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Inga [223]
3 years ago
10

A swimmer can swim at a velocity v in still water. She swims upstream a distance d against the current, which has a velocity u.

She then swims back to her starting point. A) how long does it take her to make the round-trip? B) what is her average speed for the trip? C) for what value of u u is her average speed the greatest?
Physics
1 answer:
xxTIMURxx [149]3 years ago
4 0
Upstream speed: v - u
Downstream speed: v + u
Total distance: 2d

A) total time = d / (v - u)  + d / (v + u)

B) Average speed = total distance / total time
V(av) = 2d / [ d / (v -  u) + d / (v + u) ]
V(av) = 2d / [d(v + u + v - u) / (v - u)(v + u) ]
V(av) = (v² - u²) / v

C) If u is 0, her average speed will be the greatest, equal to v
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3 years ago
1.2 m long spring with a spring force of 200.0 N/m, 12.0 kg mass is attached to it, find the length
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From Hooke's law, the length of the spring after extension and after the mass is attached is 1.788 meters. Option B is the answer.

<h3>HOOKE'S LAW</h3>

Hooke's law state that in an elastic material, the force applied is directly proportional to the extension provided that the elastic limit is not exceeded.

Given that a 12.0 kg mass is attached to 1.2 m long spring with a spring constant of 200.0 N/m.

The given parameters are:

  • mass m = 12kg
  • Initial length L_{1} = 1.2 m
  • Spring constant K = 200 N/m
  • Extension e = ?

According to Hooke's law

F = Ke

But F = mg

mg = Ke

Substitute all the parameters into the formula to get extension e

12 x 9.8 = 200e

e = 117.6 / 200

e = 0.588 m

The length of the spring after extension and after the mass is attached will be

L_{2} = L_{1} + e

L_{2} = 1.2 + 0.588

L_{2} = 1.788 m

Therefore, the correct answer is option B because the length of the spring after extension and after the mass is attached is 1.788 meters.

Learn more about Hooke's law here: brainly.com/question/12253978

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