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saw5 [17]
2 years ago
12

A less than youthful 82.6 kg physics professor decides to run the 26.2 mile (42.195 km) Los Angeles Marathon. During his months

of training, he realizes that one important component in running a successful marathon is carbo-loading, the consumption of a sufficient quantity of carbohydrates prior to the race that the body can store as glycogen to burn during the race. The typical energy requirement for runners is 1 kcal/km per kilogram of body weight, and each mole of oxygen intake allows for the release of 120 kcal of energy by oxidizing (burning) glycogen.
(a) If the professor finishes the marathon in 4:45:00 h, what is the professor's oxygen intake rate, in liters per minute, during the race if he metabolizes all of the carbo-loaded glycogen during the race and the ambient temperature is 21.5°C? 2.28 Read the problem statement again carefully. Is the air at standard temperature and pressure during the marathon? How would this affect the volume of 1 mol of oxygen? L/min
(b) The human body has an efficiency of 25.0%. Only 25.0% of the energy released from oxidizing glycogen is used as macroscopic mechanical energy, and the remaining 75.0% is used for body processes such as pumping blood and respiration, and then leaves the body through the skin via radiation, evaporative cooling, and other processes. What is the average mechanical power (in W) generated by the professor during the run? 197.561 What is the total energy required by the professor during the run? How efficient is the human body, and how long did the race last? W
(c) What is the change in entropy (in J/K) of the professor's body if his core temperature has risen to 38.3°C during the run and his skin temperature is at 36.0°C during the marathon? J/K
(d) What is the change of the entropy (in J/K) of the air surrounding the professor during the race if the ambient temperature remains constant at 21.5°C? J/K
Physics
1 answer:
o-na [289]2 years ago
4 0
Mem me e m even have. Jags. Shah. Shiv side esicjm is n meh dish so do indbbd
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Answer: C. the rod gains mass and the fur loses mass.

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4 0
3 years ago
The two types of waves are ……………………………………wave and …………………………
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Answer:

longitudinal and transverse.

Explanation:

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4 0
2 years ago
A spacecraft is moving past the earth at a constant speed of 0.60 times the speed of light. The astronaut measures the time inte
Afina-wow [57]

Answer:

the time interval that an earth observer measures is 4 seconds

Explanation:

Given the data in the question;

speed of the spacecraft as it moves past the is 0.6 times the speed of light

we know that speed of light c = 3 × 10⁸ m/s

so speed of spacecraft v = 0.6 × c = 0.6c

time interval between ticks of the spacecraft clock Δt₀ = 3.2 seconds

Now, from time dilation;

t = Δt₀ / √( 1 - ( v² / c² ) )

t = Δt₀ / √( 1 - ( v/c )² )

we substitute

t = 3.2 / √( 1 - ( 0.6c / c )² )

t = 3.2 / √( 1 - ( 0.6 )² )

t = 3.2 / √( 1 - 0.36 )

t = 3.2 / √0.64

t = 3.2 / 0.8

t = 4 seconds

Therefore, the time interval that an earth observer measures is 4 seconds

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3 years ago
An observer sees a flower pot sail up and then back down past a window 2.45 m high. If the total time the pot is in the sight is
Fynjy0 [20]

Answer: 2.55meter

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S{hieght} = U*t + {g*t²}/2

Where U is initial velocity =0m/s

g is acceleration due to gravity 10m/s²

t is time 1secs

So we have,

hieght = 0 + {g*t²}/2

hieght = {10*(1)²}/2

Total hieght travelled is 10/2

Which is 5 meter.

But we are asked to find the hieght above the window which as a hieght of 2.45meter.

So,

hieght above window would be

{5 - 2.45}meter

Which is 2.55 meter.

8 0
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Answer:B. Merkel Cells

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