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Latest 25 Comments  0  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. 
1  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. 
2  Click here to jump to the problem!  physics_guy
20180808 15:16:41  The problem is just asking for the binding energy of one of the n=1 electrons in nickel, since if this electron is removed via bombardment of fast electrons, an n=2 electron will fall into its place, releasing a photon in the xray range. This phenomenon is known as the Auger effect, and it is used in Auger Electron Spectroscopy (AES). The photon released during this transition is known as the Kalpha line in xray spectroscopy. This transition was famously studied by Moseley to provide strong evidence that atomic number is identical to nuclear charge. While it is perfectly reasonable to use Moseley\'s formula for the transition from n=2 to n=1, it is not really correct since that would give you the energy of the released photon. What we really want is the binding energy of one of the two n=1 electrons, 13.6ev * Z^2. While it is also reasonable to account for the screening of nuclear charge by the other n=1 electron, the correct way to account for screening by electrons in the same orbital is to subtract 1/2 from Z, not 1. Fortunately, most of the solutions above all work out to the same order of magnitude. However, the setting of initial n to infinity shows a misunderstanding of the role of the bombarding electrons in the Auger effect, which do not undergo transitions themselves, but merely knock out core electrons with energy less than or equal to the kinetic energy of the bombarding electrons, leaving holes to be filled by higherorbital electrons. The Kseries corresponds to the removal of an n=1 electron, causing a transition from n=2 to n=1 to fill the hole. This is analogous to the Lymanalpha transition in hydrogen. 
6  Click here to jump to the problem!  JasonHupp
20180730 11:09:59  Before the energy conversion is a hectic task to me but when i have seen here this simple tutorial on how to solve i was surprised. As i have been looking for a help and thinking is assignmenthelps.com.au legit good to help me out in this problem solving. Mean while got yours. 
7  Click here to jump to the problem!  JasonHupp
20180730 11:06:29  Seems like every one must remember the formulas to do the sum and get desired answer.\r\nhttp://www.pumpkin.com 
8  Click here to jump to the problem!  jfcdu
20180715 14:43:27  @dberger8 the idea is that the driving frequency is set by the problem. If we want to maximize the amplitude of the current, we need to pick C such that the resonant frequency *is* the driving frequency. Thus we can set and solve for C. 
9  Click here to jump to the problem!  jfcdu
20180715 14:40:57  @dberger8\\\\r\\\\n the idea, is that the driving frequency is set by the problem. If we want to maximize the amplitude of the current, we need to pick C such that the resonant frequency *is* the driving frequency. Thus we can set and solve for C. 
13  Click here to jump to the problem!  shka
20180706 00:08:13  The fraction of rays/particles detected is proportional to the solid angle subtended by the detector, so the ratio of the fractions of particles detected is equal to the ratios of the solid angles: F2 / F1 = S2 / S1. When the source is next to the detector, the solid angle is 2*Pi, since the solid angle of a whole sphere is 4*Pi, and half the rays are detected (F1 = 1/2). The formula for solid angle S is S = A / r^2 , where A is the area of the patch of sphere of radius r subtended by S. To find S2, let r = 1 meter. Since 1 meter is much larger than the radius of the detector face ( d/2 = 4 cm ), the patch of sphere subtended by S2 is almost flat, so we can use A = Pi * (d/2)^2 = 16 Pi cm^2. Thus S2 = 16 * Pi * 10^4. Thus S2 / S1 = 8 * 10^4. Multiplying this by F1 gives F2 = 4 * 10^4, choice C. 
14  Click here to jump to the problem!  shka
20180703 15:22:19  fizix has asserted the converse of the spectral theorem, which is not true in general. In fact, it\'s easy to find a nonsymmetric diagonalizable real matrix. For example,\r\n\r\n[ 5 4 ] \r\n[ 4 5 ] \r\n\r\nis ANTIsymmetric but has eigenvalues 3 and 3. The spectral theorem guarantees the diagonalizability of real symmetric matrices, but it does not force all diagonalizable real matrices to be symmetric. 




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