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
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PE=mgh
180=0.5*10*h
180=5h
h=180/5=36 m
We use Lenz's law:
emf = -(NBA)/t ; N is the number of turns, B is the magnetic field, A is the area
t = 1/f = 1/10
emf = (-100 x 0.05 x 0.1) / 0.1
emf = -5 V