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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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7 0
3 years ago
4 A magnet can exert a force of attraction or a force of repulsion on another magnet.
GenaCL600 [577]

Answer:

Push -repulsion

Pull - attraction

Explanation:

When two magnets are brought together, a push happens when a force of repulsion is experienced where the magnets move away from each other. This means their polarity is the same and this will cause the magnet to push away from each other.

When two magnets are brought together , a pull happens when a force of attraction is experienced where the magnets move close to each other. This means their polarity is different and thus causes the magnets to pull closer to each other.

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3 years ago
Compare green and orange light from the visible spectrum. You are currently in a labeling module. Turn off browse mode or quick
77julia77 [94]

Green: nm 495–570. Yellow: nm 570–590. 590–620 nm for orange. Red: 620-750 nm (400–484 THz frequency)

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The forces of attraction and repulsion in liquids are comparable. Compared to the solid state, they move a little bit more. They then assume the shape of the container while still having a fixed capacity.

The attraction forces between the molecules in gases are quite weak. They move quite freely and grow in an effort to fill as much space as they can. Consequently, their volume and shape vary (adopt the shape of the container).

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6 0
2 years ago
A 20 μF capacitor initially charged to 30 μC is discharged through a 1.5 kΩ resistor. Part A How long does it take to reduce the
Natasha_Volkova [10]

Answer:

it will take 36.12 ms to reduce the capacitor's charge to 10 μC

Explanation:

Qi= C×V

then:

Vi = Q/C = 30μ/20μ = 1.5 volts

and:

Vf = Q/C = 10μ/20μ = 0.5 volts

then:

v = v₀e^(–t/τ)  

v₀ is the initial voltage on the cap  

v is the voltage after time t  

R is resistance in ohms,  

C is capacitance in farads  

t is time in seconds  

RC = τ = time constant  

τ = 20µ x 1.5k = 30 ms  

v = v₀e^(t/τ)  

0.5 = 1.5e^(t/30ms)  

e^(t/30ms) = 10/3  

t/30ms = 1.20397

t = (30ms)(1.20397) = 36.12 ms

Therefore, it will take 36.12 ms to reduce the capacitor's charge to 10 μC.

7 0
3 years ago
A hot–air balloon is moving at a speed of 10.0 meters/second in the +x–direction. The balloonist throws a brass ball in the +x–d
Ad libitum [116K]

Answer:

Option (D) is correct.

Explanation:

The balloon lands horizontally at a distance of 420 m from a point where it as released.

Velocity of air balloon along +X axis =10 m/s

velocity of ball=4 m/s along + X axis

the velocity of balloon gets added to the velocity of ball. So the resultant velocity of the balloon=10+4 = 14 m/s

time taken= 30 s

The distance traveled is given by d= v t

d= 14 (30)

d= 420 m

Thus the balloon lands horizontally at a distance of 420 m from a point where it as released.

6 0
3 years ago
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