Recent content by shallgren

Yes, so long as the spring constant is N/m and the displacement is m.

If the ring spins fast enough to produce Earth's gravity (9.8 m/s^2) how fast must it be rotating?

Or just divide the values you already have by one since you already squared them. If you don't understand why, frequency is how many oscillations, revolutions, etc. the object makes per second, thus having the unit of hertz. Period is how many seconds it takes the object to make one revolution...

OK, thanks. First off, I assume the -25 is for the height of the cliff which isn't needed. We already proved that the shell would clear the cliff in part a, so start back at the cannon from ground level. Secondly, the shell isn't shot horizontally, as your equation as it. It was shot 43º...

I'm sorry I'm still not getting that and I'm 99% sure my work is correct. Why don't you show me your work.

Y: d_y = \frac{1}{2}at^2 + v_it X: d_x = x_it You're doing the same thing you did to make sure the shell cleared the cliff and find the velocity for that (you didn't know time or velocity) except now you don't know time or distance.

Wait, 2.51 is the time that the shell clears the cliff, we want where it lands. Set the y equation equal to 0 and then plug that time (and the same velocity from part a) into the x equation.

Hmm. d_f = \frac{1}{2}at^2 + v_it ?

Not what I got, what did you get for the time?

The y-axis is height, so you need to know at what point in time the height is 0 (that is, the object is on the ground). Since the equation is a quadratic, you will have two answers. 1 will be 0 (since it starts from ground height) and the other will be the time at which it hits the ground on the...

When the shell lands, the y distance is 0. You can find the time for this and how far out the shell lands using the same motion equations you used for part a.

Remember: friction always opposes motion. The block is going along the floor, but friction is acting upon it trying to push it backwards.

Treat the galaxy as if it was a planet that the sun was orbiting.

Continuing your work: 1631 = \frac{1.16 \times 10^7}{T} T = \frac{1.16 \times 10^7}{1631} \neq 1.89 \times 10^{10} s You have a simple arithmetic error in there somewhere.

Ah, that's it. An oscillation divided by time is a frequency. You want to use the reciprocal of that (period). I did that for the excel chart I made of your data and got both measurements of k almost identical.