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2 years ago

Jeremiah is conducting an investigation about the water cycle. He is given the following

1 answer:

4 0

The water cycle outlines the continuous water movement in liquid, solid and gaseous state between locations on the Earth's surface.

The glass jar represents the lake while the atmosphere is represented by the space above the water, and the sky is represented by the (clear) plastic wrap

Arrangement description and Processes;

The processes of the water cycle includes;

Evaporation;

Condensation

Precipitation

Sublimation

Runoff

Infiltration

The arrangement of the materials is as follows;

Place the glass jar (the lake) containing water and the lamp (the Sun) side by side, such that the lamp light shines on the water surface

Cover the glass jar by wrapping the plastic wrap (the sky) around it to prevent the escape of water vapor when the water is hot.

Switch on the lamp so that it heats the water by radiation heat transfer

Observed processes;

The processes demonstrated by the above experiment includes;

1) Evaporation: As the water in the glass jar becomes warmer, the level of the water in the jar can be observed to decrease slightly due to evaporation

2) Condensation: Fog formation, Clouds

When hotter, the water surface as seen through the clear plastic wrap becomes less clearer due to evaporation, and condensation of the vapor while floating above the water surface, similar to the clouds seen in the sky.

3. Precipitation: Rain;

The clear plastic wrap covering the top of the glass jar, prevents the movement of the vapor further away, such that the tiny condensed vapor gather together, to form big droplets under the plastic wrap that falls back into the jar, which is similar to the process of rainfall

The above processes are repeated as more water evaporates from the jar condenses on the plastic wrap and falls back into the jar, showing the process by which water is recycled from the lake into the atmosphere and back to the lake.

Learn more here:

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A 70 kg man's arm, including the hand, can be modeled as BIO 75-cm-long uniform rod with a mass of 3.5 kg. When the man raises b

Butoxors [25]

Answer:

Δy = 7.1 cm

Explanation:

The center of mass of a body is defined

$y_{cm}$ = 1 /M ∑ $m_{i} y_{i}$ i

Where M is the total mass of the body, m mass of each part and ‘y’ height

Let's apply this equation to our case

We locate the reference system on the shoulders

The height of the arms is at its midpoint

y = -75/2 = 37.5 cm

With arms down

$y_{cm}$ = 1/70 (63 y₀ - 3.5 37.5 - 3.5 37.5)

$y_{cm}$ = 1/70 (63 y)₀ - 7 37.5)

With arms up

$y_{cm}$ ’= 1/70 (63 y₀ + 3.5 y + 3.5 y)

$y_{cm}$ ’= 1/70 (63y₀ + 7 35.5)

let's subtract the two equations

$y_{cm}$ ’ - $y_{cm}$ = 1/70 2 (7 35.5)

Δy = $y_{cm}$ ’ - $y_{cm}$ = 2 7 35.5 / 70

ΔY = 7.1 cm

7 0

2 years ago

Assuming that Bernoulli's equation applies, compute the volume of water ΔV that flows across the exit of the pipe in 1.00 s . In

OLEGan [10]

Answer:

discharge rate (Q) = 0.2005 m^{3} / s

Explanation:

if you read the question you would see that some requirements are missing, by using search engines, you can get the complete question as stated below:

Water flows steadily from an open tank as shown in the figure. (Figure 1) The elevation of point 1 is 10.0m , and the elevation of points 2 and 3 is 2.00 m . The cross-sectional area at point 2 is 4.80x10-2m ; at point 3, where the water is discharged, it is 1.60x10-2m. The cross-sectional area of the tank is very large compared with the cross-sectional area of the pipe. Part A Assuming that Bernoulli's equation applies, compute the volume of water DeltaV that flows across the exit of the pipe in 1.00 s . In other words, find the discharge rate \Delta V/Delta t. Express your answer numerically in cubic meters per second.

solution:

time = 1 s

elevation of point 1 (z1) = 10 m

elevation of point 2 (z2) = 2 m

elevation of point 3 (z3) = 2 m

cross section area of point 2 = 4.8 x 10^{2} m

cross section area of point 3 = 1.6 x 10^{2} m

acceleration due to gravity (g) = 9.8 m/s^{2}

find the discharge rate at point 3 which is the exit pipe.

discharge rate (Q) = A3 x V3

where A3 is the cross sectional area at point 3 and V3 is the velocity of the fluid and can be gotten by applying Bernoulli's equation below

$\frac{P1}{ρg} + \frac{V1^{2} }{2g} + Z1 = \frac{P3}{ρg} + \frac{V3^{2} }{2g} + Z3$