2. 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 5.9 × 10^-15 m.
What is the magnitude of the repulsive force pushing these two spheres apart?
The value of the Coulomb constant is
8.98755 × 10^9N m2/C2
1. 4.12586 × 10^-11 N m
2. Unable to determine
3. None of these
4. 4.02599 × 10^34 N m/C
5. 4.75067 × 10^20 N/C
6. 3496.5 N

3.5 x 10^3

To find the magnitude of the repulsive force pushing the two spheres apart, we can use Coulomb's law. Coulomb's law 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.

In this case, the two spheres are both positively charged, and their charges can be determined by the number of protons they have. Each sphere has 46 protons, so the charge on each sphere is +46e, where e is the elementary charge (1.6 × 10^-19 C).

The distance between the centers of the two spheres is twice their radius, which is 2 × 5.9 × 10^-15 m = 1.18 × 10^-14 m.

Now we can calculate the magnitude of the repulsive force using Coulomb's law:

Force = (Coulomb's constant) × (Charge 1) × (Charge 2) / (Distance)^2

Plugging in the given values:
Force = (8.98755 × 10^9 N m^2/C^2) × (+46e) × (+46e) / (1.18 × 10^-14 m)^2

Now, we need to convert the charges from e to Coulombs:
Charge 1 = 46e = 46 × 1.6 × 10^-19 C = 7.36 × 10^-18 C
Charge 2 = 46e = 46 × 1.6 × 10^-19 C = 7.36 × 10^-18 C

Plugging in the converted charges and simplifying the equation:
Force = (8.98755 × 10^9 N m^2/C^2) × (7.36 × 10^-18 C) × (7.36 × 10^-18 C) / (1.18 × 10^-14 m)^2

Calculating the force using a calculator, the magnitude of the repulsive force is approximately 4.02599 × 10^34 N.

Therefore, the correct option is 4. 4.02599 × 10^34 N m/C.