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OLEGan [10]
2 years ago
13

The moment of water from ocean through the atmosphere and back

Physics
2 answers:
melisa1 [442]2 years ago
6 0

The water cycle is all about storing water and moving water on, in, and above the Earth. Although the atmosphere may not be a great storehouse of water, it is the superhighway used to move water around the globe. Evaporation and transpiration change liquid water into vapor, which ascends into the atmosphere due to rising air currents. Cooler temperatures aloft allow the vapor to condense into clouds and strong winds move the clouds around the world until the water falls as precipitation to replenish the earthbound parts of the water cycle. About 90 percent of water in the atmosphere is produced by evaporation from water bodies, while the other 10 percent comes from transpiration from plants.

There is always water in the atmosphere. Clouds are, of course, the most visible manifestation of atmospheric water, but even clear air contains water—water in particles that are too small to be seen. One estimate of the volume of water in the atmosphere at any one time is about 3,100 cubic miles (mi3) or 12,900 cubic kilometers (km3). That may sound like a lot, but it is only about 0.001 percent of the total Earth's water volume of about 332,500,000 mi3 (1,385,000,000 km3), If all of the water in the atmosphere rained down at once, it would only cover the globe to a depth of 2.5 centimeters, about 1 inch.

Flura [38]2 years ago
5 0

the answer is D. water cycle

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assume the suns total energy output is 4.0 * 10^26 watts, and 1 watt is 1 joule/second. assume 4.3 * 10^-12 J is released from e
Dmitry [639]

Answer:

9.3\cdot 10^{37}

Explanation:

Power is defined as the energy produced (E) per unit of time (t):

P= \frac{E}{t}

This means that the energy produced in the Sun each second (1 s), given the power P=4.0\cdot 10^{26}W, is

E=Pt=(4.0\cdot 10^{26}W)(1s )=4.0\cdot 10^{26} J

Each p-p chain reaction produces an amount of energy of

E_1 = 4.3\cdot 10^{-12} J

in order to get the total number of p-p chain reactions per second, we need to divide the total energy produced per second by the energy produced by each reaction:

n=\frac{E}{E_1}=\frac{4.0\cdot 10^{26} J}{4.3\cdot 10^{-12} J}=9.3\cdot 10^{37}

3 0
3 years ago
a machine gun fires 10 rounds per second the speed of the bullets is 300 m/s. what is the distance in the air between the flying
vovangra [49]
<span>(300 m/s)/(10 r/s) = 30 m/round.</span>
4 0
3 years ago
A rain cloud contains 5.32 × 107 kg of water vapor. The acceleration of gravity is 9.81 m/s 2 . How long would it take for a 2.0
Alja [10]

Answer:

20.85 years

Explanation:

2.61 km = 2610 m

2.07 kW = 2070 W

First we need to calculate the potential energy required to take m = 5.32 * 10^7 kg of rain cloud to an altitude of 2610 m is

E = mgh = 5.32 * 10^7 * 9.81 * 2610 = 1.4*10^{12}J

With a P = 2070 W power pump, this can be done within a time frame of

t = E/P = \frac{1.4*10^{12}}{2070} = 658037739 s

or 658037739/(60*60) = 182788 hours or 182788 / 24 = 7616 days or 7616 / 365.25 = 20.85 years

4 0
3 years ago
How many times can energy be transformed?
kykrilka [37]
Only once hope this helps
7 0
3 years ago
Read 2 more answers
The block in the diagram below is AT REST. However, the tension in the cable is not the only thing holding the block back. Stati
Vedmedyk [2.9K]

Answer:

The  tension in the rope is 229.37 N.

Explanation:

Given:

Mass of the block is, m=33.2\ kg

Coefficient of static friction is, \mu = 0.214

Angle of inclination is, \theta = 31.5°

Draw a free body diagram of the block.

From the free body diagram, consider the forces in the vertical direction perpendicular to inclined plane.

Forces acting are mg\cos \theta and normal N. Now, there is no motion in the direction perpendicular to the inclined plane. So,

N=mg\cos \theta\\N=(33.2)(9.8)\cos (31.5)\\N=277.415\ N

Consider the direction along the inclined plane.

The forces acting along the plane are mg\sin \theta and frictional force, f, down the plane and tension, T, up the plane.

Now, as the block is at rest, so net force along the plane is also zero.

T=mg\sin \theta+f\\T=mg\sin \theta +\mu N\\T= (33.2)(9.8)(\sin (31.5)+(0.214\times 277.415)\\T= 170+59.37\\T=229.37\ N

Therefore, the  tension in the rope is 229.37 N.

3 0
2 years ago
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