Recent content by jromega3

Homework Statement the ball has a mass M and is fixed to a rod having a negligible mass and length L. If it's released from rest when theta=0, determine the angle theta at which the compression force in the rod becomes zero. Homework Equations A(tangent) = dv/dt = vdv/ds A (normal) =...

Homework Statement The two problems below are related to a cart of mass M = 500 kg going around a circular loop-the-loop of radius R = 9 m, as shown in the figures. Assume that friction can be ignored. Also assume that, in order for the cart to negotiate the loop safely, the normal force...

Lovely. Wow was I stupid. Guess I got so flustered I didn't even understand I was looking for a current below the heating level instead of just solving for B with the info I had like I was doing. Thanks a lot. Truly appreciated.

It's an equation given in my book for the magnetic field inside a solenoid, where B is some special magnetic Field the author uses to make units N/C...and is basically the regular magnetic field times the speed of light. E0 is epsilon zero too if that wasn't clear.

Those numbers came from B=NI/(cLE0). Anyway, using those number of turns I'd get 196 and 488 for 18 and 26, respectively. I guess I just don't see WHAT I'm supposed to be solving for here? Using those values for N, I get a B of 1.1MN/C for 18 and 0.18MN/C for 26. Neither of which are, or even...

ah thanks. Ok so I got N of #18 to be 133, and N of #26 to be 797. So that's the number of turns. I'm now stuck with how to advance from here. I assume it has something to do with diameters and am probably missing the obvious here but not seeing how I can relate # of turns to satisfying the...

Homework Statement You want to construct a solenoid 20cm long that has an interior magnetic field Strength B (Bc) of about 1.5MN/C. The coil has to be wound as a single layer of wire around a form whose diameter is 3.0 cm. You have two spools of wire handy. #18 wire has a diameter of about...

Homework Statement A loop carries current I = 2.5 A in the x-y plane as shown in the figure above. The loop is made in the shape of a circular arc of radius R = 4 cm from qo = 60 ° to q = 360 ° . The loop is completed by horizontal and vertical sections as shown. What is BA, the magnitude...

Ok, so I get E=rho*J.....or since I don't know J J is also Vd*nq...so I get E=rho*Vd*nq... so my equation becomes T=(Vd)(Me)/(-e)(rho)(Vd)(Nq) stuff cancels and I get Me/(-e)(rho)(Nq) and I get a really wrong number. hmm Edit. Nevermind...my bad. We Have I and A can be easily solved giving J...

Homework Statement compute the average time between collisions T for conduction electrons in copper. Homework Equations Vd=-e*E*T/(Me) Where e is electron's charge E is electric Field T is what I'm solving for And Me is mass of the electron Vd=I/PA The Attempt at a Solution...

Homework Statement A conducting sphere with radius a = 0.31 m has a net charge Qa = +4 µC. A thin, non-conducting spherical shell of radius b = 0.86 m surrounds the sphere and is concentric with it. This shell has a net charge Qb = +1.5 µC distributed uniformly over its surface. The electric...

Symmetry Arguments--a finite cylindrical can Homework Statement Consider a finite cylindrical can shape that has charge uniformly distributed on its surface. Symmetry does allow us to say some things about the electric field of this distribution A) at points along the can's central axis B) At...

Yep. That would've been my problem. Without that he "lands" a few meters off the ground. Thanks for the help everyone

Yeah, I definitely erred somewhere along the road. So the initial energy was mgh + 0.5Ks(delta x)^2 90*9.8*45 + 0.5*98*15^2 = 50,715 J now to solve for x (lowest point, ie where all energy is potential)...50,715=90*9.8*X + 0.5*98*(x-15)^2 s0 50,715=882X + 49(-30x+225+X^2) 0=49X^2 +...

Ahh. The initial total Energy would be just mgh + PEs, right? Now what? Find what mgh + PEs works on the other side to equal the same total energy, and this would be the lowest point (where V=0). If so, that makes sense. Thanks...hopefully I'm on the right track ;) Well, doing this, I got X to...