Recent content by posto002

Sorry that my reply is taking awhile. I just got another assignment that I wish to get done before as soon as possible. I'll try to get an example as quick as I can after I've completed the assignment. I am interested in learning any new tips from you, and I thank you for your help.

Thank you, haruspex and DEvens for your help. I appreciate it. Do any of you have any tips on completing physics exams faster? I have great grades on my assignments, but that's mainly because we have a whole week to complete them. I take about 3 days to complete around 15-20 problems, but when...

This is the force diagram for the middle object. In order for it to be in equilibrium, it should be set in the middle, as stated that it was, between the two pulleys. The two masses at the ends should be the same height from the ground. Because both end masses are the same, that would mean the...

I got the equation --> d=mL/[2*(√(2M)^2-m^2)] where M is one of the masses at either end (multiplied by 2 because... there's two of them.) and m is the mass in the middle of the system. Something that a TA told me was that L, the horizontal distance of the system changes. I'm confused by this...

Yep, you're right. Thank you.

Okay, thank you all for your help. I reread the problem, and the entire process I went through was correct up until using the conservation of momentum when I used: "0.59v=0.05(7.65) 0.59v=0.3825 v=0.65m/s " Like hutchphd stated, the mass of the cart was 0.54 instead of 0.59 [taken from...

I understand where switching the sign for g would be appropriate, but even with g being negative I still would get the same answer. Wouldn't I? I'll ask my professor about this and see whether it's the right answer or not. Thanks for your help (and for Orodruin's help.)

I started off with vf^2=vi^2+2ad and plugged in the final velocity, acceleration (9.8) and distance which is 2.1 to get: 0=vi^2+2(9.8)(2.1) I solved this to get: vi=6.42 I then plugged this into find the x-component of vi (I labeled this as simply x)...

Oh! Right, thank you! That completely went over my head!

I know that in parallel springs, x (the displacement of the spring) is the same for both springs. However, the forces resulting for each string are different. For springs in a series, x may be different, but the force is the same on each string. I got the answer b, seeing how the weight would...

I understand where I went wrong. I was currently waiting for my Professor's Office hours to open up so I could talk to him about it. While I was waiting, I thought I'd try and solve it myself the night before and leading to this morning. Thank you once again for your help. With my Professor I...

I understand where I went wrong. I was currently waiting for my Professor's Office hours to open up so I could talk to him about it. While I was waiting, I thought I'd try and solve it myself the night before and leading to this morning. Thank you once again for your help. With my Professor I...

Oh wait! I got it! Thank you for your time and help! I greatly appreciate it!

Sorry, my mistake: KE of Pierre=0.29411m1v1^2 is what I got once I multiplied everything through including squaring the velocity term. (KE of Pierre= (1/2)(1.70m1)([v1/-1.70]^2) =(1.70m1/2)(v1^2/2.89) =(0.85m1)(v1^2/2.89)...

KE of Pierre= 1/2(0.70m1+m1)(v1/-1.70) From what we got from above (v1=-1.70v2), was it right to be able to solve for v2 to get v1/-1.70 by dividing both sides by -1.70? Then to use that and plug in v2 into the KE equation?