Recent content by Astrum

Holy crap, I never knew where that synth line came from. TIL

Why not just buy one? You need to buy a power supply and a vacuum pump anyway. You sound like a hazard to yourself, I'm expecting this thread to be closed as well. PF has pretty strict rules when it comes to people (usually people that have no idea what they're doing) playing with...

The sign of any vector is determined by your coordinate system. As long as you stay consistent with your choices, the physics is the same. When working with gravity in classical mechanics, it's standard to call it a force in the negative direction.

The energy in ##\mathbf B## comes from having to do work against the back emf, this is work that needs to be done by the battery. All the work done here is from the battery (or source of the current) rather than the fields themselves. Or at least that's the way I understand it.

I got the answer through conservation of momentum, but I think it was a little bit of an underhanded tactic. But it's all good, thanks for the help, you made me realize I could "cheat". I could have gotten an answer with 1/4 the work involved.

I was thinking that a valuable piece of information might be that the force on A and B is ##k\frac{L}{2}##, but I can't figure out how to use it. These forces from the spring oppose each other. I can't understand why we should need to do anymore work on this, we have the EoM, it seems to me...

Homework Statement Two blocks of equal mass, connected by a masslless spring of constant ##k## are on an air track. her unstreched length of the spring is ##L##. at ##t=0## block ##b## is compressed to distance from block ##a## at ##l/2## (block ##a## is pressed against a wall). Find the motion...

If they're interested in something proof based, I don't think they'd have the mathematical maturity without some experience with higher math.

Typically you study calculus before linear algebra.

You're really asking about the energy stored in E-fields and B-fields. The work to create a charge distribution (static) is ##W_e = \frac{\epsilon _0}{2} \int E^2 d\tau ## and to create a magnetic field you need to go against the back emf, so the work is ##W_m = \frac{1}{2\mu _0}\int B^2...

Do you mean if there's a current through the wire? Or are you talking about a line charge and an external magnetic field?

I'm going to go out on a limb on this one and put out what I got after quickly doing it out on paper. I didn't actually check the integral, so I'm not sure it's 100% right. You have the wrong expression for ##dm##, it should be ##dm=\rho dA = \rho dx dy##. I think this problem is easier...

I've been waiting patiently for years for someone to come out with a way to model the behavior of women. Unfortunately I think it's impossible to apply something logical such as math to the irrational world of human interactions. Dammit! Why can't woman be more like math? Instead of guessing...

I remember a group of students becoming distraught when the evolution unit was taught in HS biology. Even in a fairly educated northern state, we have our fair share of loons. I confronted them on why they thought it was BS, after telling me about "the Bible", they couldn't come up with anything...

I see, that makes a lot of sense. I was under the impression that I didn't need to place limits of integration on there. I'll be sure not to make that mistake again. Thanks for the help, I made that one way harder than it needed to be.