Recent content by jjg242

just hoping to get this double checked: 10. A grindstone spinning at the rate of 8.3 rev/s has what approximate angular speed? a. 3.2 rad/s b. 26 rad/s c. 52 rad/s d. 81 rad/s answer: C, I simply did ; 2*π*8.3 rad/s = 52.1 rad/s. is this correct?

thanks, this forum is so helpful.

maybe I am finding the momentum wrong? Final momentum = -12 Initial momentum = 8 Change (-12 - 8) = -20 since it's magnatude, just 20. I don't see how it could be anything else.

so simply: (0.2*40)-(0.2* -60) = 20 kg-m/s is this correct?

One of the questions on my test review tells me to look for the equation in the book, but i have a different book because my backpack got stolen and I can only get my hands on a old physics book at the moment. can someone help me out and let me know what equation I am looking for here? thanks...

thats what i figured, thanks so much for your help.

ok. I am getting confused now. Originnaly I figured that Kinetic friction applied to this problem considering that the road is icy. But I thought the friction would be less than w*coefficient. So are you saying you can use the static coeffient if we assume the tire is spinning without...

bump. any feedback appreciated.

so the only possible answer is D?

so... m = mass v= change in velocity? F= force in N t= time is v the initial, final, or change in velocity? if it is change in velocity, then i get: 9*1/0.002 = 4500 N Is this right?

8. A karate master strikes a board with an initial velocity of 10 m/s, decreasing to 1 m/s as his hand passes through the board. If the time of contact with the board is 0.002 s, and the mass of the coordinated hand and arm is 1 kg, what is the force exerted on the board. a. 1000 N b. 1800...

Is acceleration involving N just N/kg? so If I have a ball that is 7kg and 5 N of force is applied will the acceleration be 5 / 7 = 0.71 m/s/s ?