An electron is a subatomic particle (m = 9.11 x 10-31 kg) that is subject to electric forces. An electron moving in the +x direction accelerates from an initial velocity of +7.93 x 105 m/s to a final velocity of 2.90 x 106 m/s while traveling a distance of 0.0904 m. The electron's acceleration is due to two electric forces parallel to the x axis: = 9.68 x 10-17 N, and , which points in the -x direction. Find the magnitudes of (a) the net force acting on the electron and (b) the electric force .

To solve this problem, we'll need to use Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration.

(a) The net force acting on the electron:
We can find the acceleration of the electron using the initial and final velocities and the displacement.
Using the formula: acceleration = (final velocity - initial velocity) / time

Given:
Initial velocity (vi) = +7.93 x 10^5 m/s
Final velocity (vf) = +2.90 x 10^6 m/s
Displacement (d) = 0.0904 m

We can find the time (t) using the formula: time = displacement / average velocity
Average velocity = (vi + vf) / 2

Avg velocity = (+7.93 x 10^5 m/s + +2.90 x 10^6 m/s) / 2 = 1.4165 x 10^6 m/s

t = 0.0904 m / 1.4165 x 10^6 m/s = 6.3798 x 10^-8 s

Now, we can calculate the acceleration (a):
a = (vf - vi) / t
a = (+2.90 x 10^6 m/s - +7.93 x 10^5 m/s) / (6.3798 x 10^-8 s)
a = 3.1069 x 10^13 m/s^2

Now, to get the net force, we can use Newton's second law:
F_net = m * a
Substituting the given value of mass (m = 9.11 x 10^-31 kg) gives:
F_net = (9.11 x 10^-31 kg) * (3.1069 x 10^13 m/s^2)
F_net = 2.8278 x 10^-17 N

Therefore, the magnitude of the net force acting on the electron is 2.8278 x 10^-17 N.

(b) The magnitude of the electric force (F_1):
We have the magnitude of the second electric force (F_2) given as 9.68 x 10^-17 N, pointing in the -x direction.
Since the electron is subject to two electric forces in the x direction, we need to subtract F_2 from F_net to find F_1.

F_1 = F_net - F_2
F_1 = 2.8278 x 10^-17 N - 9.68 x 10^-17 N
F_1 = -6.852 x 10^-17 N (negative sign indicates the force is in the opposite direction to F_2)

Therefore, the magnitude of the electric force F_1 is 6.852 x 10^-17 N.