Recent content by PennyGirl

Thanks a lot! that worked and it does make sense...I was just plugging numbers into the formulas that I have and not really thinking about the problem lol. Thanks again!

the direction of the current is clockwise...but I'm not sure what you mean by the direction of the potential difference...the voltage is higher on the right side (in the picture) and lower on the left side...

I entered the value twice, once as 3.917 and the second time as 3.92...both times it had the same error message.

no...I double checked that

Homework Statement http://session.masteringphysics.com/problemAsset/1042319/4/YF-25-68.jpg" Above is a picture of the circuit! The question is: "What is the terminal voltage of the 4.00-V battery?" Homework Equations I=V/R V_ab = E - I*r The Attempt at a Solution So first, I...

So right before it is going to tip, torque = zero...maybe?

right...that makes sense. I'm still having a hard time picturing how this fact will help me solve the problem... isn't the distance between G and the contact surface going to be in the same direction as mg, otherwise the semi-cylinder would try and move back to the place where they are in the...

So then I know that the distance between G and the ramp is r*(1-4/(3*pi)), but I don't know how that helps me solve the problem...

[/tex]Homework Statement http://emweb.unl.edu/negahban/em223/sexam3/sexam3.htm" number 3...the one about the semicylinder on a ramp Homework Equations \Sigma F_x = 0 \Sigma F_y = 0 \Sigma M=0 f = \mu * N The Attempt at a Solution I got the first part to be 16.7 degrees. However...

so... integrating once... x'=.0345*x^-2*t then again... x=.0345/2*x^-2*t^2+C\ ?

I didn't think that would work because I thought the exponent on the x term would have to be 1, while in this case its -2...

thanks so much!

so it will look like a spring/slinky? because the part that isn't affected (perpendicular) with make a circle while the other one won't be affected and will therefore go in a straight line

d^2 X * X^2 = .5 dt^2 integrate both sides... dx * x^3/3=(.5*t + C) *dt at t=0, dx/dt=0 and x=10... 0*10^3/3 = .5*0 + C C=0 dx * x^3/3 = .5*t*dt integrate both sides again... x^4 / 12 = .25*t^2 + C same refs... 10^4/12 = .25*0 +C C = 833.3 x^4/12 = .25*t^2 + 833.3 algebra...

Homework Statement how will an electron move if it is launched into an magnetic field at an angle of 45o (w/ respect to the magnetic field) Homework Equations F=q*v X B The Attempt at a Solution I'm having a hard time picturing what is going on. I know that if the electron is...