Recent content by AbigailG

Oops, just a typo. Sorry. Thank you so much! I appreciate your help.

Okay, so I have: Kf - Ugf = Ugi Where: Ugi = -Gm1m2/r(between) and Ugf = -Gm1m2/2r(Jupiter) 1/2(m1+m2)-Ugf = Ugi v = sqrt((2(ugi+ugf)/(m1+m2)) = 3 x 10^4 m/s This is the correct answer. But why did I not have to divide the velocity amongst the two masses? Is it because I divided...

Homework Statement Two Jupiter sized planets are released from rest 1.0 X 10^11 m apart. What are their speeds as they crash together? I think my problem lies in figuring out which radius to use. In an equation like this are the radii of the planets included in the distance between them...

Generically, R = l which is the distance from the pivot point to the center of mass. I thought the moment of inertia would be that of a sphere about an axis through the center. ..am I just not visualizing this correctly?

Homework Statement [/B] A solid sphere of mass M and radius R is suspended from a thin rod. The sphere can swing back and forth at the bottom of the rod. Find an expression for the frequency of small angle oscillations.Homework Equations f = 1/2(pi) sqrt(MgR/I) I for a solid sphere 2/5MR^2The...

Okay! I figured it out! The equation: A(t) = A(0)e^-λt was close...but not quite there, I looked further into my notes and found: A(t) = A(0)e^-t/2tau So... 0.0027 = 0.0062 e^-83.895/2(tau) .43548 = e^-83.895/2(tau) ln(.43548)=-(83.895/2tau) tau = -83.895/2ln(.43548) = 50.46 s This is the...

Yes, that is taken into account with the equation A(t) = A(0)e^-λt As I understand it, the time used in the Amplitude function is the time taken without damping, the equation yields the amplitude with damping, and with the data we have we can solve for λ. But I could be wrong. How would you...

Okay I will. I have class tomorrow, I'll post an update tomorrow night. Thank you for your help!

Ahh, I see. That makes sense. So now all I would have to do is plug the new time (8.5 s) into my my final equation? 8.5 seconds seems much to quick to complete 60 oscillations of such a large pendulum...or am I understand the 8.5 seconds incorrectly? I'm sorry, this is really giving me trouble.

I wish I could tell you, but having a correct answer to work with would make things too easy, haha. I could click "give up" to find the answer but I don't want to lose the credit. It seemed perfectly reasonable to me, it has to just be a simple mistake somewhere. Give it a shot if you're...

The units are seconds...i'm still not sure where I went wrong. Without damping time < time with damping...Which seems to make sense because the spring is moving slower? Or would it take a longer amount of time for it to complete those oscillations...now I'm confused, I'm sorry. I know it's...

Homework Statement [/B] A spring with spring constant 10.5 N/m hangs from the ceiling. A 520 g ball is attached to the spring and allowed to come to rest. It is then pulled down 6.20 cm and released. What is the time constant if the ball's amplitude has decreased to 2.70 cm after 60.0...

Hello, my name is Abigail. I have found PF to be such an amazing resource and have learned so much from everyone here. I hope I can use the knowledge I've gain to help where I can. I also hope to continue to expand my knowledge through collaboration on specific questions. Thank you all. Abigail