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2 years ago

2. What process does water redeposit into a lake in the form of rain?

Physics

1 answer:

weqwewe [10]2 years ago

5 0

The process that water redeposit into a lake in the form of rain is precipitation.

<h3>What is water runoffs?</h3>

Water runoff occurs when there is more water than land can absorb.

The water flows across the surface of the land and into nearby creeks, streams, or lakes.

Runoff can come from both natural processes and human activity.

When rain falls to the earth from clouds and runs downhill into rivers and lakes.

During evaporation, the water turns from liquid into gas, and moves from oceans and lakes into the atmosphere where it forms clouds.

<h3>What is precipitation?</h3>

Precipitation is any liquid (rain) or frozen water that forms in the atmosphere and falls back to the Earth, for example it could fall on land or into lakes and rivers.

Thus, the process that water redeposit into a lake in the form of rain is precipitation.

Learn more about precipitation here: brainly.com/question/1783904

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Four astronauts are in a spherical space station. (a) If, as is typical, each of them breathes about 500 cm

mario62 [17]

Answer:

A) if each astronaut breathes about 500 cm³, the total volume of air breathed in a year is 14716.8m³.

B) The Diameter of this spherical space station should be 30.4m

Explanation:

The breathing frequency (according to Rochester encyclopedia) is about 12-16 breath per minute. if we take the mean value (14 breath per minute), we can estimate the total breaths of a person along a year:

$f_b=14\frac{br}{min} \cdot \frac{60min}{1hr} \frac{24hr}{1day}\frac{365day}{1year}=29433600\frac{br}{year}$

If we multiply this for the number of people in the station and the volume each breath needs, we obtain the volume breathed in a year.

The volume of a sphere is:

$V_{sph}=\frac{4\pi}{3}r^3$

So the diameter is:

$D=2r=2\sqrt[3]{\frac{3V_{sph}}{4\pi}} =30.4m$

6 0

3 years ago

I would love to stretch a wire from our house to the Shop so I can 'call' my husband in for meals. The wire could be tightened t

dezoksy [38]

Note: I'm not sure what do you mean by "weight 0.05 kg/L". I assume it means the mass per unit of length, so it should be "0.05 kg/m".

Solution:
The fundamental frequency in a standing wave is given by
$f= \frac{1}{2L} \sqrt{ \frac{T}{m/L} }$
where L is the length of the string, T the tension and m its mass. If we plug the data of the problem into the equation, we find
$f= \frac{1}{2 \cdot 24 m} \sqrt{ \frac{240 N}{0.05 kg/m} }=1.44 Hz$

The wavelength of the standing wave is instead twice the length of the string:
$\lambda=2 L= 2 \cdot 24 m=48 m$

So the speed of the wave is
$v=\lambda f = (48 m)(1.44 Hz)=69.1 m/s$

And the time the pulse takes to reach the shop is the distance covered divided by the speed:
$t= \frac{L}{v}= \frac{24 m}{69.1 m/s}=0.35 s$

7 0

4 years ago

Help with 3 please! I will give brainliest!

rusak2 [61]

The answer is point b because vertical velocity is zero at the maximum height

8 0

3 years ago

A bike travels 30m/s for 3 seconds. How far did it travel?

Elena L [17]

He traveled 90 meters

8 0

3 years ago

Suppose you want to determine the resistance of a resistor that is nominally 100 . You should be able to apply 10 V across the r

Butoxors [25]

Answer:

a) For y = 102 mA, R = 98.039 ohms

For y = 97 mA, R = 103.09 ohms

b) Check explanatios for b

Explanation:

Applied voltage, V = 10 V

For the first measurement, current $y_{1} = 102 mA = 0.102 A$

According to ohm's law, V = IR

R = V/I

Here, $I = y_{1}$

$R = \frac{V}{y_{1} } \\R = \frac{10}{0.102} \\R = 98.039 ohms$

For the second measurement, current $y_{2} = 97 mA = 0.097 A$

$R = \frac{V}{y_{2} }$

$R = \frac{10}{0.097} \\R = 103 .09 ohms$

b) $y = \left[\begin{array}{ccc}y_{1} &y_{2} \end{array}\right] ^{T}$

$y = \left[\begin{array}{ccc}y_{1} \\y_{2} \end{array}\right]$

$y = \left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3} \end{array}\right]$

A linear equation is of the form y = Gx

The nominal value of the resistance = 100 ohms

$x = \left[\begin{array}{ccc}100\end{array}\right]$

$\left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3} \end{array}\right] = \left[\begin{array}{ccc}G_{1} \\G_{2} \end{array}\right] \left[\begin{array}{ccc}100\end{array}\right]\\\left[\begin{array}{ccc}G_{1} \\G_{2} \end{array}\right] = \left[\begin{array}{ccc}102*10^{-5} \\97*10^{-5} \end{array}\right]$

3 0

3 years ago