(a) The force exerted by the electric field on the electron is given by the product between the electron charge q and the intensity of the electric field E:

Under the action of this force, the electron moves by:

And the work done by the electric field on the electron is equal to the product between the magnitude of the force and the displacement of the electron. The sign has to be taken as positive, because the direction of the force is the same as the displacement of the electron, so:

(b) The electron is initially at rest and it starts to move under the action of the electric field. This means that as it moves, it acquires kinetic energy and it loses potential energy. The change in potential energy is the opposite of the work done by the electric field:

Where Uf and Ui are the final and initial potential energy of the electron.
(c) For the conservation of energy, the sum of the kinetic energy and potential energy of the electron at the beginning of the motion and at the end must be equal:

(1)
where Ki and Kf are the initial and final kinetic energies.
The electron is initially at rest, so Ki =0, and we can rewrite (1) as

and by using the mass of the electron me, we can find the value of the final velocity of the electron:
Answer:
It's A
Explanation:
Although I'm not sure about the second one, I'm sure the first one is A
D
Because the rest of the answers are illogical
A few different ways to do this:
Way #1:
The current in the series loop is (12 V) / (total resistance) .
(Turns out to be 2 Amperes, but the question isn't asking for that.)
In a series loop, the current is the same at every point, so it's
the same current through each resistor.
The power dissipated by a resistor is (current)² · (resistance),
and the current is the same everywhere in the circuit, so the
smallest resistance will dissipate the least power. That's R1 .
And by the way, it's not "drawing" the most power. It's dissipating it.
Way #2:
Another expression for the power dissipated by a resistance is
(voltage across the resistance)² / (resistance) .
In a series loop, the voltage across each resistor is
[ (individual resistance) / (total resistance ] x battery voltage.
So the power dissipated by each resistor is
(individual resistance)² x [(battery voltage) / (total resistance)²]
This expression is smallest for the smallest individual resistance.
(The other two quantities are the same for each individual resistor.)
So again, the least power is dissipated by the smallest individual resistance.
That's R1 .
Way #3: (Einstein's way)
If we sat back and relaxed for a minute, stared at the ceiling, let our minds
wander, puffed gently on our pipe, and just daydreamed about this question
for a minute or two, we might have easily guessed at the answer.
===> When you wire up a battery and a light bulb in series, the part
that dissipates power, and gets so hot that it radiates heat and light, is
the light bulb (some resistance), not the wire (very small resistance).