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iVinArrow [24]
3 years ago
7

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

Physics
2 answers:
konstantin123 [22]3 years ago
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]3 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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If an atom has 7 valence electrons, how many electrons does it need to gain to achieve a stable electron configuration?
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It only needs one more to make it stable 
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3 years ago
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Human blood has a density of approximately 1.05×10'3 kg/m'3 .Use this to estimate the difference in blood pressure between the b
Murljashka [212]

Answer:

formula

ΔP= PgΔh

Explanation:

Calculation.

ΔP = (1.05 x 10^3kg/m^3)(9.8m/s^2)(6ft equivalent to 1.8288m)

ΔP = 1.05 x 10^3 kg/m^3 x 9.8m/s^2.

x1.8288.

ΔP = 1.88 x 10^4 Pa.

8 0
3 years ago
Please help- it’s a 15 point change to my grade.
Harman [31]

Answer: Everything except heat and density

Explanation:

6 0
3 years ago
Isla made a diagram to compare X-rays and radio waves.
pickupchik [31]

A: Has a higher frequency --> X-rays

B: Travels at the same speed --> Both

C: Has a longer wavelength --> Radio waves

Explanation:

Electromagnetic waves are waves consisting of oscillations of electric and magnetic field in a direction perpendicular to the direction of motion of the wave.

Electromagnetic waves travel at the speed of light (c=3.0\cdot 10^8 m/s) in a vacuum, and they are transverse in nature (the direction of vibration is perpendicular to the direction of propagation).

Electromagnetic waves are classified into different types according to their wavelength and frequency. In order from shortest to longest wavelength (and so, from highest to lowest frequency, since frequency is inversely proportional to the wavelength), we have:

Gamma rays

X-rays

Ultraviolet radiation

Visible light

Infrared radiation

Microwaves

Radio waves

Therefore, we can no says that:

X-rays: has a higher frequency than Radio waves

Radio waves: have a longer wavelength than x-rays

Both: they travel at the same speed

Therefore, the correct pairing is:

A: Has a higher frequency --> X-rays

B: Travels at the same speed --> Both

C: Has a longer wavelength --> Radio waves

Learn more about electromagnetic waves:

brainly.com/question/9184100

brainly.com/question/12450147

#LearnwithBrainly

6 0
3 years ago
A 0.12 g honeybee acquires a charge of +24pC while flying. The earth's electric field near the surface is typically 100 N/C, dow
shusha [124]

Answer:

150000000

\dfrac{F_e}{F_g}=0.00000203873598369

49050000 N/C

Explanation:

q = Charge = 24 pC

m = Mass of honeybee = 0.12 g

E = Electric field = 100 N/C

g = Acceleration due to gravity = 9.81 m/s²

1\ C=6.25\times 10^{18}\ electrons

Number electrons is

n=24\times 10^{-12}\times 6.25\times 10^{18}\\\Rightarrow n=150000000

The number of electrons added or removed was 150000000

Force is given by

F_e=Eq\\\Rightarrow F_e=100\times 24\times 10^{-12}\\\Rightarrow F_e=2.4\times 10^{-9}\ N

The ratio is

\dfrac{F_e}{F_g}=\dfrac{2.4\times 10^{-9}}{0.12\times 10^{-3}\times 9.81}\\\Rightarrow \dfrac{F_e}{F_g}=0.00000203873598369

The ratio is \dfrac{F_e}{F_g}=0.00000203873598369

Balancing the forces we get

Eq=mg\\\Rightarrow E=\dfrac{mg}{q}\\\Rightarrow E=\dfrac{0.12\times 10^{-3}\times 9.81}{24\times 10^{-12}}\\\Rightarrow E=49050000\ N/C

The electric field required is 49050000 N/C

4 0
3 years ago
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