Answer : The correct option is, (d) 4 times
Solution :
According to the Coulomb's law, the electrostatic force of attraction or repulsion between two charges is directly proportional to the product of the charges and is inversely proportional to the square of the distance between the the charges.
Formula used :
where,
F = electrostatic force of attraction or repulsion
= Coulomb's constant
and 
are the charges
r = distance between two charges
First we have to calculate the force exerted between S and q when the distance between the charge is 1 unit and let us assumed that the charge be 'q'
..........(1)
Now we have to calculate the force exerted between S and p when the distance between the charge is 2 unit at the same charge.
...........(2)
Equation equation 1 and 2, we get
Therefore, the force exerted between S and q is 4 times the force exerted between S and p.
Answer:
The splitting up of light into its constituent colours while passing from one medium to the other is called dispersion.
Answer:
(A) 0.8 kgm/s
Explanation:
because of the even ground it would only slow down
The energy of a single photon is given by:
where
E is the energy
h is the Planck constant
f is the frequency of the light
The light in our problem has a frequency
, so the energy of each photon of that light is:
Answer:
15.6m/s
Explanation:
V1=
because the derivate of the position is the velocity
V1=12t+3
V2=20+
-8t because the integral of the acceleration is the velocity
V2=
V1=V2 to see when the velocities of particles match
12t+3=20-4t^2
4t^2+12t-17=0 we resolve this with 
and we take the positif root
t=1.05 sec
if we evaluate the velocity (V1 or V2) the result is 15.6m/s
