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Vesna [10]
2 years ago
10

A 55 kg gymnast wedges himself between two closely spaced vertical walls by pressing his hands and feet against the walls. Part

a what is the magnitude of the friction force on each hand and foot? Assume they are all equal.
Physics
1 answer:
Aliun [14]2 years ago
5 0

solution:

Let frictional forces due to both hands and feets be "Ff" each(since its given that they all are equal), acting in upward direction( in opposite direction of supposed motion).\\
Then since there is no motion of gymnast thus net frictional force due to both hands and feets will be exactly balanced by the weight of the gymnast,\\  i.e\\
4f_{f}=weight =mg\\
f_{f}=\frac{55x9.8}{4}\\
=134.75N

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The correct answer is Option C) Sample C would be best, because the percentage of the energy in an incident wave that remains in a reflected wave from this material is the smallest.


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As shown in the figure below, Justin walks from the house to his truck on a windy day. He walks 20 m toward
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Complete Question

The complete question is shown on the first uploaded image

Answer:

The velocity is   v =0.333 \  m/s in positive x -direction

The speed is s = 0.733 \ m/s

Explanation:

From the question we are told that

The distance from the house to truck is  D =  20 m

  The distance traveled back to retrieve  wind-blown hat is  d =  15

  The distance from the wind-blown hat position too the truck is  k =  20  m

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Generally when calculating the displacement the Justin's backward movement to collect his wind - blown hat is taken as negative

Generally Justin's displacement is mathematically represented as

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=>    L  =  25 \ m

Generally the average velocity is mathematically represented as

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Generally the distance covered by Justin is mathematically represented as  

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=>      R =  20 + 15 +20

=>     R =  55 \  m

Generally Justin's average speed over a 75 s period is mathematically represented as

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=>         s = \frac{55}{ 75}

=>        s = 0.733 \ m/s

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