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xz_007 [3.2K]
4 years ago
14

1, Condensation occurs when:

Physics
2 answers:
cluponka [151]4 years ago
5 0

Answer:

1- Option (D)

2- Option (B) and (C)

3- Option (A)

Explanation:

The process of by which the water vapor is converted into liquid is known as condensation. It occurs when the hot air rises up in the atmosphere and then it gradually cools, resulting in the formation of liquid droplets. With continuous cooling of these air leads to the precipitation in the form of rainfall or snowfall.

When precipitation occurs, some of the water flows over the land surface in a gradually inclined plane, which is commonly known as the surface runoff. This water is added into the streams and rivers. In addition to this, some water in the recharge area, are percolated deep into the ground through the interconnected pores, that further flows and mixes up with the groundwater.

The plants also receive some amount of water from the roots that are transported to its entire body. The excess water is then released into the atmosphere by a process known as the transpiration process.

Hence, the correct answers are mentioned above.

andrey2020 [161]4 years ago
4 0

Answer:

1. surface water changes to vapor water

2. runs off into streams and rivers

3. transpiration

Explanation:

1. the water above the surface is being heated by the sun turning into a gas that makes up part of the cloud.

2. a runoff is the process of precipitation running down hills and into rivers and streams

3. Transpiration is essentially evaporation of water from plant leaves.

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4. Water is flowing at 12m/s in a horizontal pipe under a pressure of 600kpa
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a. 192 m/s

b. -17,760 kPa

Explanation:

First let's write the flow rate of the liquid, using the following equation:

Q = A*v

Where Q is the flow rate, A is the cross section area of the pipe (A = pi * radius^2) and v is the speed of the liquid. The flow rate in both parts of the pipe (larger radius and smaller radius) needs to be the same, so we have:

a.

A1*v1 = A2*v2

pi * 0.02^2 * 12 = pi * 0.005^2 * v2

v2 = 0.02^2 * 12 / 0.005^2

v2 = 192 m/s

b.

To find the pressure of the other side, we need to use the Bernoulli equation: (600 kPa = 600000 N/m2)

P1 + d1*v1^2/2 = P2 + d1*v2^2/2

Where d1 is the density of the liquid (for water, we have d1 = 1000 kg/m3)

600000 + 1000*12^2/2 = P2 + 1000*192^2/2

P2 = 600000 + 72000 - 1000*192^2/2

P2 = -17760000 N/m2 = -17,760 kPa

The speed in the smaller part of the pipe is too high, the negative pressure in the second part means that the inicial pressure is not enough to maintain this output speed.

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<h2>Hope it helps you</h2>
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