Answer:
<h2>
The potential difference increases </h2>
Explanation:
from the relation 
where E= electric field (force per coulomb)
V= voltage
d= distance
Hence the voltage is going to be V= E×d.
Therefore this means that increasing the distance increases the voltage.
We actually don't need to know how far he/she is standing from the net, as we know that the ball reaches its maximum height (vertex) at the net. At the vertex, it's vertical velocity is 0, since it has stopped moving up and is about to come back down, and its displacement is 0.33m. So we use v² = u² + 2as (neat trick I discovered just then for typing the squared sign: hold down alt and type 0178 on ur numpad wtih numlock on!!!) ANYWAY....... We apply v² = u² + 2as in the y direction only. Ignore x direction.
IN Y DIRECTION: v² = u² + 2as 0 = u² - 2gh u = √(2gh) (Sub in values at the very end)
So that will be the velocity in the y direction only. But we're given the angle at which the ball is hit (3° to the horizontal). So to find the velocity (sum of the velocity in x and y direction on impact) we can use: sin 3° = opposite/hypotenuse = (velocity in y direction only) / (velocity) So rearranging, velocity = (velocity in y direction only) / sin 3° = √(2gh)/sin 3° = (√(2 x 9.8 x 0.33)) / sin 3° = 49 m/s at 3° to the horizontal (2 sig figs)
Emily throws the ball at 30 degree below the horizontal
so here the speed is 14 m/s and hence we will find its horizontal and vertical components


vertical distance between them

now we will use kinematics in order to find the time taken by the ball to reach at Allison

here acceleration is due to gravity

now we will have

now solving above quadratic equation we have

now in order to find the horizontal distance where ball will fall is given as

here it shows that horizontal motion is uniform motion and it is not accelerated so we can use distance = speed * time

so the distance at which Allison is standing to catch the ball will be 5.33 m
Answer:
a = the lowest critical speed of the shaft 882.81 rad/s
b = new diameter 0.05m or 50mm
c = critical speed 1765.62rad/s
Explanation:
see the attached file