F = ma = -kx
a = 9.81 m/s²
k = 3430 N/m
m = 70 kg
x = - ma/ k = 0.2m
Answer:
The voltage increases by a factor of three.
Explanation:
Answer:
Therefore,
The potential (in V) near its surface is 186.13 Volt.
Explanation:
Given:
Diameter of sphere,
d= 0.29 cm


Charge ,

To Find:
Electric potential , V = ?
Solution:
Electric Potential at point surface is given as,

Where,
V= Electric potential,
ε0 = permeability free space = 8.85 × 10–12 F/m
Q = Charge
r = Radius
Substituting the values we get


Therefore,
The potential (in V) near its surface is 186.13 Volt.
Answer:
Explanation:
Given
frequency of wave 
We know velocity is given by

where
=wavelength


