complete question:
A child bounces a 60 g superball on the sidewalk. The velocity change of the superball is from 22 m/s downward to 15 m/s upward. If the contact time with the sidewalk is 1/800 s, what is the magnitude of the average force exerted on the superball by the sidewalk
Answer:
F = 1776 N
Explanation:
mass of ball = 60 g = 0.06 kg
velocity of downward direction = 22 m/s = v1
velocity of upward direction = 15 m/s = v2
Δt = 1/800 = 0.00125 s
Linear momentum of a particle with mass and velocity is the product of the mass and it velocity.
p = mv
When a particle move freely and interact with another system within a period of time and again move freely like in this scenario it has a definite change in momentum. This change is defined as Impulse .
I = pf − pi = ∆p
F = ∆p/∆t = I/∆t
let the upward velocity be the positive
Δp = mv2 - m(-v1)
Δp = mv2 - m(-v1)
Δp = m (v2 + v1)
Δp = 0.06( 15 + 22)
Δp = 0.06(37)
Δp = 2.22 kg m/s
∆t = 0.00125
F = ∆p/∆t
F = 2.22/0.00125
F = 1776 N
Taking specific heat of lead as 0.128 J/gK = c
We have energy of ball at 7.00 meter height = mgh = 
When leads gets heated by a temperature ΔT energy needed = mcΔT
= 
ΔT
Comparing both the equations
= ΔT
ΔT = 0.536 K
Change in temperature same in degree and kelvin scale
So ΔT = 0.536 
You do this one just like the other one that I just solved for you.
For this one ...
The density of the object is 2.5 gm/cm³.
We know that every cm³ of it we have contains 2.5 gm of mass.
We have to find out how many cm³ we have.
The question tells us: We have 2.0 cm³.
Each cm³ of space that the object occupies contains 2.5 gm of mass.
So the 2.0 cm³ that we have contains (2 x 2.5 gm) = 5 gms.
That's the mass of our object.
The answer is:
Both the distance traveled in a given time and the magnitude of the acceleration at a given instant
Hope I Helped!
