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...
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 cuz I dont have 15 posts nor image cuz 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 cuz 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 leave...