Recent content by Aubiefan

Nevermind, I just got it, sorry. I used v1i + v1f = v2i + v2f and set v2f equal to (v1i + v1f), since v2i is zero, and solved to get -4.24 m/s for the velocity after collision, then used the KE=PE rule to solve for the height, 0.92 m. Thanks for the tips! One question: the equation above, v1i...

I don't think I understand what the next step would be, I haven't been able to get as far as finding the velocity right after the collision. The initial PE (mgy) was 205.8 J, so I know that the KE at the bottom (right after collision) must equal that, but how do I tell the difference in that...

I am having a lot of difficulty with this problem: Consider a frictionless track as shown in Figure P6.48. A block of mass m1 = 4.20 kg is released from A. It makes a head on elastic collision at B with a block of mass m2 = 10.5 kg that is initially at rest. Calculate the maximum height to...

I have a problem about average power that I am stuck on: A 686 kg elevator starts from rest and moves upward for 3.10 s with constant acceleration until it reaches its cruising speed, 1.70 m/s. a) What is the average power of the elevator during this period? b) What is the average power...

I finally got it! I solved it using your method, and then tried the first method again and got the same answer, I guess I had been punching something into the calculator wrong. My final answer was 79.7 m. Thank you so much for your help and your patience, it is EXTREMELY appreciated!

I know that F=ma, so if F equals ma, so I would think that ma=0.1n, meaning acceleration is 0.1n/m. I also know that n=mg, so ma=0.1(mg), now I see where mass cancels out, thanks for the hint. So I set a=(0.1)g, and got a=.98. I plugged that into V^2=Vo^2 + a(delta X), with 12.5 m/s^2 as Vo...

The frictional force (F) is mu times the normal force, but I don't know what the mass or weight of the car is, that iis one main thing that is confusing me. I am given the coefficient, so I have F= (0.1)n, which can be rearranged to 0.1=n/F, or 0.1=applied force/frictional force. So I know...

I am having a lot of trouble with this problem: A car is traveling at 45.0 km/h on a flat highway. If the coefficient of friction between road and tires on a rainy day is 0.100, what is the minimum distance in which the car will stop? I know it uses the delta X kinematics equation, so I...

The question asked for initial speed, and I got 79.78 m/s.

I got it! Thanks, I think my main problem was using which t value (total time, falling time, or the difference) to use where. I multiplied the difference times the speed of sound, like you said, and used that as my delta X value, then used delta X = Voxt to solve for the initial horizontal...

I tried to get the line of sight distance the way you suggested by using 343 m/s times 2.525 s, to cancel out meters, and got 866.075 m. I used the pythagorean theorem with the delta Y value to get delta X, 31m^2 + delta X^2 = 866.075^2. I came up with an answer of 865.52 m for the delta X...

Net displacement should be speed times time, but I am trying to solve for the speed of the object, and what I have is the speed of sound. The vertical displacement is -31 meters, so if it hit after 2.525 seconds then it dropped with an average velocity of 12.277 m/s, but that is almost exactly...

Thanks for the responses, I appreciate it. So if there is a 0.575 second lag time between when the object hits and when the splash is heard, how do I use that to figure out the initial velocity? I'm sorry if I'm being dense with this...I've got the list of kinematic equations in front of me...

I am stumped by the following problem, I can't quite figure out what to do with the information about the speed of sound: A soccer player kicks a rock horizontally off a 31.0 m high cliff into a pool of water. If the player hears the sound of the splash 3.1 s later, what was the initial speed...

Thanks for your advice, I reworked it and found the correct components. I had gotten a little thrown by the "west of north" angle and made a mess of some negative signs when doing the trig. Thanks again!