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crimeas [40]
3 years ago
15

A 51-g rubber ball is released from rest and falls vertically onto a steel plate. The ball strikes the plate and is in contact w

ith it for 0.5 ms.
The ball rebounds elastically and returns to its original height. The time interval for the round trip is 3.00 s. What is the magnitude of the
average force that the plate exerted on the ball?
2490 N
1500 N
2000 N
3500 N
3000 N
Physics
1 answer:
Simora [160]3 years ago
4 0

Answer:

Last option 3000N

Explanation:

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3 years ago
A student falls off a cliff into the lake 54.0 m below. What is the final velocity of the student?
Vladimir79 [104]

Answer:

v_{y} = -32.53 m / s

this velocity is directed downwards

Explanation:

This is a free fall exercise, let's use the expression

         v_{y}^{2} = v_{oy}^{2} + 2 g (y -yo)

where we are assuming that there is friction with the air, as the body falls its initial velocity is zero

         v_{oy} = √ 2g (y - y₀)

let's calculate

         v_{y} = √ (2 9.8 (0-54.0))

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Q. No. 9 A body falls freely from the top of a tower and during the last second of its fall, it falls through 25m. Find the heig
HACTEHA [7]

Answer:

45.6m

Explanation:

The equation for the position y of an object in free fall is:

y=-\frac{1}{2} gt^2+v_0t+y_0

With the given values in the question the equation has one unknown v₀:

v_0=\frac{y-y_0}{t}+\frac{1}{2}gt

Solving for t=1:

1) v_0=y-y_0+\frac{g}{2}

To find the hight of the tower you can use the concept of energy conservation:

The energy of the body 1 sec before it hits the ground:

2) E=\frac{1}{2}m{v_0}^2+mgy_0

If h is the height of the tower, the energy on top of the tower:

3) E=mgh

Combining equation 2 and 3 and solving for h:

4) h=\frac{{v_0}^2}{2g}+y_0

Combining equation 1 and 4:

h=\frac{{(y-y_0+\frac{g}{2}})^2}{2g}+y_0

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