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Ostrovityanka [42]
3 years ago
15

You are working as a letter sorter in a U.S. Post Office. Postal regulations require that employees' footwear must have a minimu

m coefficient of static friction of 0.5 on a specified tile surface. You are wearing athletic shoes for which you do not know the coefficient of static friction. In order to determine the coefficient, you imagine that there is an emergency and start running across the room. You have a coworker time you, and find that you can begin at rest and move 4.38 m in 1.21 s. If you try to move faster than this, your feet slip. Assuming your acceleration is constant, does your footwear qualify for the postal regulation?
Physics
1 answer:
Serggg [28]3 years ago
7 0

Answer:

  μ = 0.66,    therefore if it compiesy with the regulations

Explanation:

Let's solve this exercise in part, let's start by finding with kinematics the acceleration of man

           y = v₀ t + ½ a t²

as it starts from rest the initial velocity is zero

          y = ½ a t²

          a = \frac{2y}{t^2}

          a =\frac{2 \ 4.38}{1.21^2}  

          a = 6.46 m / s²

Now let's use Newton's second law,

Axis y

         N- W = 0

         N = W

         N = m g

X axis

on this axis the man exerts a backward force and by the law of action and reaction the floor exerts a forward force of the same magnitude, this forward force is the friction force.

         fr = m a

     

the friction force has an expression

          fr = my N

let's substitute

      μ mg = m a

      μ = a / g

   

let's calculate

     μ = 6.46 / 9.8

     μ = 0.66

therefore if you comply with the regulations

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What would be the coefficient of performance if the refrigerator (operating between the same temperatures) was instead used as a
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Complete Question

A certain refrigerator, operating between temperatures of -8.00°C and +23.2°C, can be approximated as a Carnot refrigerator.

What is the refrigerator's coefficient of performance? COP

(b) What If? What would be the coefficient of performance if the refrigerator (operating between the same temperatures) was instead used as a heat pump? COP

Answer:

a

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b

  COP_1 = 9.49  

Explanation:

From the question we are told that

     The lower operation temperature of refrigerator is  T_1 =  -8.00^oC =  265 \  K

     The upper operation temperature of the refrigerator is   T_2 =  23.2 ^oC =  296.2 \  K

Generally the refrigerators coefficient of performance is mathematically represented as

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Generally if a refrigerator (operating between the same temperatures) was instead used as a heat pump , the coefficient of performance is mathematically represented as

            COP_1 =  \frac{T_2}{ T_2 - T_1}  

=>         COP_1 =  \frac{296.2}{ 296.2 - 265 }  

=>         COP_1 = 9.49  

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