At the point of fission, a nucleus of 235U that has 92 protons is divided into two smaller spheres, each of which has 46 protons and a radius of 5.08 multiplied by 10-15 m. What is the magnitude of the repulsive force pushing these two spheres apart?

F=kQ1Q2/d^2 coulombs law....

1.84 x 10 ^29

To determine the magnitude of the repulsive force pushing the two spheres apart, we need to use Coulomb's law, which states that the force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

Coulomb's law can be expressed as follows:
F = k * (Q1 * Q2) / r^2

Where:
F is the magnitude of the force,
k is the electrostatic constant (k = 8.988 * 10^9 N*m^2/C^2),
Q1 and Q2 are the charges of the two spheres,
r is the distance between the centers of the spheres.

In this case, each sphere has 46 protons, so the charge of each sphere is +46e, where e is the elementary charge (e = 1.6 * 10^-19 C).

Given:
Q1 = Q2 = +46e
r = 2 * (5.08 * 10^(-15) m) (since the distance between the centers is the sum of their radii)

Now we can substitute the values into Coulomb's law and calculate the magnitude of the repulsive force:

F = (8.988 * 10^9 N*m^2/C^2) * [(+46e) * (+46e)] / [2 * (5.08 * 10^(-15) m)]^2

Calculating this expression will yield the magnitude of the repulsive force pushing the two spheres apart.