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Bill N

 Discussion on answer by Superfast Jel…

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Bill N
Aug 28, 2020 10:50
@user21820 No, condensed matter isn't my specialty. I just know that crystalline (or amorphous) structure has an effect. Different unit cells will tie up different numbers of valence shell electrons changing the valence electron density.
Bill N
Aug 28, 2020 10:50
Glad someone finally distinguished between single atoms and collections (aka, solids, but also liquids). ALSO important is the specific crystalline arrangement of the atoms. The conductivity of graphite is very different from the conductivity of diamond, although both are carbon crystal structures.
 

 Discussion on answer by The Photon: I…

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Bill N
Aug 27, 2020 21:48
Also note that the English/engineering/mechanics torque unit is the foot-pound (ft-lb), not the pound-foot. I would venture to say no one ever in commercial work quotes a torque value in pound-feet. Why wouldn't ivory-tower physicists say meter-newtons? :) Nevertheless, even engineering texts use newton-meters, N$\cdot$m, when using SI.
 

 Discussion on answer by Farcher: Does…

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Bill N
Dec 2, 2019 22:44
@Thomas Because forces are always interactions of 2 things, and they are symmetric, as far as we can tell, based on symmetry and isotropy of the universe. The fields which define the interactions do not seem to have a preferred direction in the universe, so every direction is equal. The strength of the interaction "forward" is the same as it is "backward." Symmetry makes the force directions directly opposite.
Bill N
Dec 2, 2019 22:44
@AgniusVasiliauskas Consider a system with no gravitational field consisting of a spaceship and an astronaut. If the astronaut pushes against the seat in the spaceship, there arises a pair of interaction forces, which we would designate as normal forces. They arise because of the electromagnetic forces holding the molecular structures of the astronaut/spacesuit and the seat together, preventing them from passing through each other. The normal force has NOTHING to do with a gravitational field and everything to do with the electromagnetic forces.
Bill N
Dec 2, 2019 22:44
@AgniusVasiliauskas No, I don't miss the point. Farcher is correct that the 2 forces of a true N3L pair cannot target the same object. The normal force acting on the object and the force from the gravitational field interaction ARE acting on the same object and cannot be the N3L pair. They do NOT target different bodies if you do the analysis correctly.
Bill N
Dec 2, 2019 22:44
@AgniusVasiliauskas Normal forces arise because of electromagnetic (molecular) interactions between objects which prevent one structure from moving through another. If the gravitational force (weight) on an object is greater than the normal force, the object will accelerate downward. Consider a rock dropped into water. The water exerts a normal force on all surfaces of the rock with a net resultant force upward, which we call buoyant force. But the rock continues to accelerate downward because mg>F$_b$.
 

 Discussion between Jacob Manaker and …

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Bill N
May 25, 2019 21:07
Glad I could help! And I recommend that you state that the electron energy is the ratio of electron mass energy to photopeak.
Bill N
May 25, 2019 21:05
Aha! So you must use the ratio of the photopeak energy the electron mass energy as the multiplier of the (1-cos$\theta$), 661/511, or because you are dividing, 0.77. If you have a different energy primary photon, you have to get a new divisor. I think you had a misunderstanding of what the Compton edge energy came from.
Bill N
May 25, 2019 20:59
What are you using for the electron energy in your formula?
Bill N
May 25, 2019 20:57
Okay.
Bill N
May 25, 2019 20:57
Okay. Do you have more questions?
Bill N
May 25, 2019 20:55
No. The Compton edge corresponds to the energy of the maximum electron energy gained in an event. That means the secondary photon is scattered at 180 degrees. The secondary photon then has a low energy and may or may not escape. The region between the Compton edge and the backscatter peak is the region where secondary photons scattered at medium angles are displayed. And your algebra is wrong! You cannot throw a 1 in there. Do the math with a real energy and see if you get corresponding results. Plus, I did the calculation for Cs-137 and it matches the spectrum.
Bill N
May 25, 2019 20:55
The backscatter secondary photon for the Cs-137 peak will be 184 keV, and the Compton edge will be about 477 keV. Those would be channels 59 and 167. And the full photopeak matches the proper channel, too, based on the energy calibration you stated.
Bill N
May 25, 2019 20:55
Yes, that's what I said. See the edit about your calculations.
 

 Discussion on answer by spacetimeengi…

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Bill N
May 8, 2019 18:59
The relativistic relation of mass and energy is $mc^2=\sqrt{E^2-p^2c^2}$, which is frame-invariant. For the photon, $E=pc$. For particles with (rest) mass, $E=K+mc^2$, where $K$ is the kinetic energy of the particle and $p=\gamma m \beta c$, where $\gamma$ and $\beta$ are the standard relativity factors we all know and love.
 

 Discussion on question by Matt Zusy: …

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Bill N
Aug 26, 2018 09:41
1) This is too long to critique in detail. 2) My cursory reading indicates you havent brought up the important point of the differing friction behavior between steering and non-steering tires (there is a BIG difference between front, rear, and 4-wheel steering behavior). 3) The tire friction changes the rotation axis away from the center of mass. 4) You make several assertions without justification which seem to be begging the question ("I say this is the behavior, so these other guys are obviously wrong."). 5) How are you defining "yaw velocity?"
 

 Discussion on answer by probably_some…

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Bill N
Aug 14, 2018 08:51
" it gained gravitational potential energy from the inflation of the balloon. So ultimately, the energy comes from the inflation of the balloon." If the balloon was inflated with cold air or carbon dioxide or SF$_6$, the balloon would not rise, even though "the atmosphere got pushed out of the way". There is MUCH more to the story than merely the inflation.
 

 Discussion between Rory Alsop and Bill N

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Bill N
Jul 11, 2018 16:24
@RoryAlsop My question is not about a fault in the hardware, but about how to change a setting on properly working hardware. In that sense, I believe it is similar to questions like the one I cited or another which asked about where to insert a cross-over unit in the signal chain. If asking about specific brand equipment is off-topic, I can understand that, but there is nothing in the on-top help section about that limitation. It seems that general equipment questions are on-topic.
Bill N
Jul 11, 2018 07:31
@RoryAlsop Okay, I'm trying to understand the site definition of "sound design" if it's not to include operation and maintenance of live audio production (aka PA). In that light, what makes this question on-topic while mine is off-topic: sound.stackexchange.com/questions/38231/…
Bill N
Jul 11, 2018 07:31
@RoryAlsop I called both Crown and my local Crown authorized repair center. Neither of them knew how to do it. Crown said, "Even if I did know, I'm not authorized to give you details since it would probably involved getting inside the amp." The service center said, "We've never done that and we don't have any service bulletins or manual that explain how to do it." I was hoping there was someone here with experience. So, you see, I've done my homework.
Bill N
Jul 11, 2018 07:31
@RoryAlsop With all due respect, there is nothing in the help center which addresses asking about specific audio reinforcement equipment. This amplifier is NOT a home theater nor an audiophile system amp. It is part of the sound system (aka, sound design) at a moderate sized church venue. The sound design of that location depends on the proper functioning of this amplifier. It questions about specific amplifiers or troubleshooting problems of existing sound designs is off-topic, that should be detailed in the help center. This is about "live production audio," & is clearly written and specific
 

 Discussion on answer by WetSavannaAni…

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Bill N
Feb 26, 2018 15:23
Could one say that the shockwave of nuclear explosions and chemical explosions arise from the same physics principles? It's the energy release mechanism and hence the energy density that is the big difference in blast radius for a given volume of a bomb/shell.
 

 Discussion on question by alexanderd5…

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Bill N
Sep 27, 2017 00:54
I'm deleting my first comment.
Bill N
Sep 27, 2017 00:54
Oops! I interpreted the $V_g$ as the voltage of the load resistor, $V$. So, what you want to investigate is the load voltage, $V$. That's why I made my comment...You wrote an equation for the wrong thing. Try $V=IR_L$
 

 Discussion on question by dandb: The …

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Bill N
May 5, 2016 13:38
$\pi^2\simeq g$ in SI units. Coincidence?
 

 The h Bar

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Bill N
Apr 5, 2016 16:40
@ACuriousMind That happens, and we probably get mixed reactions when we brush them off.
Bill N
Apr 5, 2016 16:39
Yeah, I experience that, too, but the ones who visit once come back regularly.
Bill N
Apr 5, 2016 16:37
When I see things like that, I'm wondering why they aren't going to see their teacher to get help.
Bill N
Apr 5, 2016 16:36
Take a look at John Denker's work at av8n.com/physics/spacetime-welcome.htm
Bill N
Apr 5, 2016 16:35
I think it would be admirable if we DID try to clean up some of the confusion by having a description of why the rotation/matrix approach is better.
 

 Discussion on question by Juho: Wavef…

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Bill N
Sep 15, 2015 06:47
The wavefunction of Schroedinger has nothing to do with the nature of a wave in the physical sense. There is nothing "waving" in the SWE solutions; they are solutions to a Hamiltonian which in some cases have the same functions as classical waves. Classical wave functions are NOT probability amplitudes in the QM sense, either. For example, the QM solution to the simple harmonic oscillator is completely different from the classical simple harmonic oscillator.