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Stells [14]
2 years ago
12

2. FAt=p (remember p=mv). A 2.5Jg baseball moving at 8.5m/s to the left is slowed by a tipped

Physics
1 answer:
TiliK225 [7]2 years ago
3 0

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

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Answer:

The positively charged ball moves between both charged plates till the plates and the ball all become neutral.

Check Explanation for more.

Explanation:

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(+) [+] (-)

From the law that like charges repel and unlike charges attract.

The positive ball would go first to the negatively charged plate. After which, the ball would hold more negative charges overall than before.

Because the ball is now more negatively charged, it then travels towards the positive plate. In the same manner, the ball would transfer negative electrons to the positive plate.

So, when leaving the positive plate, the ball would be more positive and be drawn towards the negative plate once more. In doing so, it would make the negative plate more positive.

Then, the ball again holds more negative electrons and is drawn towards the positive plate once more.

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The ball hanging on the insulated thread becomes neutral too at this point.

Hope this Helps!!!

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