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3 years ago

How does water get up to the atmosphere, and how does it get back down to earth surface

Physics

1 answer:

IrinaK [193]3 years ago

6 0

Answer:

Water gets up to the Earth's atmosphere by evaporating from a body of water, which is then they become water vapor. It returns back to the surface by returning back to its water state and falling back down (as rain). The water vapor turns into clouds (clouds are really just water droplets), and when it cannot hold anymore waters, it disperses all the water (by raining).

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Moving force of air flows through areas of high pressure to areas of low pressure

vazorg [7]

The answer is true i hope this helps

7 0

3 years ago

Read 2 more answers

In which does sound travel fastest? Question 8 options: solids liquids gases plasma.

Kazeer [188]

Sound travels fastest in solids. A sound is a vibration that travels in a longitudinal direction through a medium in the form of a mechanical wave.

<h3>What is sound?</h3>

A sound is a vibration that travels in a longitudinal direction through a medium in the form of a mechanical wave.

It can propagate through a solid, a liquid, or a gas as its medium. Solids go the fastest, liquids are slower, and gases are the slowest.

Sound travels the quickest through a solid because the molecules are packed together densely, allowing sound waves to leap from one molecule to the next more quickly.

Because the molecules in solids are packed the tightest, sound travels the quickest through them, whereas sound travels the slowest through gases.

To learn more about the sound refer to the link;

brainly.com/question/927975

4 0

3 years ago

Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its te

Maksim231197 [3]

Answer:

Tt = 70 + 135e^-0.031t

13 minutes

Explanation:

Given that :

Initial temperature, Ti = 205°

Temperature after 2.5 minutes = 195°

Temperature of room, Ts= 70

Using the relation :

Tt = Ts + Ce^-kt

Temperature after time, t

When freshly poured, t = 0

205 = 70 + Ce^-0k

205 = 70 + C

C = 205 - 70 = 135°

T after 2.5 minutes to find proportionality constant, k

Tt = Ts + Ce^-kt

195 = 70 + 135e^-2.5k

125 = 135e^-2.5k

125 / 135 = e^-2.5k

0.9259 = e^-2.5k

Take In of both sides :

−0.076989 = - 2.5k

k = −0.076989 / - 2.5

k = 0.031

Equation becomes :

Tt = 70 + 135e^-0.031t

t when Tt = 160

160 = 70 + 135e^-0.031k

90 = 135e^-0.031t

90/135 = e^-0.031t

0.6667 = e^-0.031t

In(0.6667) = - 0.031t

−0.405465 = - 0.031t

t = 0.405465/ 0.031

t = 13.071

t = 13 minutes

8 0

3 years ago

(ii) Describe how the acceleration of the train at time t = 100 s differs from the acceleration

quester [9]

Explanation:

Acceleration is the rate of change of velocity with time. When acceleration increases a body moves a faster velocity.

In the graph acceleration at time t= 100s is rapidly increasing.
At t = 20s, the acceleration of the body is getting started up.

A vehicle at time 100s will have a faster velocity compared to one at t = 20s

7 0

3 years ago

Three point charges are located on the x-axis. The first charge, q1 = 10 μC, is at x = -1.0 m. The second charge, q2 = 20 μC, is

victus00 [196]

Answer:

<em>3.15 N towards the positive x-axis</em>

Explanation:

first charge has charge q1 = 10 μC = 10 x 10^-6 C

second charge has charge q2 = 20 μC = 20 x 10^-6 C

third charge has charge q3 = -30 μC = -30 x 20^-6 C

According to coulomb's law, force between two charged particle is given as

F = $\frac{-kQq}{r^2}$

Where

F is the force between the charges

k is Coulomb's constant = 9 x 10^9 kg⋅m^3⋅s^−2⋅C^−2.

Q is the magnitude of one charge

q is the magnitude of the other charge

is the distance between these two charges

For the force on q2 due to q1,

distance r between them = 0 - (-1.0) = 1 m

F = $\frac{-9*10^{9}*10*10^{-6}*20*10^{-6}}{1^2}$ = -1.8 N (the negative sign indicates a repulsion on q2 towards the positive x-axis)

For the force on q2 due to q3,

distance between them = 2.0 - 0 = 2 m

F = $\frac{-9*10^{9}*20*10^{-6}*(-30*10^{-6})}{2^2}$ = 1.35 N (the positive sign indicates an attraction on q2 towards the positive x-axis)

Resultant force on q2 = 1.8 N + 1.35 N = <em>3.15 N towards the positive x-axis</em>

3 0

3 years ago