Answer:
38 cm from q1(right)
Explanation:
Given, q1 = 3q2 , r = 60cm = 0.6 m
Let that point be situated at a distance of 'x' m from q1.
Electric field must be same from both sides to be in equilibrium(where EF is 0).
=> k q1/x² = k q2/(0.6 - x)²
=> q1(0.6 - x)² = q2(x)²
=> 3q2(0.6 - x)² = q2(x)²
=> 3(0.6 - x)² = x²
=> √3(0.6 - x) = ± x
=> 0.6√3 = x(1 + √3)
=> 1.03/2.73 = x
≈ 0.38 m = 38 cm = x
<span>that's a good question. It is a amorphous solid because it expands </span>
Explanation:
it is almost zero .this is because the distance and the electrostatic force are inversely proportional
This one is easy. You just find the rise over run for the graph. Here, its 3/2 m/s.
Answer:
v = 349.23 m/s
Explanation:
It is required to find the orbital speed for a satellite 
from the center of mass.
Mass of Mars, 
The orbital speed for a satellite is given by the formula as follows :
So, the orbital speed for a satellite is 349.23 m/s.
