Answer:
The mass of the meter stick is 
Explanation:
Here since the meter stick is in equilibrium position its net torque and net force should be equal to zero
Since at the begging the meter stick is balanced at center of the meter stick that means its center of mass should be present at 
Now lets consider the later case where stick is balanced by two 5.12 g coins .
Here torque due to two coins = 
Torque due to weight of meter stick = 
= 
where m = mass of the meter stick
Here 
.
Upon equating we will be getting mass of the meter stick =
(a)
consider the motion of the tennis ball. lets assume the velocity of the tennis ball going towards the racket as positive and velocity of tennis ball going away from the racket as negative.
m = mass of the tennis ball = 60 g = 0.060 kg
v₀ = initial velocity of the tennis ball before being hit by racket = 20 m/s
v = final velocity of the tennis ball after being hit by racket = - 39 m/s
ΔP = change in momentum of the ball
change in momentum of the ball is given as
ΔP = m (v - v₀)
inserting the above values
ΔP = (0.060) (- 39 - 20)
ΔP = - 3.54 kgm/s
hence , magnitude of change in momentum : 3.54 kgm/s
D. since the star would be larger than the sun, having more mass, its gravitational pull would be stronger thus pulling the planets into it. hope that helped!
C. It is C because it really is loose covering.
