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3 years ago

What is the relationship between temperature and altitude in the stratosphere? (2 points).

1 answer:

8 0

As altitude increases, temperature increases. The stratosphere is the part of the atmosphere that starts in the tropopause and ends in the estratopause. In the troposphere, the air is close to the Earth surface. The air surface can absorb more sunlight energy than the air, so the Earth surface heats the air. As you go higher, the distance to the Earth surface is higher, so the temperature is lower. The troposphere ends in the tropopause, where this trend changes. In the estratopause, there is a lot of ozone, which absorbs the dangerous UV radiation and converts into heat. That heat warms the air. So the air which is close to the estratopause is warm because of the heat released by the ozone reactions. The tropopause is far from the Earth surface and far from the ozone layer, that’s why it is cold. So the tropopause is cold and the estratopause is warm, which means: the air becomes warmer <span>as you rise above the tropopause until you get to the estratopause.</span>

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The mass of the earth is 5.98 1024 kg, the mass of the moon is 7.36 1022 kg, and the distance between the centers of the earth a

Setler [38]

Answer:

x_{cm} = 4.644 10⁶ m

Explanation:

The center of mass is given by the equation

$x_{cm}$ = 1 / $M_{total}$ ∑ $x_{i} m_{i}$

Where M_{total} is the total masses of the system, $x_{i}$ is the distance between the particles and $m_{i}$ is the masses of each body

Let's apply this equation to our problem

M = Me + m

M = 5.98 10²⁴ + 7.36 10²²

M = 605.36 10²² kg

Let's locate a reference system located in the center of the Earth

Let's calculate

x_{cm} = 1 / 605.36 10²² [Me 0 + 7.36 10²² 3.82 10⁸]

x_{cm} = 4.644 10⁶ m

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3 years ago

A small boat sailed straight north out of a harbor in strong east wind (blowing from west to east). After sailing for 120 minute

Amiraneli [1.4K]

it can be said that the speed of the east wind is

v=0.3608m/s

From the question we are told

A small boat sailed <u>straight </u>north out of a harbor in <em>strong </em>east wind (blowing from west to east).

After sailing for 120 minutes, it ended up hitting a buoy 60^\circ60 ∘ to the north-east of the harbor. If the straight-line distance between the buoy and the harbor is 3 km,

what is the speed of the east wind?.

<h3> the speed of the east wind</h3>

Generally the equation for the distance is mathematically given as

BA=3000sin60

BA=2598.07m

Therefore

the speed of the east wind

$V_w=\frac{BA}{120*60}\\\\V_w=\frac{2598.07}{120*60}$

v=0.3608

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2 years ago

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