Answer: 61.2N
F=kq1q2/r^2
F=8.98E9(1.6E-19)(1.6E-19)/(1.93E-15)^2
F=61.2N
Answer:
9.4 m/s
Explanation:
The work-energy theorem states that the work done on an object is equal to the change in kinetic energy of the object.
So we can write:
where in this problem:
W = -36.733 J is the work performed on the car (negative because its direction is opposite to the motion of the car)
is the initial kinetic energy of the car
is the final kinetic energy
Solving for Kf,
The kinetic energy of the car can be also written as
where:
m = 661 kg is the mass of the car
v is its final speed
Solving, we find
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)