Answer:
280 N
Explanation:
Applying Newton's third second law of motion,
F = m(v-u)/t................... Equation 1
Where F = Magnitude of the average force on the ball during contact, v = final velocity of the ball, u = initial velocity of the ball, t = time of contact of the ball and the wall.
Note: Let the direction of the initial velocity of the ball be positive
Given: m = 4 kg, u = 3.0 m/s, v = -4.0 m/s (bounce off), t = 0.1 s
Substitute into equation 1
F = 4(-4-3)/0.1
F = 4(-7)/0.1
F = -28/0.1
F = -280 N.
Note: The negative sign tells that the force on the ball act in opposite direction to the initial motion of the ball
The electric potential V(z) on the z-axis is : V = 
The magnitude of the electric field on the z axis is : E = kб 2
( 1 - [z / √(z² + a² ) ] )
<u>Given data :</u>
V(z) =2kQ / a²(v(a² + z²) ) -z
<h3>Determine the electric potential V(z) on the z axis and magnitude of the electric field</h3>
Considering a disk with radius R
Charge = dq
Also the distance from the edge to the point on the z-axis = √ [R² + z²].
The surface charge density of the disk ( б ) = dq / dA
Small element charge dq = б( 2πR ) dr
dV
----- ( 1 )
Integrating equation ( 1 ) over for full radius of a
∫dv = 
V = ![\pi k\alpha [ (a^2+z^2)^\frac{1}{2} -z ]](https://tex.z-dn.net/?f=%5Cpi%20k%5Calpha%20%5B%20%28a%5E2%2Bz%5E2%29%5E%5Cfrac%7B1%7D%7B2%7D%20-z%20%5D)
= ![\pi k (\frac{Q}{\pi \alpha ^2})[(a^2 +z^2)^{\frac{1}{2} } -z ]](https://tex.z-dn.net/?f=%5Cpi%20k%20%28%5Cfrac%7BQ%7D%7B%5Cpi%20%5Calpha%20%5E2%7D%29%5B%28a%5E2%20%2Bz%5E2%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%20%7D%20%20-z%20%5D)
Therefore the electric potential V(z) = 
Also
The magnitude of the electric field on the z axis is : E = kб 2
( 1 - [z / √(z² + a² ) ] )
Hence we can conclude that the answers to your question are as listed above.
Learn more about electric potential : brainly.com/question/25923373
21. light changes its direction when travelling through a new medium because denser mediums have a higher angle of refraction.
The right answer for the question that is being asked and shown above is that: "The image produced is virtual and of the same size as the object." the image if the object is shifted closer to the lens to a point one focal length away from it is that The image produced is virtual and of the same size as the object.<span>
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I assume that the ball is stationary (v=0) at point B, so its total energy is just potential energy, and it is equal to 7.35 J.
At point A, all this energy has converted into kinetic energy, which is:

And since K=7.35 J, we can find the velocity, v: