Answer:
It cannot reach us by conduction or convection as there is no medium such as air in most part of the space between the earth and the sun. From the sun the heat comes to us by another process known as radiation. The transfer of heat by radiation does not require any medium.
Well, first of all, a car moving around a circular curve is not moving
with uniform velocity. The direction of motion is part of velocity, and
the direction is constantly changing on a curve.
The centripetal force that keeps an object moving in a circle is
Force = (mass of the object) · (speed)² / (radius of the circle)
F = m s² / r
We want to know the radius, to rearrange the formula to give us
the radius as a function of everything else.
F = m s² / r
Multiply each side by 'r': F· r = m · s²
Divide each side by 'F': r = m · s² / F
We know all the numbers on the right side,
so we can pluggum in:
r = m · s² / F
r = (1200 kg) · (20 m/s)² / (6000 N) .
I'm pretty sure you can finish it up from here.
The initial force between the two charges is given by:

where k is the Coulomb's constant, q1 and q2 the two charges, d their separation. Let's analyze now the other situations:
1. F
In this case, q1 is halved, q2 is doubled, but the distance between the charges remains d.
So, we have:

So, the new force is:

So the force has not changed.
2. F/4
In this case, q1 and q2 are unchanged. The distance between the charges is doubled to 2d.
So, we have:

So, the new force is:

So the force has decreased by a factor 4.
3. 6F
In this case, q1 is doubled and q2 is tripled. The distance between the charges remains d.
So, we have:

So, the new force is:

So the force has increased by a factor 6.
Answer: D) The gravity on the moon is one sixth the gravity of Earth.
Explanation: