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 0 Click here to jump to the problem! russian2017-08-12 13:55:10 Charge would not be conserved in e-e- -> gamma. But yes, e+e- -> gamma is common (pair production). There are questions related to the pair production energy on the pGRE Click here to jump to the problem!
 1 Click here to jump to the problem! allenabishek2017-08-03 19:29:24 Why is it that in the solution to find dy, the first term is ignored ?? the kinematic equation is actually y = v0t + 0.5at^2. why is the first term ignored?Click here to jump to the problem!
 2 Click here to jump to the problem! allenabishek2017-07-29 15:59:05 Well you see Allen, L_rot and L_translational are two different things and not the same L_rot = IWo and L_trans = r x p.\r\n\r\nnow for this question we are looking at the total angular momentum which includes both rotational and translational.\r\n \r\nnow by conservation of laws we know \r\n L_total_initial = L_total_final \r\nsince the question asks to find the total final angular momentum it is easier to just find out the total initial angular momentum\r\n\r\nnow L_total_ini = L_rot_ini + L_trans_ini\r\nL_rot = 0.5MR^2W0 ( value of I is found out by looking at the formula sheet)\r\nL_trans = -RMvo ( we get the minus by using the right hand rule)\r\nhere v0 = 0.5RWo so L_trans = -o.5MR^2Wo\r\nthere fore adding them up we get L_tot_ini = 0\r\n\r\nhence the final total angular momentum is zero!!!Click here to jump to the problem!
 3 Click here to jump to the problem! allenabishek2017-07-29 10:40:44 is there a difference between L_trans and L_rot ?? if yes shouldnt both be taken into account ?? why is the standard only taking the translational angular momentum into account ???\r\nClick here to jump to the problem!
 4 Click here to jump to the problem! allenabishek2017-07-29 10:35:26 is there a difference between L_trans and L_rot ?? if yes shouldnt both be taken into account ?? why is the standard only taking the translational angular momentum into account ???\\\\r\\\\nClick here to jump to the problem!
 5 Click here to jump to the problem! juxtaroberto2017-07-23 08:01:08 But the boat only applied a force to the man because the man applied the force to the boat. Either way, energy is not a vector like momentum. The kinetic energies of the boat and the man don\'t cancel, the way their final momenta do. So, since the boat and the man started with zero kinetic energy, and both ended up with some kinetic energy, and the only energy source there is the man, he must have done all the work required to move himself forward and to move the boat back.\r\n\r\nThis also makes sense because as he jumps away, he\'s wasting some of his jump into pushing back against the boat, so the boat starts moving backward before his feet leave it, and that\'s some \"wasted\" energy that he expended but won\'t be propelling him forward. Thus the total work he does must be greater than his own kinetic energy.Click here to jump to the problem!
 6 Click here to jump to the problem! ilikefluids2017-07-09 15:44:05 $m_A a = F - N \\\\ m_B a = N \\\\ \\frac{m_A}{m_B} = \\frac{F}{N} - 1 \\\\ F = (\\frac{m_A}{m_B} + 1)N \\\\ \\\\ \\textrm{All of these quantities are easy to calculate;} \\\\ \\frac{m_A}{m_B}=16/4=4, \\\\\\mu N = m_B g \\Rightarrow N = \\frac{m_B g}{\\mu} = \\frac{(4)(10)}{0.5} = 80 \\\\ \\textrm{Hence, }F = 5(80) = 400, \\textrm{ choice (D)}$Click here to jump to the problem!
 8 Click here to jump to the problem! Yaroslav_Sh2017-06-28 10:27:48 Can someone explain me please why do we treat $\\mu$ and M as independent parameters?\r\n I would think that if we keep L fixed and increase M by factor of two that would essentially make $\\mu$ twice larger as well. Yet if it is indeed the case then consideration of large and small limits of M won\'t change $\\frac{\\mu}{M}$ at all.Click here to jump to the problem!
 9 Click here to jump to the problem! camarasi2017-06-10 03:29:33 Definition of expectation value of an operator O is\r\n\r\n< O > = <$\\psi$|O$\\psi$>\r\n\r\nPlugging in our state $\\psi$...\r\n\r\n<$\\psi$|O$\\psi$> = <$\\frac{1}{\\sqrt{6}} \\psi_{-1}$+$\\frac{1}{\\sqrt{2}}\\psi_{1}$+$\\frac{1}{\\sqrt{3}}\\psi_{2}$|O($\\frac{1}{\\sqrt{6}} \\psi_{-1}$+$\\frac{1}{\\sqrt{2}}\\psi_{1}$+$\\frac{1}{\\sqrt{3}}\\psi_{2}$)>\r\n\r\nNow we \'multiply out\' the inner product. This is similar to squaring a term like (a+b+c). You will get 9 terms - messy!\r\n\r\nBut remember that \r\n\r\n<$\\psi_j$|$\\psi_k$> = $\\delta_{jk}$\r\n\r\nwhere $\\delta_{jk}$ is the Kronecker delta function.\r\n\r\nSo only 3 terms survive.\r\n\r\n<$\\psi$|O$\\psi$> = <$\\frac{1}{\\sqrt{6}}\\psi_{-1}$|O$\\frac{1}{\\sqrt{6}}\\psi_{-1}$> + <$\\frac{1}{\\sqrt{2}}\\psi_1$|O$\\frac{1}{\\sqrt{2}}\\psi_1$> + <$\\frac{1}{\\sqrt{3}}\\psi_2$|O$\\frac{1}{\\sqrt{3}}\\psi_2$>\r\n\r\n=$\\frac{1}{6}$<$\\psi_{-1}$|O$\\psi_{-1}$> + $\\frac{1}{2}$<$\\psi_{1}$|O$\\psi_{1}$> + $\\frac{1}{3}$<$\\psi_{2}$|O$\\psi_{2}$>\r\n\r\nProblem tells us that \r\n\r\n<$\\psi_{-1}$|O$\\psi_{-1}$> = -1\r\n<$\\psi_{1}$|O$\\psi_{1}$> = 1\r\n<$\\psi_{2}$|O$\\psi_{2}$> = 2\r\n\r\nUsing these, \r\n\r\n<$\\psi$|O$\\psi$> = -$\\frac{1}{6}$ + $\\frac{1}{2}$ + $\\frac{2}{3}$ = 1\r\n\r\nAnswer: (C)\r\nClick here to jump to the problem!
 10 Click here to jump to the problem! camarasi2017-06-10 03:07:40 Definition of expectation value of an operator O is \r\n\r\n $\\eq$ <$\\psi$|O$\\psi$>\r\n\r\nClick here to jump to the problem!
 12 Click here to jump to the problem! rweads2017-04-30 01:28:20 since the liquid does not boil, and the heat element has been in the water for a long time, the system must be in thermal equilibrium where the rate of heat added is equal to the rate that heat is dissipated. It is then very reasonable that for small deviations from that equilibrium temperature the rate of heat loss will be the same.Click here to jump to the problem!
 14 Click here to jump to the problem! OptimusPrime2017-04-08 02:27:42 (A) and (B) are maybes so far.\r\n(C) - No, we are dealing with PROTONS instead of the usual electrons. So the mass of the electron shouldn\'t come into play at all.\r\n(D) - Maybe.\r\n(E) - No, since after this collision the photon must lose energy. Recall that higher frequency means higher energy. So if we are losing energy, that means our frequency has decreased, and since frequency and wavelength are inversely related, the wavelength has increased. \r\n\r\nIf you remember the Compton equation then you\'ll find as Yosun wrote that change in wavelength equals the Compton wavelength. But, where can we extract any numbers from that?! We can\'t, so (A) and (B) are out.\r\n\r\nThat leaves (D). On top of that, (D) has the correct dimensions. [h/mc] = (M*(L^2) / T) / (M*L / T) = L. Length as in wavelength.Click here to jump to the problem!
 15 Click here to jump to the problem! OptimusPrime2017-04-08 00:55:51 I\'m confused. Choice (E) resembles the function y = $\\frac{1}{x^2}$ which has positive concavity, not negative like the graph shows.Click here to jump to the problem!
 16 Click here to jump to the problem! OptimusPrime2017-04-08 00:36:49 For (B), plugging in $\\\\\\\\theta$ = 90 degrees gives 90/2 = 45 degrees. Then, cos(45) = $\\\\\\\\frac{\\\\\\\\sqrt{2}}{2}$. It also has the correct dimensions. How else do we eliminate (B)?Click here to jump to the problem!
 17 Click here to jump to the problem! liuyuhang5992017-04-04 14:43:16 Your answer is correct, $\\\\\\\\1s^2$ gives two possible states ( |+-> - |-+>) the singlet state, and ( |+-> +|-+> ) the triplet state. And because electron is fermion, only the antisymmetric singlet state is allowed.Click here to jump to the problem!
 19 Click here to jump to the problem! dipanshugupta2017-03-31 10:12:40 Here \\\'s my method, using \\makebf{Picking Numbers}. Pick a time, say 12, because it is divisible both by 24 and 36. In 12 mins, \\alpha decays \\frac{1}{4} and \\beta decays \\frac{1}{6} . Add them up to get \\frac{5}{12} , which is close to half. Take a time little more than 12. Hence, 14.4 minutes. Pick C. Click here to jump to the problem!
 21 Click here to jump to the problem! dipanshugupta2017-03-30 07:35:09 If you have ever studied the Harmonic Oscillator, you\'ll know that $a$ and $a^+$ are Creation and Annihilation operators, i.e, they increase or decrease one quanta of Energy. \r\nI. The commutator gives one negative quanta of energy. \r\nII. Wrong if III is correct. \r\nIII. Definitely, as per above discussion. \r\n\r\nThus, answer is C. \r\n\r\n(I have taken a QFT class so it was easier for me but it\'s in QM too). Click here to jump to the problem!
 23 Click here to jump to the problem! dipanshugupta2017-03-29 10:45:02 How do you get the $B cos(\\omega t)$ term for $B.dA$? Click here to jump to the problem!
 24 Click here to jump to the problem! dipanshugupta2017-03-29 10:02:01 Simpler answer. Beta Decay is a Weak Interaction. Weak Interactions are Parity Violating. Parity is a Reflection invariance as $P^\' (\\vec{x},t) \\Rightarrow P ( - \\vec{x},t)$. Hence, problem solved. Click here to jump to the problem!

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