Recent content by ngorecki

Alright so mgh = 1/2mv^2 Cancel mass gh = 1/2v^2 9.8(h) = .5(19)^2 h = 1768.9 Then mgh(.5h) = 1/2mv^2 Cancel mass 9.8(.5*1768.9) = 1/2v^2 8667.61 = 1/2v^2 17335.22 = v^2 131.66 = v

energy and momentum

Vf represents the total velocity at the end. The velocity relative to the ice for the girl and the plank added together should equal the Vf (1.51)

Total velocity is conserved...?

Vf = 1.51 m/s and the question is asking what the speeds are... Which I am struggling to solve. How do you suggest I go about soving for the speeds?

Homework Statement A 0.20 kg ball is thrown straight up into the air with an initial speed of 19 m/s. Find the momentum of the ball at the following locations. (a) at its maximum height (b) halfway to its maximum height Variable: ball = .2kg Vi = 19 m/s Homework Equations...

sorry for putting the blank template 2X below...didnt see that before posting

conservation of momentum?? Homework Statement A 45.0 kg girl is standing on a 156 kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat, frictionless surface. The girl begins to walk along the plank at a constant velocity of 1.51 m/s relative to the...

I have done that. And i changed gravity into the x and y components. I am not sure what to solve for first/how to solve for it

A 2.00 kg block is held in equilibrium on an incline of angle θ = 55° by a horizontal force, F. If the coefficient of static friction between block and incline is µ s = 0.300, determine (a) the minimum value of F and (b) the normal force of the incline on the block. F = ma Mu = Ff/Fn...

Can you show me the set up?

Thank you very much!

i don't know how to solve for the acceleration with only the mass of the 2 objects...

the mass of m1 * acceleration = 8.304 then i take the mass of m1 * gravity = 117.6 u = Ff/Fn = 28.128/117.6 = .239 the answer is.239

M1 = ma M1 = 12 * .692 M1 = 8.304 M1 = T- Ff 8.304 = 36.432 - Ff Ff + 8.034 = 36.432 Ff = 28.398 Is this accurate?