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Novay_Z [31]
2 years ago
13

The temperature and pressure at the surface of Mars during a Martian spring day were determined to be -50 °C and 900 Pa, respect

ively. (a)Determine the density of the Martian atmosphere for these conditions if the gas constant for the Martian atmosphere is assumed to be equivalent to that of carbon dioxide. (b) Compare the answer from part (a) with the density of the Earth’s atmosphere during a spring day when the temperature is 18 °C and the pressure 101.6 kPa
Physics
2 answers:
Elza [17]2 years ago
6 0

Answer:

(a) The density of the Martian atmosphere is 0.021kg/m^3

(b) The density of the Martian atmosphere (0.021kg/m^3) is less than the density of the Earth's atmosphere (1.217kg/m^3)

Explanation:

(a) Density of Martian atmosphere (D) = P/RT

P = 900 Pa, R = 189 J/kgK, T = -50°C = -50+273 = 223K

D = 900/189×223 = 900/42,147 = 0.021kg/m^3

b) Density of Earth's atmosphere (D) = P/RT

P = 101.6kPa = 101.6×1000 = 101,600 Pa, R = 287 J/kgK, T = 18°C = 18+273 = 291K

D = 101,600/287×291 = 101,600/83,517 = 1.217kg/m^3

The density of Martian atmosphere is less than the density of the Earth's atmosphere

Sidana [21]2 years ago
4 0

Answer:

T = 273 + (-50) = 273 – 50 = 223 K

R = 188.82 J / kg K for CO2

Density (Martian Atmosphere) = P / RT = 900 / 188.92 x 223 = 900 / 42129.16 = 0.0213 kg / m^{3}

T = 273 +18 = 291 K, R = 287 J / kg k (for air) P = 101.6 k Pa = 101600 Pa

Density (Earth Atmosphere) = P / RT = 101600 / 287 x 291 = 1.216 kg / m^{3}

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You and your family take a trip to see your aunt who lives 100 miles away along a straight highway. The first 60 miles of the tr
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Answer:

51.2 mi/h

Explanation:

Total distance, d = 100 miles

First 60 miles with speed 55 mi/h

Next 40 miles with speed 75 mi/h

Time taken for first 60 miles, t1 = 60 / 55 = 1.09 h

Time taken for 40 miles, t2 = 40 / 75 = 0.533 h

Time spent to get stuck, t3 = 20 min = 0.33 h

Total time, t = t1 + t2 + t3 = 1.09 + 0.533 + 0.33 = 1.953 h

The average speed is defined as the ratio of total distance traveled to the total time taken.

Average speed = =\frac{100}{1.953}=51.2 mi/h

Thus, the average speed of the journey is 51.2 mi/h.

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2 years ago
Consider heat transfer between two identical hot solid bodies and their environments. The first solid is dropped in a large cont
sergeinik [125]
<h2>For Second Solid Lumped System is Applicabe</h2>

Explanation:

Considering heat transfer between two identical hot solid bodies and their environments -

  • If the first solid is dropped in a large container filled with water, while the second one is allowed to cool naturally in the air than for second solid, the lumped system analysis more likely to be applicable
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Biot number = the ratio of conduction resistance within the body to convection resistance at the surface of the body

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A batter swings at a baseball. The action force is the bat hitting the ball with a force of 5N. What is the reaction force?
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The ball hitting the bat
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When wool is rubbed with a balloon, the wool is left with a positive charge as electrons have travelled from the wool to the balloon which means the balloon now has a negative charge.

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A stone is thrown vertically upward with a speed of 15.5 m/s from the edge of a cliff 75.0 m high .
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a) 2.64 s

We can solve this part of the problem by using the following SUVAT equation:

s=ut+\frac{1}{2}at^2

where

s is the displacement of the stone

u is the initial velocity

t is the time

a is the acceleration

We must be careful to the signs of s, u and a. Taking upward as positive direction, we have:

- s (displacement) negative, since it is downward: so s = -75.0 m

- u (initial velocity) positive, since it is upward: +15.5 m/s

- a (acceleration) negative, since it is downward: so a= g = -9.8 m/s^2 (acceleration of gravity)

Substituting into the equation,

-75.0 = 15.5 t -4.9t^2\\4.9t^2-15.5t-75.0 = 0

Solving the equation, we have two solutions: t = -5.80 s and t = 2.84 s. Since the negative solution has no physical meaning, the stone reaches the bottom of the cliff 2.64 s later.

b) 10.4 m/s

The speed of the stone when it reaches the bottom of the cliff can be calculated by using the equation:

v=u+at

where again, we must be careful to the signs of the various quantities:

- u (initial velocity) positive, since it is upward: +15.5 m/s

- a (acceleration) negative, since it is downward: so a = g = -9.8 m/s^2

Substituting t = 2.64 s, we find the final velocity of the stone:

v = 15.5 +(-9.8)(2.64)=-10.4 m/s

where the negative sign means that the velocity is downward: so the speed is 10.4 m/s.

c) 4.11 s

In this case, we can use again the equation:

s=ut+\frac{1}{2}at^2

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s is the displacement of the package

u is the initial velocity

t is the time

a is the acceleration

We have:

s = -105 m (vertical displacement of the package, downward so negative)

u = +5.40 m/s (initial velocity of the package, which is the same as the helicopter, upward so positive)

a = g = -9.8 m/s^2

Substituting into the equation,

-105 = 5.40 t -4.9t^2\\4.9t^2 -5.40 t-105=0

Which gives two solutions: t = -5.21 s and t = 4.11 s. Again, we discard the first solution since it is negative, so the package reaches the ground after

t = 4.11 seconds.

5 0
2 years ago
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