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...
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...
I'm gonna 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 to...
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.
I tried to solve it with respect to ##t##, because that's what we're after, why would be integrate ##r##?
Doing it the way you said, we find ##\omega = r + C##, and that just isn't right.
My attempt is here: Solving this, we end up with $$\omega (t) = ce^{\frac{2v}{r}t}$$, solving for...
Sorry, I always seem to miss terms when typing them =/
I realize that, ##r\omega = v##. What does this do for us though? All it does is eliminate the very thing we're interested in finding.
I don't really understand the direction you're pointing me in.
Because there's no force in the radial...
Yes, this needs to be worked out through using force, rather than conservation of angular momentum.
I'm not sure what you mean by tangential, are you talking about ##\hat{\theta}##?
##a_r =( \ddot{r}-r\omega ^2 )\hat{r}## and ##a_{\theta} = (r\dot{\omega}+\dot{r}\omega )\hat{\theta}##, these...
Homework Statement
A mass ##m## whirls around on a string which passes through a ring. Neglect gravity. Initially the mass is at distance ##r_0## from the center and is revolving at angular velocity ##\omega _0##. The string is pulled with constant velocity ##v## starting at ##t=0## so that...
Welcome to PF
Have you checked out the basics kinematics equations? Link
If you're neglecting air resistance, it doesn't really get much more complicated. Of course you'll need to break the motion into it's components with basic trigonometry.
I realize that you're probably looking for quick...
What? Did you just call that grade F meat, scavenged from the local dump FOOD?!
I've read too many disgusting stories about fast-food chains to ever attempt to eat there again. I'd rather eat boiled potatoes.
It appears that page is wrong. You're right, the correct relation is
##\theta = \tan ^{-1}\left( \frac{\sqrt{x^2 +y^2 }}{z}\right)##.
You could also write it as ##\theta = \cos ^{-1} \left(\frac{z}{\sqrt{x^2 +y^2 +z^2 }}\right)##
Edit: Darn you, jtBell, you beat me by seconds!
I think this thread is starting to derail with all this talk of dog brains.
Assuming you're being 100% truthful, I can fully understand that you want a physics mentor, it really does make things easier, but you still need to put in a lot of work on your own. Work that can't be done but...
I can't tell if you're being sarcastic, or just stupid.
In regards to what? You already have a job set-up, and you say that you don't want to go to grad school. So what do you want us to tell you? Here's a two step solution for you.
1. Find a textbook
2. Study said textbook
What? You need to find the centripetal acceleration, yeah. What you have written ##ma_g-\omega^2 R## is, as you say, for when ##\theta = \pi /2##.
Generalize everything, and you'll have your answer. ##a=a_g + \frac{\mathbf F _r }{m}= a_g - \frac{(\mathbf \omega \times \mathbf r )^2}{r}##...
Oh, I see what you're asking about. What you'd need to find is the centripetal acceleration at the surface of the earth and hence, ##\mathbf v##.
Use ##\mathbf v = \mathbf \omega \times \mathbf r ## to find this, then you can plug it into the equation for radial acceleration.
Your post was difficult to follow, but I understand that you're confused about the "component of gravity" here.
When we say ##g=9.8 \frac{m}{s^2}##, we're talking about the average acceleration due to gravity across the earth. The Earth is an oblate spheroid, which means that the distance...
I've never actually scrolled to the bottom of a thread list for the entire year I've been here. Thanks, that helps a lot.
What do you mean when you mention the availability of the requested books? Amazon has pretty much any book you can think of.