Suppose you spend 30.0 minutes on a stair-climbing machine, climbing at a rate of 85 steps per minute, with each step 8.00 inches high. If y

Question

Suppose you spend 30.0 minutes on a stair-climbing machine, climbing at a rate of 85 steps per minute, with each step 8.00 inches high. If you weigh 150 lb and the machine reports that 690 kcal have been burned at the end of the workout, what efficiency is the machine using in obtaining this result?

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Nem 4 years 2021-08-22T05:27:17+00:00 1 Answers 21 views 0

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  1. minhkhoi
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    2021-08-22T05:28:38+00:00
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    Answer:

    The efficiency of the machine  is  \eta =12%

    Explanation:

    From the question we are told that

        The time available to climb the stairs is  t = 30 \ minutes

        The rate at which the stairs is  climbed is  v = 85 \ steps /minute

         The height of each step is  h = 8.00 \ inch = 8 * (0.0254 \frac{m}{inh} ) = 0.2032 \ m

          The mass of the person is  m = 150 lb

          The amount of calories burned is  E = 690 kcal = 690 *1000 cal = 690000 * (4.186 J/cal) = 2888340 J

     

    Generally the workdone is taking a step is mathematically represented as

            W = mgh

    Here  g (acceleration due to gravity is  4.448\ N/lb)

    substituting values

             W = 150 * 4.44 * 0.2032

            W = 135.58 \ J

    Now the total workdone during the course of the workout is mathematically represented as

            W_T = W * v * t

    substituting values

            W_T = 135.58 * 85 * 30

            W_T = 345716.4 \ J

     The efficiency of the machine is mathematically represented as

                 \eta = \frac{W_T}{E} * \frac{100}{1}

    substituting values  

                 \eta = \frac{34716.4}{2888340} * \frac{100}{1}

                \eta =12%

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