C. The Densities are equal.
<h3>
What is density?</h3>
Density is mass per unit volume or mass of a unit volume of a material substance.
If m1, V1 and D1 = mass, volume and density respectively of ball C
m2, V2 and D2 = mass, volume and density respectively of ball D
According to the Question ,

Therefore,

Hence, D1 = D2
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Answer:
(a) 2.542 cm
(b) 272.7°C
Explanation:
diameter, d = 2.540 cm
T1 = 20°C
α = 11 x 10^-6 /°C
(a) Let d' be the diameter.
T2 = 87°C
Use he formula for the areal expansion
A' = A ( 1 + βΔT)
where, β is the coefficient of areal expansion and ΔT is teh rise in temperature, A' be the area at high temperature and A be the area at low temperature.
β = 2 α = 2 x 11 x 106-6 = 22 x 10^-6 /°C
So,

D'^2 = 2.54^2 ( 1 + 22 x 10^-6 x 67)
D' = 2.542 cm
(b) Let the change in temperature is ΔT.
Use the formula for the volumetric expansion
ΔV = V x γ x ΔT
Where, γ = 3 x α = 3 x 11 x 10^-6 = 33 x 10^-6 /°C
0.9/100 = 33 x 10^-6 x ΔT
ΔT = 272.7°C
Velocity is displacement over time, so do 60/8 to get 7.5 yards per second, and remember to add the direction the ball is traveling in.
1) You can find the velocity with which the jumper leaves the ground by using the formula for the final velocity of an upward vertical motion.
Vf^2 = Vo^2 - 2gd
The data known are: Vf = 0, g = 9.81 m/s^2, and d = 60 cm = 0.6 m
=> Vo^2 = 2gd = 2*9.81m/s^2 * 0.6 m = .11.772 m^2/s^2 =>
Vo = 3.4 m/s
2) You can obtain the acceleration of the jumper during the jump by using the equivalent formula, Vf^2 = Vo^2 + 2ad
Where, Vf^2 is the velocity with which he leaves the ground (3.4 m/s), Vo is zero because he jumps from the rest, d is the distance that his body rises from knees,which is 50 cm = 0.5 m
Then, a = Vf^2 / 2d = [3.4m/s]^2 / (2*0.5m) = 11.6 m/s^2
And now, you can find the force exerted by the jumper as:
F = m*a, and if you know his weight, W = m*g => m = W/g
Replace m in F = m*a => F = W*a/g = W*11.6m/s^2 / 9.81 m/s^2 = 0.10 W
F = 0.10 W