<span> applies to any object moving more slowly than the speed of light</span>
This is a distance not a displacement
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:
See explanation
Explanation:
Since the original count rate is 600 Bq,
i) after 1 half life, the count rate decreases to 1/2 of 600 Bq = 300 Bq
ii) after 2 half lives, the count rate decreases to 1/4 of 600 = 150 Bq
iii) after 3 half lives, the count rate decreases to 1/6 of 600 = 100 Bq
iv) after 4 half lives, the count rate decreases to 1/8 of 600 = 75 Bq