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Novay_Z [31]
4 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]4 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]4 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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The change in potential energy when the block falls to ground is -480J.

The maximum change in kinetic energy of the ball is 480 J.

The initial kinetic energy of the ball is 0 J.

The final  kinetic energy of the ball is 0.148J.

The initial potential energy of the ball is 0.187 J.

The final  potential energy of the ball is 0 J.

The work done by the air resistance is 0.039 J.

<h3>Change in potential energy when the block falls to ground</h3>

ΔP.E = -mgh

ΔP.E = -Wh

ΔP.E = - 40 x 12

ΔP.E = -480 J

<h3>Maximum change in kinetic energy of the ball</h3>

ΔK.E = - ΔP.E

ΔK.E = - (-480 J)

ΔK.E = 480 J

<h3>Initial kinetic energy of the ball</h3>

K.Ei = 0.5mv²

where;

  • v is zero since it is initially at rest

K.Ei = 0.5m(0) = 0

<h3>Final kinetic energy</h3>

K.Ef =  0.5mv²

K.Ef = 0.5(0.0091)(5.7)²

K.Ef = 0.148 J

<h3>Initial potential energy of the ball</h3>

P.Ei = mghi

P.Ei = 0.0091 x 9.8 x 2.1

P.Ei = 0.187 J

<h3>Final potential energy</h3>

P.Ef = mghf

P.Ef = 0.0091 x 9.8 x 0

P.Ef = 0

<h3>Work done by the air resistance</h3>

W = ΔE

W = P.E - K.E

W = 0.187 J - 0.148 J

W = 0.039 J

Learn more about potential energy here: brainly.com/question/1242059

#SPJ1

<h3 />
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