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enyata [817]
3 years ago
8

Identify the stages of the water cycle where water moves downward from the force of gravity

Physics
2 answers:
hoa [83]3 years ago
8 0

Umm I think that it's called a run off! Because it's running off of a surface into water.

Flura [38]3 years ago
6 0

Answer:

Precipitation, runoff and percolation

Explanation:

The water cycle refers to the movement of water from one place to another by undergoing certain processes such as evaporation, transpiration, condensation. This plays an important role in maintaining the total amount of water on earth.

From the figure, it can be concluded that the process of precipitation, runoff and percolation are the three stages in which the water moves downward due to the influence of gravity.

The precipitation occurs in the form of snowfall, rainfall, and so on. This allows the water to fall freely because of gravity.

The surface water runoff also occurs under the influence of gravity, that rapidly falls downward along the slope of the area and gets mixed up with the rivers, streams and lakes.

Lastly, during the water cycle process, some of the water gets percolate downward through the voids present within the rocks. Here the water moves because of the effect of gravity but at a much slower rate. This water then slowly adds to the groundwater.

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Given vectors D (3.00 m, 315 degrees wrt x-axis) and E (4.50 m, 53.0 degrees wrt x-axis), find the resultant R= D + E. (a) Write
Eva8 [605]

Answer:

  • R = ( 4.831 m , 1.469 m )
  • Magnitude of R = 5.049 m
  • Direction of R relative to the x axis= 16°54'33'

Explanation:

Knowing the magnitude and directions relative to the x axis, we can find the Cartesian representation of the vectors using the formula

\vec{A}= | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )

where | \vec{A} | its the magnitude and θ.

So, for our vectors, we will have:

\vec{D}= 3.00 m \ ( \ cos(315) \ , \ sin (315) \ )

\vec{D}=  ( 2.121 m , -2.121 m )

and

\vec{E}= 4.50 m \ ( \ cos(53.0) \ , \ sin (53.0) \ )

\vec{E}= ( 2.71 m , 3.59 m )

Now, we can take the sum of the vectors

\vec{R} = \vec{D} + \vec{E}

\vec{R} = ( 2.121 \ m , -2.121 \ m ) + ( 2.71 \ m , 3.59 \ m )

\vec{R} = ( 2.121 \ m  + 2.71 \ m , -2.121 \ m + 3.59 \ m )

\vec{R} = ( 4.831 \ m , 1.469 \ m )

This is R in Cartesian representation, now, to find the magnitude we can use the Pythagorean theorem

|\vec{R}| = \sqrt{R_x^2 + R_y^2}

|\vec{R}| = \sqrt{(4.831 m)^2 + (1.469 m)^2}

|\vec{R}| = \sqrt{23.338 m^2 + 2.158 m^2}

|\vec{R}| = \sqrt{25.496 m^2}

|\vec{R}| = 5.049 m

To find the direction, we can use

\theta = arctan(\frac{R_y}{R_x})

\theta = arctan(\frac{1.469 \ m}{4.831 \ m})

\theta = arctan(0.304)

\theta = 16\°54'33''

As we are in the first quadrant, this is relative to the x axis.

3 0
3 years ago
If you fire a gun horizontally, and at the same time drop a bullet/slug from the same height, what will happen?
Elena-2011 [213]

Answer:

There you have it. The y-component of air resistance for the fired bullet still depends on the fired speed of the bullet (since it is proportional to v2). A fired bullet (with air resistance) does not hit the ground at the same time as a dropped bullet.

Explanation:

hope this helped ✨

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The distance from the Earth to the Sun equals 1 AU. Neptune is 30 AU from the Sun. How far is Neptune from the Earth? AU
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Answer:

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evablogger [386]
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F equals 3N with respect to the circle's center, moving in the same direction as the centripetal acceleration.

<h3>How much centripetal force is there in a centrifuge?</h3>

Centripetal force is the force that pushes an item in the direction of its center of curvature. It is fundamental to how a centrifuge operates.

<h3>On a roller coaster, what is centripetal force?</h3>

An item travelling in a circle is pushed inward toward what is known as the center of rotation, which is essentially what a roller coaster accomplishes when it travels through a loop. The force that maintains an object moving along a curved route is this pull toward the center, or centripetal force.

To know more about centripetal force visit:-

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