Recent content by AmagicalFishy

I know that scales measure weight (a force) via displacement of a spring. I also know the difference between a force and a mass. I know that, on the moon, the numbers on a scale might be different (depending on the answer to this question). I've read through the few threads on this matter here...

I'm aware that it helps keep warm things warm—but does it similarly help keep cool things cool?

If I could delete this thread, I would. Instead, I will answer it: The \sqrt{2} and "rms" are very related. The RMS value for some periodic current is the DC current that delivers the same average power. For sinusoidal waves of the form a \sin{(2 \pi f t)}, the corresponding RMS value is...

This is just a quick question: A problem I'm working on says "a sinusoidal current of .5 amps (rms) and 5 kHz." Later, in the problem solution, I(t) is written as .5 \sqrt{2} \cos{(10^4 \pi t)}. I think I'm simply misunderstanding something about the construction of a current function when...

It took me a while, man, but thank you very much. I was getting my... well, my everything pretty badly mixed up. I was mixing up the \hat{x} with the x component of the E-field and the variable x when converting from Cartesian Coordinates to Cylindrical. I don't know what in God's name went...

My coordinate transformations come from the fact that the E-field only moves in the x-direction; thus, when we integrate it (to get a voltage), we only get something dependent on the x-coordinate. This means that the potential function is only a function of x. It has no y or z dependence. So...

Hey, folks. I've typed out the question here. I hope you don't mind my just linking this (if that's against the rules or something, I'll type it out). I'm still pretty confused about what to do. I assume, as usual, it's probably something... well... stupid—but I can't figure out what. Thanks...

Hey, everyone. I am trying to find the power series of secant from the known power series of cosin, but my answer does not match up with Wolfram and Wikipedia. I know: cos(\theta) = 1 - \frac{1}{2}x^2 + \frac{1}{4!}x^4 + ... So, using the first two terms (assuming a small angle)...

Hey, everyone. I am going to post a question—but it's not the question I need help with. It's something deeper (and way more troubling). Consider a particle of mass m subject to an isotropic two-dimensional harmonic central force F= −k\vec{r}, where k is a positive constant. At t=0, we...

What do you mean a "side to side" force? If the axle were actually moving side to side (in which case, wouldn't the axis not coincide with the axle anyway?) or if the axle itself were tilted w/ respect to gravity so there was some normal force or something?

Ah! Ok. I was mixing up "axle" and "axis" (I know, a stupid mistake; an axis can't have... well... physical properties like those that create friction). I think I understand now. So, say the axle weren't frictionless. If I were to grab one end and twist, the axle would exert a torque on the...

So there will be torque when the putty hits the turntable? If that is the case, then how do we still know angular momentum is conserved? Or is it that—there is torque only when the putty hits the table? The moment the putty actually hits, the torque is gone and we can assume that, from that...

I'm currently reading John Taylor's Classical Mechanics. Near the end of Chapter 3, he states "Because the table is mounted on a frictionless axle, there is no torque in the z direction. Therefore, the z component of the external torque on the system is zero and the [system's angular momentum in...

Whoa. There's that \frac{\sqrt{2}}{2} Geeze. That... that makes the problem so much easier. And one could use two different coordinate systems for the different parts because the value of the integral should come out to be the same anyway. Also, approaching the problem as I was (w/o...

How did you get the \alpha = \frac{\sqrt{2}}{4} and v = \alpha u? I keep ending up with v = -u + 2\sqrt{2}a—which makes my calculation of an infinitesimal length of wire really... uh... ugly.