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AURORKA [14]
3 years ago
10

Consider a hot-air balloon. The deflated balloon, gondola, and two passengers have a combined mass of 315 kg. When inflated, the

balloon contains 845 m^3 of hot air. Find the temperature (∘C) of the hot air required to lift the balloon off the ground. The air outside the balloon has a temperature of 20.2∘C and a pressure of 0.916×105 Pa. The pressure of the hot air inside the balloon is the same as the pressure of the air outside the balloon. The molar mass of air is 29 g/mol. Ignore the buoyant force on the gondola and the passengers. Hint: First find the density of the air outside the balloon. Then use Archimedes' principle to find the density required for the air inside the balloon. Then find the temperature of the air inside the balloon.
Physics
1 answer:
vladimir2022 [97]3 years ago
3 0

This will help you.

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All I can say is it is decreasing
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3 years ago
What is the density of an object that has a mass of 30 g and a volume of 20cm cubed/ to the third power?
Ksenya-84 [330]

Answer:

d= 1.5 g/cm3

Explanation:

datos

m= 30g

v= 20cm3

d=?

formula

d= m / v

solución

d= 30g / 20cm3 = 1.5g/cm3

6 0
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When an object like a tree is illuminated by the sun, and you are looking toward the tree, light rays leave the object _________
Dima020 [189]

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The correct answer is A. From every point on the surface of the tree, and in every direction

6 0
3 years ago
What happens to the magnetic domains in a material when the material is placed in a strong magnetic field?
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If it is diamagnetic then it magnetise opposite to magnetic field
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6 0
3 years ago
A gas is compressed from an initial volume of 5.75 L to a final volume of 1.23 L by an external pressure of 1.00 atm. During the
nataly862011 [7]

Answer:

<em>The internal energy change is 330.01 J</em>

Explanation:

Given

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v_{2} the final volume = 1.23 L

P_{e} is the external pressure = 1.00 atm

q the heat energy removed = -128 J (since is removed from the system)

expansion against a constant external pressure is an example of an irreversible pathway, here pressure in is greater than pressure out and can be obtained thus;

W = -P_{e}ΔV

W = -1.00 x(1.23 - 5.75)

W =  -1.00 x -4.52

W = 4.52 L atm

converting to joules we have

W = 4.52 L atm x 101.33 J/ L atm = 458.01 J

The internal energy change during compression can be calculated thus;

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ΔU = -128 J + 458.01 J

ΔU = 330.01 J

Therefore the internal energy change is 330.01 J

8 0
3 years ago
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