To solve this problem it is necessary to apply the concepts related to the Moment. The moment in terms of the Force and the time can be expressed as

F = Force

At the same time the moment can be expressed in terms of mass and velocity, mathematically it can be given as

Where
m = Mass
Change in velocity
Our values are given as

By equating the two equations we can find the Force,



Therefore, the net average force will be:

The negative symbol indicates that the direction of the force is upwards.
Answer:
The point at which the electrical potential is zero is x = +0.33 m.
Explanation:
By definition the electrical potential is:

Where:
K: is Coulomb's constant = 9x10⁹ N*m²/C²
q: is the charge
r: is the distance
The point at which the electrical potential is zero can be calculated as follows:

(1)
q₁ is the first charge = +3 mC
r₁ is the distance from the point to the first charge
q₂ is the first charge = -6 mC
r₂ is the distance from the point to the second charge
By replacing r₁ = 1 - r₂ into equation (1) we have:
(2)
By solving equation (2) for r₂:

Therefore, the point at which the electrical potential is zero is x = +0.33 m.
I hope it helps you!
Answer:
they do some of the work for you, they get it done faster, and they make the work more accurate
Explanation:
The junction rule says that the sum of the currents going into a junction must equal the sum of the currents leaving a junction. This describes the conservation of current.
The sun. lol. It is closer to Earth, therefore bigger to us, and not just a small point of light.