Two charged particles are placed 2.0 meters apart. the first charge is +2.0 e-6 c and the second charge is +4.0 e-6
c. what is the electrical force between the two charges?

The Coulomb force between two charges is defined as: Fc=(k*Q₁*Q₂)/r² where k=9*10^9 N m² C⁻², Q₁ and Q₂ are charges and r is the distance between those charges. 
In our case:

Q₁=2*10^-6 C
Q₂=4*10^-6 C
r=2 m

Now we simply plug in the numbers into the equation:

Fc={(9*10^9)*(2*10^-6)*(4*10^-6)}/2^2=0.018 N

The Coulombs force between the two positive charges is Fc=0.018 N and it is a repulsive force because both charges are positive.

It may possibly be c i think |-/

The objects do not reach terminal velocity         there is not enough time from the time the object falls from when it hits the ground to achieve terminal velocity. an object must be placed higher to achieve terminal velocity in order to be in free fall.