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EleoNora [17]
3 years ago
8

A hot-air balloon plus cargo has a mass of 308 kg and a volume of 2910 m3 on a day when the outside air density is 1.22 kg/m3. T

he balloon is floating at a constant height of 9.14 m above the ground.
Required:
What is the density of the hot air in the balloon?
Physics
1 answer:
lesya [120]3 years ago
6 0

9514 1404 393

Answer:

  1.114 kg/m³

Explanation:

The total mass of the air in the balloon and the balloon + cargo will be the mass of the displaced air. If d is the density of the air in the balloon, then we have ...

  2910d +308 = 2910×1.22

Solving for d, we find ...

  2910d = 2919(1.22) -308

  d = 1.22 -308/2910

  d ≈ 1.114 . . . kg/m³

The density of the hot air is about 1.114 kg/m³.

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Answer:

Part A

Coriolis effect is used to describe how objects which are not fixed to the ground are deflected as they travel over long distances due to the rotation of the Earth relative to the 'linear' motion of the objects

Due to the Coriolis effect the wind flowing towards the Equator from high pressure belts in the subtropical regions in both the Northern and Southern Hemispheres are deflected towards the western direction because the Earth rotates on its axis towards the east

Part B

In the Northern Hemispheres, the winds are known as northeasterly trade winds and in the Southern Hemisphere, they are known as the southeasterly trade wind. Therefore, Coriolis effect has the same effect on the direction of the Trade Winds in the Southern Hemisphere as it does in the Northern Hemisphere

Explanation:

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Review. As an astronaut, you observe a small planet to be spherical. After landing on the planet, you set off, walking always st
anyanavicka [17]

To find the mass of the planet we will apply the relationship of the given circumference of the planet with the given data and thus find the radius of the planet. From the kinematic equations of motion we will find the gravitational acceleration of the planet, and under the description of this value by Newton's laws the mass of the planet, that is,

The circumference of the planet is,

\phi = 25.1m

Under the mathematical value the radius would be

\phi = 2\pi r

r = \frac{25}{2\pi}

r = 3.9788km

Using second equation of motion

x = \frac{1}{2} at^2

Replacing the values given,

1.4 = \frac{1}{2} a (29.2)^2

Rearranging and solving for 'a' we have,

a = 0.003283m/s^2

Using the value of acceleration due to gravity from Newton's law we have that

a = \frac{GM}{r^2}

Here,

r = Radius of the planet

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M = 7.79201*10^{14}kg

Therefore the mass of this planet is 7.79201*10^{14}kg

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