Recent content by Schu

I went through it again after rearranging the formula, I'm not sure why I got what I did before but I came out with about 3.88 mm THanks for the help.

How much will the aluminum stretch a 3.57 kg mass is supported by an aluminum wire with a length of 2.43 m and a diameter of 2.01mm. How much will the wire stretch? Young's Modolus for Al= 6.9 * 10^10 First get the cross sectional area of the Al then plug it into the formula Y =...

I need help ASAP Is anyone out there? I would appreciate the help :confused:

Here's the question: the rotor of a helicopter starts from REST and reaches it's operating speed in 7.28s If the angular acceleration of the rotor is 92.5 rad/s^2, what is the angular displacement during that time? I'm not sure where to start. If I multiply the time by the acceleration I...

[FONT=Tahoma]Particulars: ball has a radius of 2.5 cm a mass of .125 and is rolling across a table with a speed of .547 m/s, this table is 1.04 m off the ground. It rolls to the edge and down a ramp How fast will it be rolling across the floor? First I found the Gravitational Potential...

Ok check to see if I am OK on this; we know T = I α I = 1/2 mr^2 and α = a / r so plug it in I = (1/2) 13.13 * .081^2 = .043072965 α = (29.9425 / 5.57)/.081 = 66.36633641 Multiply the two together for T and you get 2.8585 Nm How'd I do??

Now ya'll have me confused again, I thought I was all set with the first formula, T = f * r. Now where does Newtons Second Law fit into it to get the rotational movement. Take me step by step if you would. Thanks for the help so far

I am attempting to the following question: How large is the torque that act on an armature? Particulars: (can be approximated as a solid cylinder) radius of .081 m length of .124 m mass of 13.13 kg Accelerated from REST to operating speed of 3530 rpm in 5.57 second I know T = F*r...

I think I am all set now. The F net = m*a correct? THen a = Sum F / m Thanks for the help.

The car was moving on a level plane in a straight direction which would be a 0 degree direction if looking on an x,y grid. The mass of the car is 1250 kg. It took the car 24.9 m to stop. The final velocity is 0 m/s the coefficient of friction was 0.836. See if I am making any sense...

gravitational force = mass * gravity normal force = gravitational force friction force = coefficient of friction * Normal Force sum of the forces = mass * acceleration?? What forces do you add up, the vertical or the horizontal or both? I had gone down this road but this last part confused me.

It's been a while since I have gone through my physics classes so help me out if you would. While investigating a traffic accident we found skid marks that were 24.9 m in length. The mass of the car was around 1250 kg. After the accident reconstructionist used his "sled" the coefficient of...