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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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According to the nebular theory, what early event eventually led to the formation of our solar system?​
qaws [65]

Answer:Solar system formed about 4.6 billion year ago, when gravity pulled together low-density cloud of interstellar gas and dust (called a nebula)(movie). The Orion Nebula, an interstellar cloud in which star systems and possibly planets are forming. Initially the cloud was about several light years across.

Explanation:

4 0
2 years ago
The moon has a mass of 7.34 . 102 kg and a radius of 1.74 . 106 meters. If you have a mass of 66 kg,
Ierofanga [76]

Answer:

F=1.06\times 10^{-18}\ N

Explanation:

Given that,

Mass of the Moon, M=7.34\times 10^2\ kg

Mass of the person, m = 66 kg

The radius of Moon, r=1.74\times 10^6\ m

We need to find the force between the person and the Moon. The formula for the gravitational force is given by :

F=G\dfrac{Mm}{r^2}\\\\F=6.67\times 10^{-11}\times \dfrac{7.34\times 10^2\times 66}{(1.74\times 10^6)^2}\\\\=1.06\times 10^{-18}\ N

So, the required force is 1.06\times 10^{-18}\ N.

3 0
3 years ago
A machinist turns on the power on to a grinding wheel at time t= 0 s. The wheel accelerates uniformly from rest for 10 s and rea
Ugo [173]

Answer:

θt = 514.3 revolutions

Explanation:

(1)The wheel accelerates uniformly from rest for 10 s and reaches the operating angular speed of 58rad/s.

The uniformly accelerated circular movement  a circular path movement in which the angular acceleration is constant.

We apply the equations of circular motion uniformly accelerated

ωf = ω₀ + α*t  Formula (1)

θ = ω₀*t + (1/2)*α*t² Formula (2)

ωf² = ω₀² +2*α*θ Formula (3)

Where:

θ : angle that the body has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

t : time interval (s)

ω₀ : initial angular speed ( rad/s)

ωf : final angular speed ( rad/s)

Number of revolutions made by the wheel from t = 0 to t = 10 s

Data

ω₀ = 0

t = 10 s

ωf = 58 rad/s

We replace data in the formula (1) to calculate α

ωf = ω₀ + α*t

58 = 0 + α*(10)

α = 58 /10

α = 5.8 rad/s²

We replace data in the formula (2) to calculate θ

θ = ω₀*t + (1/2)*α*t²

θ = 0 + (1/2)*( 5.8)*(10)²

θ₁ = 290 rad

(2)The wheel is run at that angular velocity for 30 s, and then power is shut off.

The movement of the wheel is circular with constant angular speed and the formula to calculate θ is:

θ = ω*t

ω = 58 rad/s  , t= 30s

θ = (58 rad/s)*(30)

θ = (58 rad/s)*(30)

θ ₂= 1740 rad

(3)The wheel slows down uniformly at 1.4 rad/s² until the wheel stops.

ω₀ = 58 rad/s

α = -1.4 rad/s²

ωf = 0

We replace data in the formula (3) to calculate θ

(ωf)² = (ω₀)² + (2)*(α )*θ

0 = (58)² + (2)*(-1.4)*θ

(2)*(1.4)*θ = (58)²

θ = (58)² / (2.8)

θ = (58)² / (2.8)

θ₃ = 1201.42 rad

Total number of revolutions made by the wheel (θt)

θt =θ₁+θ₂+θ₃

θt  = 290 rad+ 1740 rad + 1201.42 rad

θt  = 3231.42 rad

1 revolution = 2π rad

θt = 3231.42 rad* ( 1revolution/2π rad)

θt = 514.3 revolutions

7 0
3 years ago
hree identical resistors are connected in series. When a certain potential difference is applied across the combination, the tot
pav-90 [236]

Answer:

The power dissipated if the three resistors were connected in parallel across the same potential difference is 405 W

Explanation:

Given;

three identical resistors connected in series

let the first resistor = R₁

let the second resistor = R₂

let the third resistor = R₃

Rt = R₁ + R₂ + R₃

Since the resistors are identical, thus, R₁ = R₂ = R₃ = R

Rt = 3R

Power is given as;

P = IV = V² / R

P = \frac{V^2}{R_t} = \frac{V^2}{3R} \\\\3P = \frac{V^2}{R} ------equation(i)

If the 3 identical resistor connection were changed to parallel, then the equivalent resistance in the circuit will be;

\frac{1}{R_t} = \frac{1}{R_1} +\frac{1}{R_2} + \frac{1}{R_3} \\\\But, R_1 = R_2 = R_3\\\\\\frac{1}{R_t} = \frac{1}{R} +\frac{1}{R} + \frac{1}{R} \\\\\frac{1}{R_t} =\frac{3}{R} \\\\R_t = \frac{R}{3} \\\\P = \frac{V^2}{R_t} = \frac{3V^2}{R} \\\\P_{parallel} = \frac{3V^2}{R} ---------equation (ii)\\\\From \ equation \ (i), 3P_{series} = \frac{V^2}{R}, Substitute \ this \ into \ equation \ (ii)\\\\P = 3(\frac{V^2}{R} )\\\\P = 3(3P)\\\\P_{parallel} = 9P_{series}\\\\P_{parallel} = 9(45)\\\\

P_{parallel} = 405 \ W

Therefore, the power dissipated if the three resistors were connected in parallel across the same potential difference is 405 W

4 0
3 years ago
Two uniform solid spheres of the same size, but different mass, are released from rest simultaneously at the same height on a hi
SpyIntel [72]

Answer:

options A and C

Explanation:

Since, the spheres are of same size and rotational speed of the sphere are not dependent on their masses. So, both the sphere will reach the bottom of the at the same time with the same speed. But their kinetic energies are different.

So, options A and C are correct.

4 0
3 years ago
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