Answer:
The distance of separation is 
Explanation:
The mass of the each ball is 
The negative charge on each ball is 
Now we are told that the lower ball is restrained from moving this implies that the net force acting on it is zero
Hence the gravitational force acting on the lower ball is equivalent to the electrostatic force i.e

=> 
here k the the coulomb's constant with a value 
So
![0.01 * 9.8 = \frac{ 9*10^9 *[1*10^{-6} * 1*10^{-6}]}{d}](https://tex.z-dn.net/?f=0.01%20%2A%209.8%20%20%3D%20%20%5Cfrac%7B%209%2A10%5E9%20%2A%5B1%2A10%5E%7B-6%7D%20%2A%201%2A10%5E%7B-6%7D%5D%7D%7Bd%7D)

Initial speed, u = 15 m/s
Final speed, v = 10 m/s
Distance traveled, s = 6.0 m
The acceleration, a, is determined from
u² + 2as = v²
(15 m/s)² + 2*(a m/s²)*(6.0 m) = (10 m/s)²
225 + 12a = 100
12a = -125
a = -10.4167 m/s²
The time, t, for the velocity to change from 15 m/s to 10 m/s is given by
(10 m/s) = (15 m/s) - (10.4167 m/s²)*(t s)
10 = 15 - 10.4167t
t = 0.48 s
The average speed is
(6.0 m)/(0.48 s) = 12.5 m/s
Answer: 12.5 m/s
W=ΔKE , W=-5000j
KEinitial=(1/2)mv² , KEfinal=0j
ΔKE=-(1/2)mv²
-5000=-(1/2)(100kg)v²
v=10 m/s
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