Answer:
Protons and neutrons are all attracted to each other as a result - the strong nuclear force. This is an attractive force that only has an effect over a very short range in the nucleus.
Answer:
Coefficient of friction will be 0.587
Explanation:
We have given mass of the car m = 500 kg
Distance s = 18.25 m
Initial velocity of the car u = 14.5 m/sec
As the car finally stops so final velocity v = 0 m/sec
From second equation of motion



We know that acceleration is given by



So coefficient of friction will be 0.587
Answer:
(A) We are using them faster than they are replenished by nature
Answer:
A. The target nucleus split into two nuclei, each with fewer nucleons than the original.
Explanation:
Answer:
-54.12 V
Explanation:
The work done by this force is equal to the difference between the final value and the initial value of the energy. Since the charge starts from the rest its initial kinetic energy is zero.

The change in electrostatic potential energy
, of one point charge q is defined as the product of the charge and the potential difference.
