Recent content by bigredd87

well i put in 0.57 and it was correct, so hopefully my teacher won't see this... what is this factor of 2 discretion?

Yes, it was definitely intimidating at first, but once u realize that they don't give u enough information along with the numbers that are already there, then u start to simplify the problem a little.

yeah it is, i tried to simplify the question as much as possible. I think i got an answer of 0.57cm

for m1=bullet and m2=chair: ok, so i do m1v1 + m2v2 (which is 0 since the chair is at rest)=(m1+m2)vf which gives me 0.15m/s i can then say that the kinetic friction force is equal to the mass of both the bullet and the chair. so if friction force is equal to -Fk=mu_k(m1 +m2)g then the...

im sorry, i left the mass of the bullet out, it is 10 grams

There is a gun in which the barrel length is 62cm and has a muzzle velocity of 450m/s. A wooden chair of 30 kg sits on a wooden floor. A bullet penetrates the chair and the bullet lodges itself into the chair 9 centimeters. How far back, in centimeters, did the chair fly back? The...

ok, so I think I got it now. a2 = (-1/2)a1 so u have for m2 in y direction: 2T-m2g=m2a2 ; since there are two ropes pulling on m2 i already know that T =m1a1 and the acceleration constraint, so plugging this in i get: 2m1a1-m2g=-m2a1/2 4m1a1-2mg=-m2a1 a1(4m1 + m2)= 2mg...

so a2 =-a1? I didn't understand if the mass on the pulley made it different from other problems. So there is tension T from the rope attached to the wall, the weight force m2g, and the tension from m1 which is m1a1?

here is what I have so far: Fnetx for m1 is T=m1a Fnety for m1 is N-m1g=0 I guess my real dilemma is i don't really know what the forces on m2 are. I believe there is no acceleration constraint since m2 is on the second pulley. Does that mean that the tension and weight are equal to m2a...

I have a picture provided. You can assume that the table is frictionless. I need an expression for the acceleration of m1.

well i guess since it is a steel table it is rotating along the tangential axis, and here is the pic that i have. this pic along with what the problem is all the info I have. I also think that u have kinetic friction, so if that is the case, then the coefficient for steel on steel without...

I guess I need to find a tangential acceleration or something and and velocity at which the tube breaks and also a time it takes to get to that velocity, but I'm not sure. Can anybody help please? A 600 g steel block rotates on a steel table while attached to a 1.20 m-long hollow tube...