Recent content by :::JMANN:::

I realize that there is no tension = 146N in this problem. I took the values from this problem and plugged them into another problem. T from this problem would = 2T from the other problem, that's how I got 292N in the first place. I understand that I didn't really use the correct formula, but...

I found the formula for the rocket and fuel. Vf-Vi= Vrel(ln(Fi/Ff) Where F is the fuel initial/final Vf-0=1500(ln(1/.2) Vf=2414m/s I don't know if I followed your method correctly or not, but this is not the answer that I got.

for this particular problem 2T is actually not 292N T=292N, however for the problem that I used, the equation was for an accerating block on a vertical plane, pulling a block on a horizontal plane, where T=146N. Then if the it were a hanging pulleye system as describe, T= 2T in respect to the...

This answer was correct, however I left of then - for C, I hope I remebered it on the test, otherwise I might lose some points. :eek:

Well I turned in the exam. I got all of them right except the rocket fuel question. We did not go over it, so I don't know what the answer was, but I am going to check, a class mate found a problem in the book exactly like it, so I'm a plug a chug and see what comes up. He said he got like...

Well I turned in the test, and so far I got all of them right, with the exception of #2, the rocket fuel question, which we didn't go over, so I don't know the answer, but I don't think that I got it right. The Tnet=292N like we got for the final answer, but for the way I did it, Tnet for the...

Yes, it is open book open notes. We are allowed to seek help too. He wants us to understand the concepts and apply them, even if we are aided, rather than stay confused and not know what we are doing.

calculator was in radians for this one too. I got 23.78*. I appreciate all the help. I have to turn it in now, but I know I have done better than I would of without the help. Thanks a lot.

The book says on a pulley system that the tension is 2T. Follow the equation for acceleration on the blocks due to mass, the original T=145.7N then 2T=291.4N roughly 292N like your equation resulted. As far as significant figures, I don't think that the professor will be that picky, as long as...

Calculator was in radians. C=3057.94N T=2209.01N Still seems awefully out there.

okay, for #8, there's no change in kenetic energy, so it would be 0 work, so there's no watts if there's no work. #4 I got 291.667N # I got 16.37*

The ball. Okay, I get the forces acting on the ball. Given that ma=mgtan@ a/g=tan@ tan@= 4.32/9.8= 0.441 tan-1(0.441)= 0.415* angle Doesn't seem like much of an angle, but I guess the radius would have to be smaller of the speed quicker to force the ball to move more right?

For the fuel question, the ships velocity equals 6000m/s, but I am still having trouble gripping the concept of it. If it's 4 times the fuels velocity. The book says Mvi=Pi Pf=(0.80M)v(fuel) +(0.20M)v(ship) Vship= Vrel +Vfuel Vfuel= Vship - Vrel Mvi= 0.80M(Vship - Vrel) + 0.20MVship I...

Yeah it did, thanks. It is weird explaining the diagram without being able to show it to you, but I guess you had it right. Okay, so here I go. I came up with the values c=-2670.92N and t=-2075.96N For some reason these figures don't seem right, but I'm not sure. I solved for C and T...

Quote: 8. A person holds an 80N weight, 2 meters above the floor for 30 seconds, what is the power required in watts to do this? My answer: 4800watts Wg= -mgd Wg= -80N*2m Wg=-160J Wf=-Wg=160J 1watt=1J/s 30*160J/s=4800J/s=4800watts Does it take any more energy to hold it at 10 m...