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Shtirlitz [24]
2 years ago
13

a 2.0 kg ball is dropped from a height of 20 m onto a soft surface and rebounds to a height of 5.0 m . what is the magnitude of

the impulse exerted on the ball by the floor?
Physics
1 answer:
Yuki888 [10]2 years ago
5 0

Based on the data provided, the impulse of the floor on the ball is 59.4 Ns.

<h3>What is the impulse of the floor on the ball?</h3>

Using the equation of motion to determine the velocity at the end of the fall

  • v^2 = u^2 + 2gh

Where v is velocity at the end of fall

u is initial velocity = 0

g is acceleration due to gravity = 9.81 m/s^2

h is height = 20

  • Taking downward velocity as negative and up as positive

v^2 = 0 + 2 (9.81)(20)

v^2 = 392.4

v = - 19.8 m/s

The velocity, v after bouncing is calculated also:

u = 0

g = 9.81 m/s^2

h = 5.0 m

v^2 = 0 + 2(9.81)(5)

v^2 = 98.1

v = 9.904 m/s

  • Impulse = change in momentum
  • Impulse = m(v- u)

Impulse = 2.0 × (9.9 -(-19.8)

Impulse = 59.4 Ns

Therefore, the impulse of the floor on the ball is 59.4 Ns.

Learn more about impulse at: brainly.com/question/904448

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Which of the following is same dimension quantity?
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Answer:

Find the dimension of each and every quantity in all the options to check whether they are the same or not. We can use any one formula of each identity to find its dimension.

Complete step by step solution:

To find the dimension of a quantity, we can use any formula related to that quantity but we will use the easiest ones to save time.

Force-

from Newton’s law of motion,

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Dimension of force =[M][LT−2]=[MLT−2]=[M][LT−2]=[MLT−2]

Work done-

W=F×sW=F×s

Dimension of work=[MLT−2][L]=[ML2T−2]=[MLT−2][L]=[ML2T−2]

Momentum-

p=mvp=mv

Dimension of momentum=[M][LT1]=[MLT−1]=[M][LT1]=[MLT−1]

Impulse-

I=F×tI=F×t

Dimension of impulse=[MLT−2][T]=[

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Explanation:

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field flow is directly proportionate to the charge found inside it, therefore if we place a Gaussian surface outside the plastic spherical shell.  the flow must be zero since the charge of the sphere is equal  induced in the shell, for which the net charge is zero. we see with this analysis that this shell meets the requirement to block the elective field

From the same Gaussian law it follows that if the sphere is not in the center, the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow , so no matter where the sphere is, the total induced charge is always equal to the charge on the sphere.

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