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2 years ago

What is the main determining factor in defining boundaries between layers of earth's atmosphere?

2 answers:

5 0

The main determining factor in defining boundaries between layers of earth's atmosphere would be temperature changes in these layers. Temperature is one essential property that varies in the atmosphere. Based from this variation, the atmosphere is divided into four major layers and further to three smaller layers - troposphere, tropopause, the stratosphere, stratopause, the mesosphere, mesopause, and the thermosphere.The troposphere is the layer that is nearest to the surface of the Earth. It is the part where humans, plants and animals survive. Also, it is the warmest layer of the atmosphere. And as we go higher the atmosphere, the temperature would drop making it much cooler.

lakkis [162]2 years ago

4 0

Answer:

temperature changes.

Explanation:

Temperature changes with altitude determine the boundaries between layers of Earth’s atmosphere

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The circuit has a 3 volt EMF and two ohm resistors. How much power in watts does this circuit draw? A) 4.5 , B) 24, C) 1.13 D) 2

dlinn [17]

Answer:

P = 4.5 watts

Explanation:

Given that,

EMF of the circuit, E = 3 volt

The resistance of the resistors, R = 2 ohms

We need to find the power of this circuit. The relation between power, emf and resistance is given by the formula as follows :

$P=\dfrac{V^2}{R}$

Substitute all the values,

$P=\dfrac{3^2}{2}\\\\P=4.5\ W$

So, the power of this circuit is equal to 4.5 watts.

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3 years ago

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Anika asks Eva to roll a basketball and then a bowling ball to her. Which requires more force to roll, and why?

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The bowling ball will require more force to roll because it is more massive.

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2 years ago

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An adaptation of an animal in the desert might be to

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Keep cool by being active at night, whereas some other desert animals get away from the sun's heat by digging underground burrows.

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2 years ago

An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the lighter fragment is equal to 2.90 ✕ 1

motikmotik

Answer:

The speed of the heavier fragment is 0.335c.

Explanation:

Given that,

Mass of the lighter fragment $M_{l}=2.90\times10^{-28}\ kg$

Mass of the heavier fragment $M_{h}=1.62\times10^{-27}\ Kg$

Speed of lighter fragment = 0.893c

We need to calculate the speed of the heavier fragment

Let v is the speed of the second fragment after decay

Using conservation of relativistic momentum

$0=\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}-\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}$

$\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}=\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}$

$\dfrac{2.90\times10^{-28}\times0.893c}{\sqrt{1-(0.893)^2}}=\dfrac{1.62\times10^{-27}v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}$

$\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=\dfrac{2.90\times10^{-28}\times0.893c}{1.62\times10^{-27}\times0.45}$

$\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=0.355c$

$\dfrac{v_{2}}{1-\dfrac{v_{2}^{2}}{c^2}}=(0.355c)^2$

$\dfrac{1-\dfrac{v_{2}^2}{c^2}}{v_{2}^2}=\dfrac{1}{(0.355c)}$

$\dfrac{1}{v_{2}^2}-\dfrac{1}{c^2}=\dfrac{1}{(0.355c)^2}$

$\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}+\dfrac{1}{0.126c^2}$

$\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}(1+\dfrac{1}{0.126})$

$\dfrac{1}{v_{2}^2}=\dfrac{8.93}{c^2}$

$v_{2}^2=\dfrac{c^2}{8.93}$

$v_{2}=0.335c$

Hence, The speed of the heavier fragment is 0.335c.

7 0

3 years ago