Asked by Anonymous
a 70-kg man climbs a mountain 1200 m high (measured from the base) in 4 h and uses 9.8 kcal/min. calculate his power consumption in watts. what is his power output in useful work? what was the efficiency of this man during the climb?
Answers
Answered by
Elena
9.8 kcal min = 41 030.64 J/min
41 030.64 J/min•60 • 4 = 9847353.6 J.
Useful work
mgh = 70•9.8•1200 = 823200 J.
mgh/t=823200/4•60 = 3430 J/min
η =(3430/41030.64) •100% = =0.086%
41 030.64 J/min•60 • 4 = 9847353.6 J.
Useful work
mgh = 70•9.8•1200 = 823200 J.
mgh/t=823200/4•60 = 3430 J/min
η =(3430/41030.64) •100% = =0.086%
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