Recent content by Blubla

A 215 g particle is released from rest at point A inside a smooth hemispherical bowl of radius 30.0 cm. Point B to C have Height= 2/3(R) Calculate the following: A. The gravitational potential energy at A relative to B B. The particle's kinetic energy at B...

Ok I will

PE = KE vf^2= 2mgh/m=2gh Vf^2=2*40*9.81 Vf^2=784.8 Vf=28.0m/s ---------------- PE = KE I changed 15cm into m so it will be = 0.15 PE=mgh PE=1(9,81)(0.15) PE=1.5 KE=1/2mgv^2 KE= 1/2(1) KE= 0.5 KE/PE= Vf^2 0.5/1.5=Vf^2 0.3=Vf^2(Apply square root to...

Errn.. Well, I cannot post any link because I don't have 15 posts nor image because it exceeds the required size. I can tell the answers, 1. Vf= 28m/s 2. Vf=0.55m/s Thats what I have at the moment, is it correct?

Hello all, I have a little problem solving these 2 problems, guess you all can help me because its piece of cake for you all. Ok here we go: 1. Old faithful geyser in yellowstone national park shoots water every hour to a height of 40.0m. With what velocity does the water...