The record time for a Tour de France cyclist to ascend the famed 1100-m-high Alpe d'Huez was 37.5 min, set by Marco Pantani in 1997. Pantani and his bike had a mass of 65 kg. Assume the body works with 25 % efficiency.

A)How many Calories did he expend during this climb?

B)What was his average metabolic power during the climb?

Marco Pantani set the ‘record’ when he averaged 6.63 W/kg
during the 1997 Tour. So the amount of calorie marco pantani expended can be
calculated by multiplying it by its mass and dividing by its efficiency

Calorie = (6.63 W/ kg) (65 kg) ( 1 cal/s/ 4.19 W) ( 37.5
min)( 60 s/ 1 min) / 0.25

Calorie = 925, 668 cal

Answer Link

obasola1 obasola1 · Answer 2 · 2019-12-12T07:01:19-05:00

Answer:

Pe=2682275.334J ,51012j/s

Explanation:

The record time for a Tour de France cyclist to ascend the famed 1100-m-high Alpe d'Huez was 37.5 min, set by Marco Pantani in 1997. Pantani and his bike had a mass of 65 kg. Assume the body works with 25 % efficiency

Definition of terms

potential energy is energy stored due to position

power is rate of doing work

mass is the quantity of mater in a body,65kg

potential energy of the cyclist will be

Pe=mgh

m=65kg

g=9.81m/s^2

h=1100

the body works with 25% efficincy

.25P=659.811100

Pe=701415/0.25

Pe=11222640J

recall

4184 Joules = 1000 calories

Pe=11222640/4.184

Pe=2682275.334 calories. the cyclist will expend tat amount of calories during the climb

2.Power=energy/time

time= 37.5 60=2250s

power=11222640J/2250s

power=51012j/s

His average metabolic power during the climb is 51012j/s