Hi there! Lets see!
- m is mass, and its units are kg
- k is the elastic constant measured in newtons per meter (N/m), or kilograms per second squared kg/s²
Therefore:
![\sqrt{\dfrac{m}{k}} =\sqrt{\dfrac{[kg]}{[\dfrac{kg}{s^2}]}} =\sqrt{\dfrac{[kg]}{[kg]}\cdot s^2} = \sqrt{[s]^2} = s](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cdfrac%7Bm%7D%7Bk%7D%7D%20%3D%5Csqrt%7B%5Cdfrac%7B%5Bkg%5D%7D%7B%5B%5Cdfrac%7Bkg%7D%7Bs%5E2%7D%5D%7D%7D%20%20%3D%5Csqrt%7B%5Cdfrac%7B%5Bkg%5D%7D%7B%5Bkg%5D%7D%5Ccdot%20s%5E2%7D%20%3D%20%5Csqrt%7B%5Bs%5D%5E2%7D%20%3D%20s)
The period is given in seconds so the formula is dimensionally correct.
Acceleration is given by:

where
v is the final velocity
u is the initial velocity
t is the time interval
Let's apply the formula to the different parts of the problem:
A) 
Let's convert the quantities into SI units first:


t = 4.0 min = 240 s
So the acceleration is

B) 
As before, let's convert the quantities into SI units first:


t = 94 s
So the acceleration is

C) 
For this part we have to use a different formula:

where we have
v = 0 is the final velocity
u = 89.2 m/s is the initial velocity
a is the acceleration
d = 75 m is the distance covered
Solving for a, we find

Friction , as the angle gets huger and higher , this is less and less normal force into the inclined plane .
Newton's second law of motion. F = m a .
solution
In This question we have given ,
Charge on each ball=
force of repulsion between balls is 
Let distance between ball be x
We know by Coulombs law,
..............(1)
here,
Put values of k and charges in equation 1





therefore distance between two given charges is =.701m