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umka21 [38]
3 years ago
5

In an experiment, an object is released from rest from the top of a building. Its speed is measured as it reaches a point that i

s a distance d from the point of release. If this distance was doubled, what would the new speed be, assuming air resistance is negligible?
Physics
1 answer:
NikAS [45]3 years ago
4 0

Answer:

so the speed will increase by 1.44 times then the initial speed if the distance is increased to double

Explanation:

As we know that the air friction or resistance due to air is neglected then we can use the equation of kinematics here

v_f^2 - v_i^2 = 2 a d

since we released it from rest so we have

v_i = 0

so here we have

v_f = \sqrt{2gd}

now if the distance is double then we have

v_f' = \sqrt{2g(2d)}

now from above two equations we can say that

v_f' = \sqrt2 v_f

so the speed will increase by 1.44 times then the initial speed if the distance is increased to double

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4 0
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The product nuclei may or may not be _______
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3 years ago
A 45 kg bear slides, from rest, 11 m down a lodgepole pine tree, moving with a speed of 5.8 m/s just before hitting the ground.
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Answer:

Explanation:

a )

change  in the gravitational potential energy of the bear-Earth system during the slide  = mgh

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b )

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