A 0.140-kg baseball is dropped and reaches a speed of 1.20 m/s just before it hits the ground. It rebounds with a speed of 1.00 m/s. What is the change of the ball's momentum

Explanation:

An impulse describes a change in momentum. The change in momentum of an object is its mass times the change in its velocity.

The change in moment is given by the expression below

Δp=m⋅(Δv)=m⋅(vf−vi) .

Given data

mass m= 0.140-kg

initial velocity vi= 120 m/s

final velocity vf= 1 m/s

substituting we have

Δp=m⋅(Δv)=0.14⋅(1−1.2)

Δp=m⋅(Δv)=0.14⋅(-0.2)

Δp= 0.028 kg m/s

The change in momentum was found to be Δp= 0.028 kg m/s

explanation:

v=uf/(u-f)

= 5.-10/(5+10) convex lens has negative focal length

this could be newton#'s law of cooling. the cooling curve is exponential