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umka21 [38]
3 years ago
6

Describe the water cycle process starting from an afternoon thunderstorm. There are many different variations that could happen.

Physics
1 answer:
spin [16.1K]3 years ago
3 0

Explanation:

The water cycle basically involves five steps:

  • evaporation and transpiration ⇄
  • condensation, ⇄
  • precipitation, ⇄
  • runoff, ⇄
  • infiltration ⇄

So when a <u>thunderstorm </u>occurs it <em>helps in completing the precipitation process </em>by enabling the release of water vapor stored up in the atmosphere to fall on the ground as rain.

After this, the water <em>runoffs </em><em>to the surface of the ground, on plants, into rocks, rivers, and lakes.</em>

Next, the <em>Infiltration process</em> enables the water on the ground surface to enter the soil some of which becomes groundwater.

The cycle begins again as the<em> </em><em>evaporation and transpiration</em> <em>process </em>begins, where the groundwater as a result of heat from the sun is taken back into the atmosphere, while water in plants by means of transpiration goes back <em>into the atmosphere</em>.

It then <em>condenses </em>and falls back as precipitation again.

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A rocket starting from its launch pad is subjected to a uniform acceleration of 100 meters/second2. Determine the time needed to
gizmo_the_mogwai [7]

Answer:

10s

Explanation:

Acceleration is a measure of a rate of change of velocity, or in other words, a measure of how quickly the velocity is changing.

If acceleration is constant, then the velocity is changing by a constant amount.

With an acceleration of 100 m/s^2, starting from the launching pad (and thus, an initial velocity of zero), we can calculate how long it will take to reach a final velocity of 1000m/s with the following formula:

v=at+v_o where "v" is the final velocity at some later time "t", "a" is the constant acceleration, and "v" sub-zero is the initial velocity.

v=at+v_o

(1000\text{ [m/s]})=(100 \text{ } [\text{m/s}^2] )t+(0\text{ [m/s]})

1000\text{ [m/s]}=100 \text{ } [\text{m/s}^2] *t

\dfrac{1000\text{ [m/s]}}{100 \text{ } [\text{m/s}^2]}=\dfrac{100 \text{ } [\text{m/s}^2] *t}{100 \text{ } [\text{m/s}^2]}

10\text{ [s]}=t

So, it will take 10 seconds for the rocket to reach 1000m/s when starting from the launching pad, with a constant velocity of 100m/s^2.

<u>Verification:</u>

In this situation, it is quick to verify that 10 seconds is correct by looking at what the velocities will be each second.

Recognizing that the acceleration is a=\dfrac{100 [\frac{m}{s}]}{1[s]}, the velocity increases by 100 units [m/s] every second.

At time 0[s], the velocity is 0[m/s]

At time 1[s], the velocity is 100[m/s]

At time 2[s], the velocity is 200[m/s]

At time 3[s], the velocity is 300[m/s]

At time 4[s], the velocity is 400[m/s]

At time 5[s], the velocity is 500[m/s]

At time 6[s], the velocity is 600[m/s]

At time 7[s], the velocity is 700[m/s]

At time 8[s], the velocity is 800[m/s]

At time 9[s], the velocity is 900[m/s]

At time 10[s], the velocity is 1000[m/s]

So, indeed, after 10 seconds, the velocity reaches 1000 m/s

5 0
2 years ago
A driver notices an upcoming speed limit change from 45 mi/h (20 m/s) to 25 mi/h (11 m/s). If she estimates
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Answer:

-2.79 m/s²

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Given:

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(11 m/s)² = (20 m/s)² + 2a (50 m)

a = -2.79 m/s²

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