Recent content by pingpong240

Thanks for the help. : ) I computed the answer to be 1.177 m. This checked out. I understand the methodology behind it too, thanks again.

Homework Statement The figure below shows the kinetic energy K of a simple pendulum versus its angle θ from the vertical. The pendulum bob has mass 0.260 kg. What is the length of the pendulum? Homework Equations K=(1/2)mv2 U=mgh x(t)=xm*cos(w*t+phi) *where w is the angular frequency, and...

ah i see now. thank you very much i got it.

yeah that's what i have i think, i double checked the format and what i have is equivalent to (L^2+L^2)^(1/2) or 2(L)^(1/2) that's what you are talking about right? i know the other charge doesn't get involved in the x coordinate...so what might i be doing wrong?

Question Charges at three corners of a square are shown in Figure P26.34, in which n = 9. To answer the following questions, use the letters k, Q, q, L, and m. IMAGE IS LOCATED HERE: http://www.webassign.net/knight/p26-34.gif a) Write the electric field at point P in component form. b)A...

Yeah it's 2.8 m/s^2. Thank you for all of your help, I really appreciate everyone's help! The clarifications helped out greatly Ahmed. I understand how to do these types of problems now. Someday maybe I'll help all of you out ;) ...after I actually learn how to do all of these things that is...

OK here's what I did. For the record I am not at all confident in my answer :(...yet the accelerations seem to agree well... Following the format above, I did this: 28 - 7.84 - 3.92 - T = 2* a_2 7.84 - T = 1* a_1 I messed around with the equations to get T=10.64 N a_2 = 2.8 m/s^2 and a_1 = -2.8...

so when you say a_2, you say that is the acceleration of the block on the floor? I'm working it out right now...

i've got one more stab at this question...so i have to make sure that i get it right... so what is F_a, is that the tension force? I'm not sure i understand. also, when you say the force of 2 on 1, you mean block 2 on 1 right? and the lower block is moving to the left, correct? I'm quite...

Can anyone offer their insight?

OK so that gives me a force of -26.4 N in the x direction. I divided by the mass of the bottom block, 2 kg, to get an acceleration of -13.2 m/s^2. I tried dividing by the total mass to get -8.8 m/s^2, which didn't work either. Is there some way that I set up Newton's second law wrong other...

The lower block in Figure P8.28 is pulled on by a rope with a tension force of F = 28 N. The coefficient of kinetic friction between the lower block and the surface is 0.40. The coefficient of kinetic friction between the lower block and the upper block is also 0.40. What is the acceleration of...

Yeah that was the problem. My mistake, you were doing everything right, I'm sorry for making it confusing in the first post with differing quantities for the height. Thank you very much for your help, I understand how to do these types of problems now.

I'm entering that answer, 1228. I tried entering it more exact, but that doesn't work either. Everything before that seems right doesn't it? Hmm nevermind I think I might have it...I just noticed a very stupid mistake, the vertical distance is 495, not 485.

Ok so I should factor in the motion of the plane. I tried using that as an answer but it isn't coming out right either. I think my work is right up to this point...Well, if anyone else has any suggestions, please feel free to help out. Shouldn't I consider the plane moving right to left, as...