Two protons are moving directly toward one another. When they are very far apart, their initial speeds are 2.20 x 105 m/s. What is the distance of closest approach

r = 2.85 * 10^(-12) m

Explanation:

The distance of closest approach is the minimum distance at which both protons can be. This is due to the repulsive force that acts on both protons.

We know that the total kinetic energy of both protons is equal to the potential energy due to one proton acting on another, hence:

½mv² + ½mv² = kq²/r

mv² = kq²/r

=> r = kq²/mv²

Given that:

k = Coulumbs constant

q = 1.6023 * 10^(-19) C

m = 1.673 * 10^(-27) kg

v = 2.20 * 10^5 m/s

Then, r is:

r = (9 * 10^9 * [1.6023 * 10^(-19)]²) / (1.673 * 10^(-27) * [2.2 * 10^5]²)

r = 2.85 * 10^(-12) m

This is the distance of closest approach.

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2

there are no atoms