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Romashka-Z-Leto [24]
3 years ago
8

If the effort distance = 8 meters and the resistance distance = 1/2 meter. How much force does this man have to apply to lift th

e 1 ton (2000 N) weight?
Physics
1 answer:
Delicious77 [7]3 years ago
5 0

Answer:

Explanation:

Effort x effort distance = load x resistance  distance

effort distance = 8 m ,

load = 2000N

resistance distance = 1/2 m = 0.5 m

Putting the values in the equation above

effort x 8m = 2000N x .5

effort = 2000 x 0.5 / 8

= 125 N

force required = 125 N .

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An 4.0-kg fish pulled upward by a fisherman rises 1.9 m in 2.4 s, starting
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Answer:

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Explanation:

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3 0
3 years ago
Read 2 more answers
8. A 2 kg flower pot weighing 20 N falls from a window ledge.
Alina [70]

The force of the air resistance is 4 N.

The given parameters;

  • mass of the flower pot, m = 2 kg
  • weight of the flower pot, W = 20 N

Let the air resistance = F

Apply Newton's second law of motion to determine the force of the air resistance acting upward to oppose the motion of the pot falling downwards.

\Sigma F = ma\\\\W - F = ma\\\\a = \frac{W - F}{m} \\\\8 = \frac{20 - F}{2} \\\\20 - F = 16\\\\F = 20 - 16\\\\F = 4 \ N

Thus, the force of the air resistance is 4 N.

Learn more here: brainly.com/question/19887955

8 0
3 years ago
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