Let l = Q/L = linear charge density. The semi-circle has a length L which is half the circumference of the circle. So w can relate the radius of the circle to L by
<span>C = 2L = 2*pi*R ---> R = L/pi </span>
<span>Now define the center of the semi-circle as the origin of coordinates and define a as the angle between R and the x-axis. </span>
<span>we can define a small charge dq as </span>
<span>dq = l*ds = l*R*da </span>
<span>So the electric field can be written as: </span>
<span>dE =kdq*(cos(a)/R^2 I_hat + sin(a)/R^2 j_hat) </span>
<span>dE = k*I*R*da*(cos(a)/R^2 I_hat + sin(a)/R^2 j_hat) </span>
<span>E = k*I*(sin(a)/R I_hat - cos(a)/R^2 j_hat) </span>
<span>E = pi*k*Q/L(sin(a)/L I_hat - cos(a)/L j_hat)</span>
Explanation:
It is given that,
Mass of the ball, m = 0.06 kg
Initial speed of the ball, u = 50.4 m/s
Final speed of the ball, v = -37 m/s (As it returns)
(a) Let J is the magnitude of the impulse delivered to the ball by the racket. It can be calculated as the change in momentum as :

J = -5.24 kg-m/s
(b) Let W is the work done by the racket on the ball. It can be calculated as the change in kinetic energy of the object.


W = -35.1348 Joules
Hence, this is the required solution.
The force of the tripped catch exerted on the 2.5 Kg ball moving at 8.5 m/s to the Left is 160 N
<h3>Data obtained from the question </h3>
- Initial velocity (u) = 8.5 m/s
- Final velocity (v) = 7.5 m/s
- Time (t) = 5 ms = 0.25 s
- Mass (m) = 2.5 Kg
- Force (F) = ?
<h3>How to determine the force</h3>
The force exerted on the ball can be obtained as follow:
F = m(v + u) / t
F = [2.5(7.5 + 8.5)]/ 0.25
F = 40 / 0.25
F = 160 N
Thus, the force exerted on the ball is 160 N
Learn more about momentum:
brainly.com/question/250648
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