Recent content by pitaaa

Hmm I'll try it :/ Unless my diagram is totally wrong :P

Okay, I redrew the vector diagram. But what do I do once I have my components? :/ I don't know what to do with them.

Homework Statement Two rolling gold balls of the same mass collide. The velocity of one ball is initially 2.70 m/s [E]. After the collision, the velocities of the balls are 2.49 m/s [62.8 degrees N of W] and 2.37 m/s [69.2 degrees S of E]. What are the magnitude and direction of the unknown...

Collisions and Determining the Original Speed Homework Statement Two spacecraft s from different nations have linked in space and are coasting with their engines off, heading directly toward Mars. The spacecraft s are thrust apart by the use of large springs. Spacecraft 1, of mass 1.9 x...

Andddddd I finalllllllllllly got it haha - acceleration is 1.4 m/s2. The thing with a question like this, is that I would have never thought to do what was supposed to be done - which troubles me big time! :/ I get to a certain spot in the procedure and then I just get stuck because I don't know...

Er, and how do you go about doing that... heh. I just finished grade eleven, and I'm taking grade twelve physics in summer school, so as of yet, I've never been faced with a three unknown situation.

Acceleration is in the direction FT2 is pulling, so, m2a = FT2 - FT1? So for m3a, couldn't it = FT2 - Fg3? That's what I had initially, but you said I had a sign error?

Okay, in that case, m1a = FT1 - FG1 m3a = Fg3 - FT2 So, would m2a = FT1 + FT2?

Okkkay, so: m1: ma = FT1 - FG1 m2: mya = FT1 - FT2 mxa = FN - m2g mxa = FN - FG2 m3: ma = FT2 - FG3

m2 (38 kg) is on a table, with m1 suspended from the left side from a pully, and m3 hanging from the right side from a pully ... the pully's are attached to the corners of the table. Here's a link to the photobucket file: http://i347.photobucket.com/albums/p475/piiita/PHYS1.jpg The equations I...

Homework Statement Three blocks, of masses m1 =26 kg, m2 = 38 kg, and m3 = 41 kg, are connected by two strings over two pulleys, as in the given attachment. Friction is negligible. Determine (a) the magnitude of the acceleration of the blocks and (b) the magnitude of the tension in each of the...

Thanks so much! I know where my mistake was coming from (I wasn't adding the first 214 m long shot into the x-component, I was simply adding it to the end, and that's why I wasn't getting the 7.29 degrees as my direction - which, now I get problemlessly). That really clarified everything for me...

Woot! I managed to solve it using sine/cosine laws. The answer was 11 km [18 degrees N of W] Thanks for keeping me on the right track, alphysicist :)

Okaaay, that definitely explains it. It's 71 degrees :) And now I get 7.0 km. Thanks so much! Now I must get cracking on Part B.

Well I used the cosine law, so: c^{2} = a^{2} + b^{2} - 2abcosC c^{2} = 5.1^{2} + 6.8^{2} - 2(5.1)(6.8)cos123 c= 10.48 I attached the diagram I made... I'm guessing something's wrong with that, I just don't know how else to draw it.