(2.0 x 10² N/Coul) x (-1.6 x 10⁻¹⁹ Coul) = -3.2 x 10⁻¹⁷ N
The magnitude is 3.2 x 10⁻¹⁷ Newton.
The best option is C. This is due to friction.
Q: ken, 0.75 kg, moves toward a wall (his path normal to the wall) at 52 m/s. 13.0 ms after he touches the wall he pushes himself off in the opposite direction at 60 m/s. What is the magnitude of the average force the wall exerts on Ken during this rapid maneuver
Answer:
-6461.54 N
Explanation:
From Newton's Fundamental equation,
F = m(v-u)/t.................... Equation 1
Where F = Force exerted in sonic, m = mass of ken, v = final velocity, u = initial velocity, t = time.
Given: m = 0.75 kg, v = - 60 m/s (opposite direction), u = 52 m/s, t = 13 ms = 0.013 s
Substitute into equation 1
F = 0.75(-60-52)/0.013
F = 0.75(-112)/0.013
F = -84/0.013
F = -6461.54 N
Note: The negative sign tells that the force act in opposite direction to the initial motion of ken.
Hence the magnitude of the average force of the wall = -6461.54 N
Complete question
The complete question is shown on the first and second uploaded image
Answer:∈
Answer to first question is shown on the second uploaded image.
Part B the Answer is:
The ratio 
is evaluated to be 49.99
Explanation:
The explanation is shown on the third ,fourth and fifth image.
Answer:
law of gravity
Explanation:
cause the ball was still moving
