Recent content by tal444

Wow... that was silly of me. That makes a lot more sense. So I would integrate 2\pi\int(x-1)(x2)dx from 1 to 2, which gives me an answer of 17/2. Thanks for the help.

Homework Statement Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the given axis: y = x2, y = 0, x = 1, x = 2, about x = 1 Homework Equations V = 2\pi\intxf(x) The Attempt at a Solution I am assuming based off the...

Nicely done, and glad to be of help.

a) KEoriginal must equal the workfriction + KEafter b) You must find the a that the block causes on the bullet first. Use v22 = v12 + 2ad to solve for a.

This involves a lot of formula manipulation... We know that W = Fd and F = ma. ∴ W = mad Solving for a from v22 = v12 + 2ad gives us a = (v22 - v12)/2d plugging that back into W = mad gives us W = m[(v22 - v12)/2d]d = \frac{1}{2}m(v22 - v12) = \frac{1}{2}mv22 - \frac{1}{2}mv12...

All you need to do is plug in your known values into s=ut+\frac{1}{2}gt2. Then you move everything to one side and solve the quadratic equation. What are you having issues with?

Friction has nothing to do with what the last question is asking.

As far as I can see, it looks right to me.

Sorry, I don't have AIM.

a) I believe you would calculate the number of combinations with 7 as the leading number first, then add the number of combinations with the leading number as 6 or lower. b) You have the last number wrong. There are only 2 possibilities for the last number, not 3. Change that and you should get...

Probably, because the problem is unsolvable the way it is right now.

Possibly you made an error in your calculations. As jedishrfu said, if you plug in all the known values in, you should get 1.5kg no problem.

That's what I'm getting at.

Find the horizontal component of the force that the cable is pulling on the beam. The force that the pin provides will be equal to that.

For the second question, the cable is acting at an angle right? This means that the cable has a horizontal force. Since the beam is not moving, therefore the pin must be supplying a horizontal force that either pushes out against the beam or pulls it toward the wall to stop it from moving sideways.