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Monica [59]
3 years ago
14

A rock is dropped from rest. How fast is it going after it has been falling for 9.2 s?

Physics
1 answer:
natka813 [3]3 years ago
6 0

90.2 m/s. A rock dropped from rest after it has falling for 9.2 seconds will have a velocity of 90.2m/s.

The bodies left in free fall increase their speed (downwards) by 9.8 m/s² every second. The acceleration of gravity is the same for all objects and is independent of the masses of these. In the free fall the air resistance is not taken into account.

v = g*t

v = (9.8m/s²)(9.2s) = 90.2 m/s

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7nadin3 [17]

Answer:

<h2>C. <u>0.55 m/s towards the right</u></h2>

Explanation:

Using the conservation of law of momentum which states that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision.

Momentum = Mass (M) * Velocity(V)

BEFORE COLLISION

Momentum of 0.25kg body moving at 1.0m/s = 0.25*1 = 0.25kgm/s

Momentum of 0.15kg body moving at 0.0m/s(body at rest) = 0kgm/s

AFTER COLLISION

Momentum of 0.25kg body moving at x m/s = 0.25* x= 0.25x kgm/s

<u>x is the final velocity of the 0.25kg ball</u>

Momentum of 0.15kg body moving at 0.75m/s(body at rest) =

0.15 * 0.75kgm/s = 0.1125 kgm/s

Using the law of conservation of momentum;

0.25+0 = 0.25x + 0.1125

0.25x = 0.25-0.1125

0.25x = 0.1375

x = 0.1375/0.25

x = 0.55m/s

Since the 0.15 kg ball moves off to the right after collision, the 0.25 kg ball will move at <u>0.55 m/s towards the right</u>

<u></u>

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