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expeople1 [14]
3 years ago
13

A rope exerts a 280 N force while pulling an 80 Kg skier upward along a hill inclined at 12o. The rope pulls parallel to the hil

l. The coefficient of friction between the skier and the hill is 0.15. If the skier starts from rest, determine her speed after moving 100 m up the slope.
Physics
1 answer:
user100 [1]3 years ago
4 0

Answer:

The speed of the skier after moving 100 m up the slope are of V= 25.23 m/s.

Explanation:

F= 280 N

m= 80 kg

α= 12º

μ= 0.15

d= 100m

g= 9,8 m/s²

N= m*g*sin(α)

N= 163 Newtons

Fr= μ * N

Fr= 24.45 Newtons

∑F= m*a

a= (280N - 24.5N) / 80kg

a= 3.19 m/s²

d= a * t² / 2

t=√(2*d/a)

t= 7.91 sec

V= a* t

V= 3.19 m/s² * 7.91 s

V= 25.23 m/s

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Explanation:

Value of the cross-sectional area is as follows.

        A = 1.5 \times 2.30

           = 3.45 in^{2}

The given data is as follows.

          Allowable stress = 14,500 psi

          Shear stress = 7100 psi

Now, we will calculate maximum load from allowable stress as follows.

           P_{max} = \sigma_{a}A

                       = 14500 \times 3.45

                       = 50025 lb

Now, maximum load from shear stress is as follows.

           P_{max} = 2 \times \tau_{a} \times A

                      = 2 \times 7100 \times 3.45

                      = 48990 lb

Hence, P_{max} will be calculated as follows.

       P_{max} = min((P_{max})_{\sigma}, (P_{max})_{\tau})

                  = 48990 lb

Thus, we can conclude that the maximum permissible load P_{max} is 48990 lb.

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A 500 kg roller coaster moving at 15 m/s suddenly comes to a stop at the end of the ride. How much work energy was needed to sto
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If I rub a polythene piece with a wool , is there a transfer of mass from wool to polythene?​
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A 2 kg ice cube is released from point A and slides on a frictionless track as shown in the figure. Determine the speed of the c
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According to the conservation of energy

  • Potential energy at any given instance is equal to the Kinetic energy as energy can neither be created nor be destroyed

Mass is 2kg=m

#A

h=5m

  • PE=mgh

PE

  • 2(9.8)(5)
  • 10(9.8)
  • 98J

Now

  • KE=98J
  • 1/2mv²=98J
  • 1/2×2v²=98J
  • v²=98J
  • v=√98
  • v=9.4m/s

#B

h=3.2m

PE:-

  • 2(3.2)(9.8)
  • 6.4(9.8)
  • 62.7J

Now

  • KE=62.7J
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  • v²=62.7
  • v=√62.7
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#C

h=2m

PE

  • 2(2)(9.8)
  • 4(9.8)
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Now

  • v²=39.2
  • v=√39.2
  • v=6.3m/s
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2 years ago
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