Two protons are released from rest when they are 0.750 {\rm nm} apart.

a) What is the maximum speed they will reach?

b) What is the max acceleration they will reach?

Explanation:

Given:

m = 1.673 × 10^-27 kg

Q = q = 1.602 × 10^-19 C

r = 0.75 nm

= 0.75 × 10^-9 m

A.

Energy, U = (kQq)/r

Ut = 1/2 mv^2 + 1/2 mv^2

1.673 × 10^-27 × v^2 = (8.99 × 10^9 × (1.602 × 10^-19)^2)/0.75 × 10^-9

v = 1.356 × 10^4 m/s

B.

F = (kQq)/r^2

F = m × a

1.673 × 10^-27 × a = ((8.99 × 10^9 × (1.602 × 10-19)^2)/(0.075 × 10^-9)^2

a = 2.45 × 10^17 m/s^2.

find the sum of the two horizontal vectors and the sum of the two vertical vectors

isoceles right angled triangle