An object is dropped from rest from a height of 4.4 107 m above the surface of the Earth. If there is no air resistance, what is
1 answer:
You might be interested in
Answer:
The distance between the two spheres is 914.41 X 10³ m
Explanation:
Given;
4 X 10¹³ electrons, and its equivalent in coulomb's is calculated as follows;
1 e = 1.602 X 10⁻¹⁹ C
4 X 10¹³ e = 4 X 10¹³ X 1.602 X 10⁻¹⁹ C = 6.408 X 10⁻⁶ C
V = Ed
where;
V is the electrical potential energy between two spheres, J
E is the electric field potential between the two spheres N/C
d is the distance between two charged bodies, m
where;
K is coulomb's constant = 8.99 X 10⁹ Nm²/C²
d = (8.99 X 10⁹ X 6.408 X 10⁻⁶)/0.063
d = 914.41 X 10³ m
Therefore, the distance between the two spheres is 914.41 X 10³ m
Answer:
Explanation:
As we know that system of two boxes are moving on frictionless surface
So here if two boxes are considered as a system
then we have
Now since we know that both the boxes are moving together so force applied by first box on other box is given as