Answer:
C
Explanation:
When an object has a greater mass than the drag force pushing it back then it is moving forward therefore making car C faster than car D. If car D was much smaller it wouldn't have enough mass to push past the drag force.
Answer:
n = 756.25 giga electrons
Explanation:
It is given that,
If the charge on the negative plate of the capacitor, 
Let n is the number of excess electrons are on that plate. Using the quantization of charges, the total charge on the negative plate is given by :

e is the charge on electron

or
n = 756.25 giga electrons
So, there are 756.25 giga electrons are on the plate. Hence, this is the required solution.
3. In a uniform electric field, the equation for the magnitude of the magnetic field is E=(V/d). V= voltage d= distance. If the magnetic field magnitude is
constant , as stated in your problem, then the voltage must stay the same otherwise the value of "E" would change". And the problem already told us the "E" is uniform and so, not changing. Does that make sense?
4a. If the magnetic field lines are equally spaced apart, in other words share the same
density. Then we know that the magnitude of the magnetic field is unchanging. This is because the density of of the magnetic field lines(how many are in a certain area) is related to the magnitude being expressed by the electric field. Greater magnitude is expressed by the presence of more lines (higher line density)
4b. The electric potential is measured in Volts(V) and is uniform along the same equipotential line. What is an equipotential line(gray)? It is a line drawn perpendicular(forms a right angle with) to the magnetic field lines(black) to show the changes in electric potential. One space where electric potential will always be the same because it will always be equal to 0 Volts is exactly in between a positive and negative charges of equal charge value I have pointed to this line with a purple arrow in my picture.
I really hope this makes sense to you and that my pictures help! :)
The total work done by the electric field on the charge is given by the scalar product between the electric force acting on the charge and the displacement of the charge:

where the force is F=qE, d=0.556 and

. Using the value of q and E given by the problem, we find