Answer with Explanation:
We are given that


Charge on proton,q=
a.We have to find the electric potential of the proton at the position of the electron.
We know that the electric potential

Where 


B.Potential energy of electron,U=
Where
Charge on electron
=Charge on proton
Using the formula


Answer:
The horizontal component of displacement is d' = 1422.7 m
Explanation:
Given data,
The distance covered by the truck, d = 1430 m
The angle formed with the horizontal, Ф = 5.76°
The displacement is a vector quantity.
The horizontal component of displacement is given by,
d' = d cos Ф
= 1430 cos 5.76°
= 1422.7 m
Hence, the horizontal component of displacement is d' = 1422.7 m
Answer:
hypothesis , hope it helps
Explanation: