Recent content by kiarrahannice

I hope u are right about the first problem The second problem although I can set it up I can't solve I am having difficulty plugging the numbers in I know m1 is mass of the ball m2 is mass of car v1 is speed of ball after collision and v2 is speed of car. But I still can't solve. algebra...

Fine I'll bite: for the first problem here is what I got Ui + Ki= Uf+Kf mgh + 0 = 0 +1/2 mv^2---v^2= 2gh sigma Fy=macp N-mg=m v^2/r N=mg+m 2gh/r= mg(1+2h/r) (57kg)(9.8 m/s^2)(2(4m)/4m)--1.68 kN second problem:Elastic collision means both momentum and total KE are conserved: mv + MV...

I have the answers already as you can see. What I don't understand is how to get the answers. and if I thought it was an answer machine then I would have posted all 25 questions. If you can't honor the request do not reply to my question Thanks. No time for smart remarks.

Question Details: 1) A skateboard track has the form of a circular are with a 4.00 m radius, extending to an angle of 90 degrees relative to the vertical on either side of the lowest point. A 57.0 kg skateboarder starts from rest at the top of the circular arc. What is the normal force exerted...

no it wouldn't people do it all the time within the physics forum. I have LEARNED so much by individuals writing the steps out including when my professor writes the equation for our class. It's okay if you choose to be difficult I will just ask someone else who is willing to help and so I can...

I was looking for how to solve the equation You are confusing me. Thanks for your help anyways

so would it be (cos 65)(2.0)? since its on the horizontal X axis?

okay so force would be (9.81)(2.0)? Is that correct? and if it is if I divide that by 295 I do not get 6.6

okay so the weight is 2.0kg so that's force right and the upper end is directed down the incline. and right the angle is 65 degrees still lost

A 2.0kg box rests on a plank that is inclined at a angle of 65 degrees above the horizontal. The upper end of the box is attached to a spring with a force constant of 295 N/m. If the coefficient of the static friction between the box and the plank is 0.19, what is the maximum amount the spring...

That is the problem I don't know how to find Force. I am assuming the additional force is .19 I put .22 in the problem. I was looking at something else. How do I find force so I can solve.

still confused I'M thining I will get the answer if I do the following F/295 is that right and to find force would I do F=295/2.0? Please fill in the blanks if you can

I know that I have to find the required force and divide that by 295 N/m--I know I said 360 but it is 295 in order to find the maximum amount how would the 2.0kg box factor in.

A 2.0kg box rests on a plank that is inclined at a angle of 65 degrees above the horizontal. The upper end of the box is attached to a spring with a force constant of 360 N/m. If the coefficient of the static friction between the box and the plank is 0.22, what is the maximum amount the spring...