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raketka [301]
3 years ago
8

If one were 230 kilometers above sea level, one would be in the part of the atmosphere known as the ________.

Physics
1 answer:
stellarik [79]3 years ago
8 0
The Earth's atmosphere consists of 6 layers going up. 

Troposhere is the closest layer. From the Earth's surface it extends up into the atmosphere about 8-15 meters.

Stratosphere extends above the troposphere about 50 km from the ground. 

Above the stratosphere is the Mesosphere which goes up to 85 km above the surface of the Earth.

Then we have the Thermosphere which goes up to 500 to 1,000 km above ground. 

The Exosphere, which extends from the thermosphere to about 100,000km to 190,000 km.

The ionosphere is a layer that is not exactly as distinct as the others. It overlaps the mesosphere and exosphere. 

Now if you were 230 kilometers above sea level, then you would be in the 4th layer of the atmosphere, or the Thermosphere because it starts at about 86 km above sea level and extends to about 500 to 1,000 km above sea level.
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A 1000 kg car moving a 10 m/s collides with a stationary 2000 kg truck. The two vehicles interlock as a result of the collision.
IgorLugansk [536]

Answer:

v₃ = 3.33 [m/s]

Explanation:

This problem can be easily solved using the principle of linear momentum conservation. Which tells us that momentum is preserved before and after the collision.

In this way, we can propose the following equation in which everything that happens before the collision will be located to the left of the equal sign and on the right the moment after the collision.

(m_{1}*v_{1})+(m_{2}*v_{2})=(m_{1}+m_{2})*v_{3}

where:

m₁ = mass of the car = 1000 [kg]

v₁ = velocity of the car = 10 [m/s]

m₂ = mass of the truck = 2000 [kg]

v₂ = velocity of the truck = 0 (stationary)

v₃ = velocity of the two vehicles after the collision [m/s].

Now replacing:

(1000*10)+(2000*0)=(1000+2000)*v_{3}\\v_{3}=3.33[m/s]

7 0
3 years ago
A glider with mass 0.24 kg sits on a frictionless horizontal air track, connected to a spring of negligible mass with force cons
shepuryov [24]

Answer:

v=2.556m/s

Explanation:

From the conservation of mechanical energy

K_{E1}+U_1=K_{E2}+U_2

\frac{1}{2}m*v_1^2+\frac{1}{2}*K*x_1^2=\frac{1}{2}m*v_2^2+\frac{1}{2}*K*x_2^2

x_2=0.08m

v_1=0 m/s

Solve to velocity v2

m*v_2^2=k*x_1^2-k*x_2^2

v^2=\frac{k}{m}*(x_1^2-x_2^2)

v^2=\frac{5.5N/m}{0.24kg}*(0.54m^2-0.080^2)

v=\sqrt{6.54m^2/s^2}=2.556m/s

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3 years ago
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Answer:

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Explanation:

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