Scenario 2: Use the following information to answer questions 3 and 4: Your client, Jim, is interested in weight control. He weighs 75

Question

Scenario 2: Use the following information to answer questions 3 and 4:
Your client, Jim, is interested in weight control. He weighs 75kg.
3. If Jim walks 3.3 mph (0% grade), how long must he walk to expend 300 kcal total?
A. 52 min
B. 42 min
C. 65 min
D. 99 min
4. If Jim exercises at an intensity of 6 kcal/min, what is the leg ergometer work rate?
A. 47 watts
B. 90 watts
C. 61 watts
D. 71 watts

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Thái Dương 5 years 2021-08-31T16:47:21+00:00 1 Answers 57 views 0

Answers ( )

  1. Gerda
    0
    2021-08-31T16:49:20+00:00
    Reply

    Answer:

    A. 52 min

    .A. 47 watts

    Explanation:

    Given that;

    jim weighs 75 kg

    and he walks 3.3 mph; the objective here is to determine how long must he walk to expend 300 kcal.

    Using the following relation to determine the amount of calories burned per minute while walking; we have:

    \dfrac{MET*weight (kg)*3.5}{200}

    here;

    MET = energy cost of a physical activity for a period of time

    Obtaining the data for walking with a speed of 3.3 mph From the  standard chart for MET, At 3.3 mph; we have our desired value to be 4.3

    However;

    the calories burned in a minute = \dfrac{4.3*75 (kg)*3.5}{200}

    = 5.644

    Therefore, for walking for 52 mins; Jim  burns approximately 293.475 kcal which is nearest to 300 kcal.

    4.

    Given that:

    mass m = 75 kg

    intensity = 6 kcal/min

    The eg ergometer work rate = ??

    Applying the formula:

    V_O_2 ( intensity ) = ( \dfrac{W}{m}*1.8)+7

    where ;

    V_O_2 ( intensity ) = \dfrac{1 \ kcal min^{-1}*10^{-3}}{5}

    V_O_2 ( intensity ) = \dfrac{6*1 \ kcal min^{-1}*10^{-3}}{5}

    V_O_2 ( intensity ) = 0.0012

    ∴0.0012 = (\dfrac{W}{75}*1.8)+7 \\ \\ W = \dfrac{0.0012-7}{1.8}*75 \\ \\ W = \dfrac{7*75}{1.8} \\ \\ W = 291.66 \ kg m /min

    Converting to watts;

    Since;  6.118kg-m/min is =  1 watt

    Then 291.66 kgm /min will be equal to 47.67 watts

    ≅ 47 watts

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