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Vlad [161]
3 years ago
14

A weightlifter lifts a 125-kg barbell straight up 1.15 m in 2.5 s. What was the power expended by the weightlifter?

Physics
1 answer:
labwork [276]3 years ago
7 0

Answer:

560

Explanation:

Round your answer to two significant figures.

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Romashka-Z-Leto [24]

Answer:

F = \frac{-Gm_{1}m_{2} }{r^{2} }.

Explanation:

Gravitational force between two objects of masses m_{1},  m_{2} kept at a distance r is given by the formula

F = \frac{-Gm_{1}m_{2} }{r^{2} }

Here ,m_{1} = 2m

         m_{2} = \frac{m}{2}

         

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          F = \frac{-Gm_{1}m_{2} }{r^{2} }.

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Suspend a heavy weight by two pieces of rope. Tension is greatest when the ropes
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Assuming that the angle is the same for both ropes, then D. is the answer.  You have to consider also if the ropes are close together or far apart and if the force to move the object is in line with the ropes or perpendicular to them.
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What can you tell from comparing these waves? Please help
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5.0 kg, with a bullet of mass 0.1 kg. The target was mounted on
kari74 [83]

Answer:

306 m/s

Explanation:

Law of conservation of momentum

m1v1 + m2v2 = (m1+m2)vf

m1 is the bullet's mass so it is 0.1 kg

v1 is what we're trying to solve

m2 is the target's mass so it is 5.0 kg

v2 is the targets velocity, and since it was stationary, its velocity is zero

vf is the velocity after the target is struck by the bullet, so it is 6.0 m/s

plugging in, we get

(0.1 kg)(v1) + (5.0 kg)(0 m/s) = (0.1 kg + 5.0 kg)(6.0 m/s)

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8 0
3 years ago
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