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spin [16.1K]
3 years ago
10

Region that contains the majority of molecules in the atmosphere

Physics
1 answer:
Anna71 [15]3 years ago
4 0

Answer:

Trophosphere

Explanation:

The troposphere is the atmospheric layer closest to the planet and is characterized because contains the largest percentage of the mass of the total atmosphere.

On special characteristic is that in this layer the temperature and water vapor content decrease a lot respect to the altitude. Also on this layer the Water vapor is important in order to regulate the air temperature since on this zone we have absorption of the solar energy.

The troposphere contains almost all the water vapor in the atmosphere. And specially on the tropics we have an accumulation of the water vapour.

All weather phenomena occur within the troposphere. Tropos means "change" and Troposphere means "region of mixing".

Above this layer, we have the tropopause, ranges in height from 5 miles near the poles up to 11 miles above the equator. And the height depends of the seasons, with an special characteristic: the is highest height occurs in the summer and lowest height occurs in the winter.

The troposphere contains almost 75% of the mass of the entire atmosphere. The air on this layer is composed by 78% nitrogen, 21% oxygen and 1% is made of argon, water vapor, and carbon dioxide.

So for this reason this is the Region that contains the majority of molecules in the atmosphere.

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Answer:

1000 kgm²/s, 400 J

1000 kgm²/s, 1000 J

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Explanation:

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d = Diameter

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v = Velocity of astronauts = 2 m/s

Angular momentum of the system is given by

L=mvr+mvr\\\Rightarrow L=2mvr\\\Rightarrow L=2\times 100\times 2\times 2.5\\\Rightarrow L=1000\ kgm^2/s

The angular momentum of the system is 1000 kgm²/s

Rotational energy is given by

K=I\omega^2\\\Rightarrow K=\frac{1}{2}(mr^2)\left(\frac{v}{r}\right)^2\\\Rightarrow K=mv^2\\\Rightarrow K=100\times 2^2\\\Rightarrow K=400\ J

The rotational energy of the system is 400 J

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L_i=L_f\\\Rightarrow 2mv_ir_i=2mv_fr_f\\\Rightarrow v_f=\frac{v_ir_i}{r_f}\\\Rightarrow v_f=\frac{2\times 2.5}{0.5}\\\Rightarrow v_f=10\ m/s

Energy

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             W = -nRT ln(\frac{V_{2}}{V_{1}})

Putting the given values into the above formula as follows.

        W = -nRT ln(\frac{V_{2}}{V_{1}})

             = - 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})

             = -12280.82 \times ln (0.09)

             = -12280.82 \times -2.41

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or,         = 29.596 kJ       (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

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or,       P_{2} = \frac{P_{1}V_{1}}{V_{2}}

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          P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}

or,       P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}

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Therefore, required pressure is 28.7 atm.

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