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
#SPJ1
Answer:
Explanation: find the attached solution below
Answer:
The zero field location has to be on the line running between the two point charges because that's the only place where the field vectors could point in exactly opposite directions. It can't be between the two opposite charges because there the field vectors from both charges point toward the negative charge.
the answer to this question is y
