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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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<u>Explanation:</u>

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3 years ago
A block is held at rest against a wall by a force of magnitude F exerted at an angle theta from the horizontal, as shown in the
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Answers:

B.) F cos\theta=F_{n}

C.) F sin\theta=F_{g} \pm F_{f}

Explanation:

The image attached shows the way the force F is acting on the block. Now, if we draw a free body diagram of the situation and write the equations for the Net Force in X and Y, we will have the following:

Net Force in X:

-F_{n}+F cos\theta=0 (1)

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F_{n} is the Normal force

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Net Force in Y:

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Where:

F_{f} is the Friction force (it is expresed with the \pm sign because this force may be up or down, we cannot know because the block is at rest)

F_{g} is the gravity force

Rewrittin (1):

F cos\theta=F_{n} (3) This is according to option B

Rewritting (2):

F sin\theta=F_{g}\pm F_{f} (3) This is according to option C

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2 years ago
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Answer: find the answer in the explanation

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