Two point charges, Q1 and Q2, are separated by a distance R. If the magnitudes of both charges are tripled and their separation is also tripled, what happens to the magnitude of the electrostatic force that each charge exerts on the other

The force remains the same.

Explanation:

Let the magnitude of the forces in each case be F1 and F2 respectively.

The charges are Q1 and Q2.

The distance of separation between them is R.

Hence, for F1;

F1 = KQ1Q2/R^2

For F2:

F2 = K * 3Q1 * 3Q2/(3R)^2

F2 = 9KQ1Q2/9R^2

F2 =KQ1Q2/R^2

Hence F1=F2

The force is the same in both cases!

u=13

v=30

s/t=u+1\2at

=13+431/2

=13+265.5

=278.5

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6