Physics College

1. A 1.30 kg ball strikes a wall with a velocity of -10.5 m/s. The ball bounces off with a velocity of 6.50 m/s. If the ball is in contact with the wall for 0.0210 seconds, what is the force exerted
on the ball by the wall?
O 1050 N
22.1 N
0 402 N
0 651 N
CAN SOMEONE PLS PLS ACTUALLY HELP ME

Answers

Answer 1

Let F be the magnitude of the force. The impulse of this force while the ball is in contact with the wall is

Ft = F (0.0210 s)

and this impulse is equal to the change in the ball's momentum,

m ∆v = (1.30 kg) (6.50 m/s - (-10.5 m/s)) = (1.30 kg) (17.0 m/s)

Solve for F :

F (0.0210 s) = (1.30 kg) (17.0 m/s)

F = (1.30 kg) (17.0 m/s) / (0.0210 s)

F ≈ 1050 N

Related Questions

You are standing 30 m due east of a 50 kg person who is running at a speed of 20 m/s due west. What is the magnitude of that person's angular momentum about you (in units of )

Answers

Hi there!

We can use the following equation for angular momentum:

[tex]\large\boxed{L = mrv}[/tex]

m = mass (kg)

r = distance from reference point (m)

v = velocity (m/s)

We can simply plug-and-chug the values provided in the question.

L = (50)(30)(20) = 30000 kgm²/s

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Answers

Answer:

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A. It would increase.

B. It would decrease.

C. It would remain constant.

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Answers

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an object is traveling with a constant velocity of 5m/s. How far will it have gone after 7s

Answers

Answer:

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Explanation:

Answer:

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Explanation:

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What is the force exerted on a selectively permeable membrane because water has moved from an area of higher concentration to an area of lower concentration of water?
Diffusion
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Answers

Question

what is the force exerted on a selectively permeable membrane beacuse water has moved from an area of higher concentration to an area of Lower concentration of water?

Answer

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Explanation

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