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2 years ago

Why are water vapor and aerosols important constituents of earth's atmosphere? choose all that apply.

Physics

1 answer:

RoseWind [281]2 years ago

7 0

They are important because numerous serve as areas where water vapour can condense.

<h3></h3><h3>What is condensation?</h3>

The transformation of water vapour into liquid is known as condensation. The process is the opposite of evaporation, in which liquid water turns into a vapour. Either the air is chilled to its dew point or it gets too saturated with water vapour to retain any more water, causing condensation to occur.

to learn more about condensation go to -

brainly.com/question/1268537

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The mean diameters of planets A and B are 8.1 × 103 km and 1.4 × 104 km, respectively. The ratio of the mass of planet A to that

lisabon 2012 [21]

Answer:

dA/dB = 4.955

Approximately, the ratio is 5/1

(Where dA is mean density for planet A while dB is mean density for planet B)

Explanation:

Mass of A = mA

Mass of B = mB

mA/mB = 0.96

Mean radius for A = mA = (8.1 × 10^3)/2 = 4.05 × 10^3 km

Mean radius for B = mB = (1.4 × 10^4)/2

= 7×10^3km

Density = mass/volume

Volume of a sphere = 4/3Πr3

Mean volume for A = (4/3) × Π × (4.05 × 10^3)^3

= 2.784 × 10^11 km3

Mean volume for B = 4/3×Π×(7×10^3)^3

= 1.437 × 10^12km3

Since m/v = d ( where m = mass, v = volume and d = density)

mA = 2.784 × 10^11 km3 × dA ...equation 1

mB= 1.437 × 10^12km3 × dB... equation 2

but mA/mB= 0.96

mA = 0.96 × mB

substitute for mA in equation 1

0.96 × mB = 2.784 × 10^11 x dA equation 3

Substitute for mB in equation 3..

(refer to equation 2)

0.96×1.437×10^12 × dB = 2.784 × 10^11 × dA .....equation 4

divide through by the coefficient of dA

dA = (0.96×1.437×10^12×dB)/(2.784 × 10^11)

divide through by dB

dA/dB = 4.955

therefore, the ratio of dA to dB is 5/1

Therefore, the mean density of A is almost five times that of B

7 0

3 years ago

9. Complete the following statement:

mario62 [17]

Answer:

E) the flow of energy due to a temperature difference.

Explanation:

Heat can be described as the flow of energy due to a temperature difference.

Which is expressed mathematically as;

H = MCΔT

Where;

H is the quantity of heat in a body, measured in Joules

M is the mass of the body, measured in kg

C is the specific heat capacity of the body, J/kg.K

ΔT is change in temperature or temperature difference.

So, heat energy in any system flows from a hotter region to a colder region due to temperature difference.

E) the flow of energy due to a temperature difference.

7 0

3 years ago

A hand pump is being used to inflate a bicycle tire that has a gauge pressure of 59.0 lb/in2. If the pump is a cylinder of lengt

bagirrra123 [75]

Answer:

∴The air cannot be made to flow in with the given pump at the given conditions.

Explanation:

Given:

gauge pressure of bicycle tyre, $P_g=59\ lb.in^{-2}$

length of cylinder of the pump, $l=17.4\ in$

area of the the cylinder of the pump, $a= 3\ in^2$

we have the density of air at STP, $\rho=4.4256\times 10^{-5}\ lb.in^{-2}$

The piston must be pushed more than the pressure inside the tyre:

$P_g=\rho\times V\div a$

$59=4.4256\times 10^{-5}\times a\times h\div a$

$h=13.33\times 10^5\ in$

∴The air cannot be made to flow in with the given pump at the given conditions.

4 0

3 years ago

what frequency of light must an electron in hydrogen absorb to jump from the n=2 state to the n=5 state? (a) 2.86 Hz (b) 4.08 Hz

Ostrovityanka [42]

Answer:

Explanation:

Find the level energy of n=2 and n=5, using the formula:

$E = -E_0/n^2$

where $E_0=13.6eV$

$E_2 =\frac{-13.6}{2^2}=-3.4eV$

$E_5 =\frac{-13.6}{5^2}=-0.544eV$

To jump from n=2 to n=5 the electron absorbs a photon with energy equal to $(-0.544) - (-3.4)=2.856eV$ , using the next formula to find specific wavelength $\lambda$ to that energy