E = <u>kQ</u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>
(r + h)²
where,
k = 9 × 10^9Nm²C^-2
Q = total charge, 300uC = 300 × 10^ -6C
r = 8 × 10^ -2m
h = 16 × 10^ -2m
then,
E = <u>9</u><u>e</u><u>9</u><u> </u><u>*</u><u> </u><u>3</u><u>0</u><u>0</u><u>e</u><u>^</u><u>-</u><u>6</u><u> </u><u> </u><u> </u><u> </u>
(8e^-2 + 16e^-2)²
E = 4687500N/C
The gravitational potential energy of the object is 100 J.
Gravitational potential energy stored in an object is the work done in raising the object to a height <em>h</em> against the gravitational force acting on it.
The gravitational force acting on a body is its weight mg, where m is its mass and g, the acceleration due to gravity.
Work done by a force is equal to the product of the force and the displacement made by the point of application of the force.

The weight of the object is given as 20 J and it is raised to a height of 5 m.

The gravitational potential energy of the object is 100 J.
Answer:
In physics and chemistry, the law of conservation of energy states that the total energy of an isolated system remains constant; it is said to be conserved over time. ... For instance, chemical energy is converted to kinetic energy when a stick of dynamite explodes.
Answer:
the kinetic energy lost due to friction is 22.5 J
Explanation:
Given;
mass of the block, m = 0.2 kg
initial velocity of the block, u = 25 m/s
final velocity of the block, v = 20 m/s
The kinetic energy lost due to friction is calculated as;

Therefore, the kinetic energy lost due to friction is 22.5 J
Answer:

Where
represent the force for each of the 5 cases
presented on the figure attached.
Explanation:
For this case the figure attached shows the illustration for the problem
We have an inverse square law with distance for the force, so then the force of gravity between Earth and the spaceship is lower when the spaceship is far away from Earth.
Th formula is given by:

Where G is a constant 
represent the mass for the earth
represent the mass for the spaceship
represent the radius between the earth and the spaceship
For this reason when the distance between the Earth and the Spaceship increases the Force of gravity needs to decrease since are inversely proportional the force and the radius, and for the other case when the Earth and the spaceship are near then the radius decrease and the Force increase.
Based on this case we can create the following rank:

Where
represent the force for each of the 5 cases
presented on the figure attached.