan0013
20181012 15:05:09  I think the better way is to just use the \r\nand again combine both 2 waves to make the vector for the electric field.\r\n\r\nnow if we calculate the vector\'s magnitude we easily find the solution for choice (A)\r\n\r\nthis is recombination after portions : \r\n\r\n\r\nnow it needs just to calculate the magnitude of E to get the answer...\r\n 
8  Click here to jump to the problem!  deneb
20181010 05:14:29  Agreed, just a simple ratio problem. \r\n\r\nT1 = 4/3 hours\r\nT2 = 24 hours\r\n\r\nR1 = 1 Re\r\nR2 = a * Re\r\n\r\nsolve for a\r\n\r\n18^2 = a^3\r\n\r\na = 18^(2/3)\r\n\r\nwe can approximate 18^(1/3) to be around 2.5, which squared is 6.25. Answer is B. 
12  Click here to jump to the problem!  cquinone
20181004 19:26:08  The area estimations I\'m seeing here and in other solutions are wrong, you can easily calculate the range of V to be 25, so area should be estimated as 1/2(b)*(h) = 1/2(52)(500200)= 1/2(900) = 450. This is equidistant to 300 and 600, but you know that your estimate was slightly high as the isotherm BC dips down. So you pick 300. 
13  Click here to jump to the problem!  Batman
20180913 09:06:44  Rotation of electron in its orbital creates a magnetic moment that interacts with the inhomogeneous magnetic field. The magnetic moment of a proton is very small compared to that of electron and hence negligible in this case. \\\\r\\\\nHowever, we have methods to work with the nuclear magnetic moment too. 
17  Click here to jump to the problem!  physics_guy
20180815 03:03:11  Alternatively, letting a or b = 0 gives you a particleinabox stationary state times a phase factor. Since bound stationary states in 1d are nondegenerate, the probability current must vanish. Again the only solutions that work are A and E, and A is not true in general. 
18  Click here to jump to the problem!  physics_guy
20180815 03:00:04  If b = i*a then you have the wavefunction of a free particle with wavevector k, which does have nonzero probability current. It is true that if a and b are real  or if a is any complex number and b is simply zero  then the probability current is zero. But it is not simply because psi is a stationary state. The condition for a state to be stationary is that the divergence of the probability current must vanish, not the probability current itself. However, if a stationary state is also nondegenerate, then the probability current vanishes. It is not clear that psi is a bound state here, indeed it looks more like a superposition of free particle wavefunctions, which are degenerate even in one dimension. However, if you are really itching for a shortcut, it follows from the definition of probability current that if psi is of the form (possibly timedependent phase factor) x (real function of position) then the probability current vanishes. This is why letting a and b be real (or just letting b=0) gives you an instant solution. It\'s a nice trick for this problem, but I would argue it\'s hard to convince yourself it\'s true unless you knew the definition of J in the first place. 
19  Click here to jump to the problem!  physics_guy
20180815 02:45:41  I think ETS really does want you to know the definition of current density. Otherwise, you should know that the probability current vanishes for nondegenerate stationary states, i.e. wavefunctions of the form (possibly timedependent phase factor) x (realvalued function). In this problem, if we let a and b be real, then the probability current should vanish. Only choices A and E satisfy this, and choice A does seem suspiciously stringent. To see why A is false, let b = i * a. Then we have the wavefunction of a free particle of wavenumber k, whose probability current is not zero (even though its probability density is constant everywhere in space), but is instead equal to the probability density times the velocity of the the particle. 
20  Click here to jump to the problem!  physics_guy
20180815 01:23:47  Clarification: the last three components of the fourvector are zero for the positron/electron system in their CM frame. The last three components of the single photon fourvector are nonzero in this frame because the photon always travels at the speed of light. 
21  Click here to jump to the problem!  physics_guy
20180815 01:09:11  In the center of momentum frame of the positron/electron system, their energymomentum fourvectors sum to zero. But a single photon has nonzero energymomentum in every reference frame. So this process would simultaneously violate energy conservation and linear momentum conservation. 
23  Click here to jump to the problem!  physics_guy
20180808 21:23:00  The field of a magnetic dipole is proportional to the magnetic dipole moment, which is the product of the current and the area of the loop. 
24  Click here to jump to the problem!  physics_guy
20180808 21:13:34  There are two kinds of impedance matching. One minimizes reflection at the boundary between materials, and the other maximizes power transfer between two materials. To minimize reflection, the impedance of the generator must equal the impedance of the load. To maximize power transfer, the impedance of the generator must equal the complex conjugate of the impedance of the load. If we wanted to minimize reflection, (B) would be the correct answer. Since we instead want to maximize power transfer, (C) is the correct answer. 




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