The physical fitness of athletes is measured by \"VO2max,\" which is the maximum

Question

The physical fitness of athletes is measured by "VO2max," which is the maximum volume of oxygen consumed by an individual during incremental exercise (for example, on a treadmill). An average male has a VO2max of 45 mLO2/kg body mass/min, but a world-class male athlete can have a VO2max reading of 88.0 mLO2/kg body mass/min.

Part A

Calculate the volume of oxygen, in mL, consumed in 1 hr by an average man who weighs 188 lbsand has a VO2max reading of 48.0 mL O2/kg body mass/min.

SubmitMy AnswersGive Up

Part B

If this man lost 16 lbs , exercised, and increased his VO2max to 65.0 mL O2/kg body mass/min, how many mL of oxygen would he consume in 1 hr?

The physical fitness of athletes is measured by "VO2max," which is the maximum volume of oxygen consumed by an individual during incremental exercise (for example, on a treadmill). An average male has a VO2max of 45 mLO2/kg body mass/min, but a world-class male athlete can have a VO2max reading of 88.0 mLO2/kg body mass/min.

Part A

Calculate the volume of oxygen, in mL, consumed in 1 hr by an average man who weighs 188 lbsand has a VO2max reading of 48.0 mL O2/kg body mass/min.

SubmitMy AnswersGive Up

Part B

If this man lost 16 lbs , exercised, and increased his VO2max to 65.0 mL O2/kg body mass/min, how many mL of oxygen would he consume in 1 hr?

Explanation / Answer

Just use conversion factors and dimensional analysis to get a final answer in mL/hour.

188 lbs x (1 kg / 2.2046 lbs) x (48 mL O2 / kg) x (60 min/ 1 hour) = 245595.57 or 2.4559557x105 mL in one hour.

B. Do the same for part B, but subtract the weight he lost (188-x) and change the volume.

(188 -16) lbs x (1 kg / 2.2046 lbs) x (65 mL O2 / kg) x (60 min/ 1 hour) = 304272.88 mL = 3.0427288 x 105 mL in one hour.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.