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physics_guy
2018-08-08 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.
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physics_guy
2018-08-08 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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physics_guy
2018-08-08 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 x-ray 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 K-alpha line in x-ray 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 higher-orbital electrons. The K-series 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 Lyman-alpha transition in hydrogen.
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Benjamin
2018-08-08 07:15:11
Hi there tim\r\n\r\n[url=http://www.pumpkin.com]harry[/url] \r\n\r\nhttp://www.essaywritinglab.co.uk/
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Benjamin
2018-08-08 07:12:18
brillassignment review
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JM777
2018-08-06 07:07:27
Nice job!
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JasonHupp
2018-07-30 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.
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JasonHupp
2018-07-30 11:06:29
Seems like every one must remember the formulas to do the sum and get desired answer.\r\nhttp://www.pumpkin.com
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jfcdu
2018-07-15 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 \\omega_d = \\omega_0 = \\frac{1}{\\sqrt{LC}} and solve for C.
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jfcdu
2018-07-15 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 \\\\\\\\omega_d = \\\\\\\\omega_0 = \\\\\\\\frac{1}{\\\\\\\\sqrt{LC}} and solve for C.
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shka
2018-07-09 18:02:05
nice typo btw
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shka
2018-07-09 17:59:48
*reads Yohni Kahn once* *becomes Einstein*
12 Click here to jump to the problem!
shka
2018-07-08 19:07:34
carle257 is using symmetry instead of explicitly writing out the equations for p_y and p_x conservation. His answer is fine and consistent with Yosun\'s, and the fact that Quark is confused by the notation v_x = v^\\prime cos\\theta makes me think he doesn\'t understand how right triangles work. You can ALWAYS solve for the components of the velocity, by definition of linear momenta p_x = mv_x and p_y = mv_y, and since p_x and p_y are conserved separately. Whether or not you choose to fuss with the angles is completely up to you. Since the angles and masses are identical, if you paid attention in PHYS 100 you can literally just read off the answer. Thanks Quark, for obfuscating a perfectly good solution with your utter misunderstanding of first principles. Maybe you should crack open Giancoli a few more times before applying to CERN, since I\'m pretty sure they like to conserve momentum linear momentum there.
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shka
2018-07-06 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
2018-07-03 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 non-symmetric diagonalizable real matrix. For example,\r\n\r\n[ 5 -4 ] \r\n[ 4 -5 ] \r\n\r\nis ANTI-symmetric 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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lukenlow
2018-06-22 12:31:32
I\'m glad that I found this useful information in such a simple explanation, this knowledge will be very useful to me 192.168.l.l
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lukenlow
2018-06-22 12:29:34
An excellent and very simple idea, like all the ingenious in this world, thank you very much for your help 192.168.1.1
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lukenlow
2018-06-22 12:27:08
Excellent info, it is very useful to learn such things in trivial discussions 192.168.O.1
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lukenlow
2018-06-22 12:22:34
Incredibly useful post! I read a lot of topics on this site, but this only helped me 192.168.l.254
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jedymark
2018-06-07 07:20:21
they are not necessarily each others anti particle: https://happywheelsnew.com
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aprilrussell
2018-05-30 03:51:45
run 3
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aprilrussell
2018-05-30 03:50:43
This seems the best solution. Yosun\'s is rather ambiguous. run 3
22 Click here to jump to the problem!
aprilrussell
2018-05-30 03:48:38
At the very least we can eliminate choice (C) by units. \r\nhttp://fivenightsat-freddys.com
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aprilrussell
2018-05-30 03:45:18
A quick way I did this is similar to chemicalsoul\'s solution. https://run2.online/run-3 Thanks
24 Click here to jump to the problem!
aprilrussell
2018-05-30 03:43:14
Relativity is so glamorous! I love it. \r\nhttps://run2.online/run-3
 
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