Verified answer

Set the kineitc energy at the end equal to the potential energy of the electric potential and charge at the beginning:

1/2mv^2 = qV ---> solve for V ---> V = 1/2 m/qv^2 = 0.5*0.004kg*(2 m/s)^2/(20x10^-6 Coul) = 400 V

The equation V = I*R tells you that for a given resistance, cutting-edge is precisely proportional to voltage. this means that better cutting-edge means better voltage. i'm assuming that some variety of graph or diagram is going alongside with this question, so seem at it to verify at which of those circumstances the present (I) is the only right to verify which era has the only right skill distinction.