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EleoNora [17]
3 years ago
6

A high diver of mass 74.0 kg jumps off a board 9.00 m above the water. If his downward motion is stopped 2.50 seconds after he e

nters the water, what average upward force did the water exert on him​
Physics
1 answer:
stiv31 [10]3 years ago
6 0

Answer:

1120 N

Explanation:

The velocity with which he hits the water can be found with kinematics:

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (-9.8 m/s²) (-9.00 m)

v = -13.3 m/s

Or it can be found with conservation of energy.

PE = KE

mgh = ½ mv²

v = √(2gh)

v = √(2 × -9.8 m/s² × -9.00 m)

v = -13.3 m/s

Sum of forces on the diver after he hits the water:

∑F = ma

F − mg = m Δv/Δt

F − (74.0 kg) (9.8 m/s²) = (74.0 kg) (0 m/s − (-13.3 m/s)) / (2.50 s)

F = 1120 N

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Spaceship 1 and spaceship 2 have equal masses of 200 kg. Spaceship 1 has a speed of 0 m/s, and spaceship 2 has a speed of 10 m/s
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Answer:

2000 kg m/s

Explanation:

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3 years ago
Find the work w1 done on the block by the force of magnitude f1 = 60.0 n as the block moves from xi = -3.00 cm to xf = 1.00 cm
Norma-Jean [14]
By definition, the work done by a force is given by:
 W1 = F1 * d

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 d: distance traveled.
 Substituting values we have:
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W1 = 60 * (0.01 - (-0.03))

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W1 = 60 * (0.04)

W1 = 2.40 J

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 the work w1 done on the block by the force of magnitude f1 = 60.0 n is:
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The number of cell phone users increases every year.

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