While leaning out a window that is 6.0 m above the ground, you drop a 0.60-kg basketball to a friend at ground level. Your friend catches the ball at a height of 1.6 m above the ground. Determine the following.(a) the amount of work done by the force of gravity on the ball.(b) the gravitational potential energy of the ball-earth system, relative to the ground when it is released.(c) the gravitational potential energy of the ball-earth system, relative to the ground when it is caught.

a) W = 25.872 J

b) - 35.28 J

c) - 9.408

Explanation:

a) The amount of work done by the force of gravity on the ball = Change in potential energy between the two vertical points = - mg (H₂ - H₁)

F = - mg (gravity is acting downwards)

F = - 0.6 × 9.8 = - 5.88 N

(H₂ - H₁) = (1.6 - 6) = - 4.4 m

W = (-5.88)(-4.4) = 25.872 J

b) Gravitational-potential energy of the ball when it was released relative to the ground = (- mg) H₁ = (- 0.6 × 9.8) × 6 = - 35.28 J

c) Gravitational-potential energy of the ball when it is caught relative to the ground = (-mg)(H₂) = -0.6 × 9.8 × 1.6 = - 9.408 J

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm