Recent content by orb

The key is that the horizontal component of motion is at constant velocity, and that the vertical component is at constant acceleration (due to gravity). So separate your two components. Using x = v_{x} t solve for t and then use that in the equation for uniform acceleration.

\pi\int_0^2 [ (8)^2 - (8 - 2x^2)^2 ] \,dx I think that works, because if you use the washer method, the outer radius is just the part that has a y-length of 8, and the inner radius is the part above the function and under y=8, so using pi (R^2 - r^2) integrated, that's what I get. Hope that...

Yeah, just factor out the 1/sqrt[3] and treat the sqrt[x] on its own.

Thanks :) I've worked it out.

So by my understanding, that's what you mean, right? If so, I understand what you're saying up to actually using the RHR#2 as your last instruction. I can't see how that makes sense. How do you know where the red arrow points? The thumb is the force, right? How can I get the magnetic field to...

Homework Statement There seem to be three right hand rules for dealing with magnetic field, current, and force, and I don't quite understand how to use them. Namely, I'm having trouble understanding why two parallel wires with current flowing in each of them attract if the currents are in...