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3 years ago

How does the sun affect the water cycle?

2 answers:

7 0

The answer is B!

Reasoning:
- A is just wrong.
- B happens after the water has evaporated and cooled. The sun cannot cool water vapor. In fact, the sun can’t cool anything.
- C is also just plainly wrong because the sun cannot cool.
- Therefore, D is the answer.

If you could mark me as brainiest, that’d help me level up :D

Savatey [412]3 years ago

4 0

i think the answer is D.

heating surface water causes it to evaporate and enter the air as a gas.

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Charlotte throws a paper airplane into the air, and it lands on the ground. Which best explains why this is an example of projec

lubasha [3.4K]

Explanation:

yes it is the force that is subjected by the force of gravity only

4 0

3 years ago

Read 2 more answers

What best describes the motion of the medium particles in a longitudinal wave? The medium does not move. The medium moves in all

Citrus2011 [14]

In the motion of the medium particles in a longitudinal wave, the medium vibrates parallel to the direction of the wave.

<h3>What is a longitudinal wave?</h3>

A longitudinal wave is a wave that is transversing along the length. When the displacement of medium and travel of wave is the same in that condition wave is known as the longitudinal wave.

It requires some medium to travel. A mechanical and sound wave is an example of a longitudinal wave.

Hence in the motion of the medium particles in a longitudinal wave, the medium vibrates parallel to the direction of the wave.

To learn more about the longitudinal wave refer to the link;

brainly.com/question/8497711

3 0

2 years ago

The specialty of an athlete on the women's track team is the pole vault. She has a mass of 48 kg and her approach speed is 8.9 m

Lyrx [107]

Answer:

H = 3.9 m

Explanation:

mass (m) = 48 kg

initial velocity (initial speed) (U) = 8.9 m/s

final velocity (V) = 1.6 m/s

acceleration due to gravity (g) = 9.8 m/s^{2}

find the height she raised her self to as she crosses the bar (H)

from energy conservation, the change in kinetic energy = change in potential energy

0.5m(V^{2} - [test]U^{2}[/tex]) = mg(H-h)

where h = initial height = 0 since she was on the ground

the equation becomes

0.5m(V^{2} - [test]U^{2}[/tex]) = mgH

0.5 x 48 x (1.6^{2} - [test]8.9^{2}[/tex]) = 48 x 9.8 x H

-1839.6 = 470.4 H (the negative sign indicates a decrease in kinetic energy so we would not be making use of it further)

H = 3.9 m

4 0

3 years ago

A sample of helium gas has a volume of 900. milliliters and a pressure of 2.50 atm at 298 K. What is the new pressure when the t

Sladkaya [172]

Answer:

The new pressure = 5.64 atm.

Explanation:

Using The general gas equation,

P₁V₁/T₁ = P₂V₂/T₂..................... Equation 1

making P₂ the subject of the equation

P₂ = P₁V₁T₂/V₂T₁ ...................... Equation 2

Where P₁ = initial pressure, V₁ = initial Volume, T₁ = Initial temperature, P₂ = final pressure, V₂ = final volume, T₂ = final Temperature.

Given: P ₁= 2.5 atm, T₁ = 298 K, V₁= 900 milliliters, T ₂= 336 K, V₂ = 450 milliliters

Substituting these values into equation 2,

P₂ = (2.5×900×336)/(298×450)

P₂ = 756000/134100

P₂ = 5.64 atm.

Thus the new pressure = 5.64 atm.

5 0

3 years ago

A supersonic nozzle is also a convergent–divergent duct, which is fed by a large reservoir at the inlet to the nozzle. In the re

Lady_Fox [76]

Answer:

155.38424 K

2.2721 kg/m³

Explanation:

$P_1$ = Pressure at reservoir = 10 atm

$T_1$ = Temperature at reservoir = 300 K

$P_2$ = Pressure at exit = 1 atm

$T_2$ = Temperature at exit

$R_s$ = Mass-specific gas constant = 287 J/kgK

$\gamma$ = Specific heat ratio = 1.4 for air

For isentropic flow

$\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=T_1\times \frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=00\times \left(\frac{1}{10}\right)^{\frac{1.4-1}{1.4}}\\\Rightarrow T_2=155.38424\ K$

The temperature of the flow at the exit is 155.38424 K

From the ideal equation density is given by

$\rho_2=\frac{P_2}{R_sT_2}\\\Rightarrow \rho=\frac{1\times 101325}{287\times 155.38424}\\\Rightarrow \rho=2.2721\ kg/m^3$

The density of the flow at the exit is 2.2721 kg/m³

4 0

3 years ago