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.
Answer:
a) 
b) 
Explanation:
given,
n =1.5 for glass surface
n = 1 for air
incidence angle = 45°
using Fresnel equation of reflectivity of S and P polarized light

using snell's law to calculate θ t


a) 

b) 

The charge of the copper nucleus is 29 times the charge of one proton:

the charge of the electron is

and their separation is

The magnitude of the electrostatic force between them is given by:

where

is the Coulomb's constant. If we substitute the numbers, we find (we can ignore the negative sign of the electron charge, since we are interested only in the magnitude of the force)
Answer:

Explanation:
According to question,
Charge 1 and charge 2 are 
The distance between charges is 2 m
We need to find the force with which two positive charges repel. It is called electrostatic force of repulsion. It can be given by :

So, the electric force of repulsion is
.
Efficiency = Work Output / Work Input
92% = Work Output / 100
0.92 = Work Output / 100
Work Output = 0.92 * 100
Work Output = 92 joules.