Answer:
Charge on B is 12 uC.
Explanation:
Initial charge on A = 32 uC
Initial charge on B and C = 0
Now A touches to B, so the charge on A and B both is
q = (32 + 0) / 2 = 16 uC
Now A touches to C, so the charge on A and C both is
q' = (16 + 0) / 2 = 8 uC
Now again A touches to B so the charge on B is
q''= (8 + 16) / 2 = 12 uC
True because molecules don't have to be compounds but compounds have to be molecules
Period = (1) / (frequency)
= (1) / (20/sec)
= 1/20 sec
Answer:
The Electric flux will be 
Explanation:
Given
Strength of the Electric Field at a distance of 0.158 m from the point charge is

We know that the flux of the Electric Field can be calculated by using Gauss Law which is given by

Let consider a sphere of radius 0.158 m as Gaussian Surface at a distance of 0.158 m from the point charge and Let
be the flux of the Electric Field coming out\passing through it which is given by

It can be observed that same amount of flux which is passing through the Gaussian sphere of radius 0.158 is also passing through the Gaussian sphere of radius 0.142 m at a distance of 0.142 m from its centre.
Also it can be observed that the charge inside the two Gaussian Sphere mentioned have same value so the Flux of electric field through them will also be same.
So the electric flux through the surface of sphere that has given charge at its centre and that has radius 0.142 m is 
Answer:
<h3>
30.66m</h3>
Explanation:
Using the equation of motion formula
where;
S is the height to which the ball rises
u is the initial velocity of the ball = 0m/s
a is the acceleration due to gravity = 9.81m/s²
t is the time taken by the ball in air = 5.0s
Note that the time to rise to the peak is one-half the total hang-time = 5.0/2 = 2.5s
Substituting the given parameters into the formula above to get S:

This means that the ball rises 30.66m before it reaches its peak.