Which two processes allow water to enter the atmosphere

Answers

Answer 1

Answer:

The two processes that allow water to enter the atmosphere are: Evaporation and Transpiration.

1. **Evaporation:** Evaporation is the process by which liquid water on the Earth's surface (such as oceans, lakes, rivers, and even moist soil) is heated by the sun and turns into water vapor. This water vapor then rises into the atmosphere. Evaporation is a key component of the water cycle, where water constantly moves between the surface of the Earth and the atmosphere.

2. **Transpiration:** Transpiration is the process by which water is released into the atmosphere from plants. Plants take up water from the soil through their roots, and this water travels through the plant and eventually evaporates from small openings called stomata on the leaves. Transpiration serves various functions in plants, including cooling the plant and transporting nutrients.

Together, evaporation and transpiration contribute to the overall movement of water from the Earth's surface into the atmosphere, where it eventually condenses to form clouds and participates in various atmospheric processes before returning to the surface as precipitation through processes like rain, snow, sleet, or hail.

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Answer 2

Answer:

Answer:

sublimation and transpiration

Explanation:

The water can enter the atmosphere from snow and ice with the process of sublimation where they also make water vapors. Last water can get in the atmosphere from plants through transpiration which means that the water is evaporated through the pores of the leaves

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A certain x-ray tube requires a current of 7 mA at a voltage of 80 kV. The rate of energy dissipation is:

Answers

Answer:

560 watts

The rate of energy dissipation is 560 W

Explanation:

Rate of energy dissipation is the rate of energy consumption in the x-ray tube.

P = VI

Given;

Voltage V = 80 kV = 80,000

Current I = 7mA = 0.007 A

Substituting the given values;

P = 80,000 V × 0.007A

P = 560 Joules per second

P = 560 watts

The rate of energy dissipation is 560 W

The International Space Station (ISS) is a space station orbiting the earth above the ground. If the radius of the earth is 3,958.8 miles, mass of earth is 5.972 x 10 24 kg, the period of the ISS at the orbit around the earth is 7.84 hours, can you calculate what is the distance from the ISS to the surface of the earth, in unit of miles

Answers

Answer:

8488 miles

Explanation:

The orbital period around an earth is given as:

$T=2\pi \sqrt{(r^3)/(Gm) }$

Where G = constant = 6.67 x 10ˉ¹¹ N m² kgˉ², m = mass of object, T = period taken to round the earth, r = distance from the center of the earth to the orbiting object = radius of earth + orbital altitude.

Given that T = 7.84 hours = 28224 seconds, m = 5.972 x 10²⁴ kg, radius of earth = 3,958.8 miles = 6371071 m

$T=2\pi \sqrt{(r^3)/(Gm) }\n\nsquaring:\n\nT^2=4\pi^2 ((r^3)/(Gm) )\n\nr^3=(GmT^2)/(4\pi^2) \n\nr=\sqrt[3]{(GmT^2)/(4\pi^2) } \n\nr=\sqrt[3]{(6.67*10^(-11)*5.972*10^(24)*(28224)^2)/(4\pi^2) } \n\nr=20031232.62\ meters$

r = radius of earth + distance from the ISS to the surface of the earth

distance from the ISS to the surface of the earth = r - radius of earth

distance from the ISS to the surface of the earth = 20031232.62 meters - 6371071 meters = 13660161.62 meters

distance from the ISS to the surface of the earth = 13660161.62 meters = 8488 miles

A glass plate 2.95 mmmm thick, with an index of refraction of 1.60, is placed between a point source of light with wavelength 600 nmnm (in vacuum) and a screen. The distance from source to screen is 1.25 cm. How many wavelengths are there between the source and the screen?

Answers

Answer:

$N_T=2086285.67$

Explanation:

Given;

Thickness of the glass plate, $x=2.95* 10^(-3)\ m$

refractive index of the glass plate, $n=1.6$

wavelength of light source in vacuum, $\lambda=600* 10^(-9)\ m$

distance between the source and the screen, $d=1.25\ m$

Distance travelled by the light from source to screen in vacuum:

$d_v=d-x$

$d_v=1.25-0.00295$

$d_v=1.24705\ m$

So the no. of wavelengths in the vacuum:

$N=(d_v)/(\lambda)$

$N=(1.24705)/(6* 10^(-7))$

$N\approx2.0784* 10^(6)$ .......................(1)

Now we find the wavelength of the light wave in the glass:

$n=(\lambda)/(\lambda')$

where:

$\lambda'=$ wavelength of light in the medium of glass.

$1.6=(600* 10^(-9))/(\lambda')$

$\lambda'=375* 10^(-9)\ m=375\ nm$

Now the no. of wavelengths in the glass:

$N'=(x)/(\lambda')$

$N'=(2.95* 10^(-3))/(375* 10^(-9))$

$N'=7.8667* 10^(3)$ ............................(2)

From (1) & (2):

total no. of wavelengths are there between the source and the screen:

$N_T=N+N'$

$N_T=2086285.67$

A student lifts their 75 N backpack 0.50 m onto their chair. How much work is done?

Answers

Answer:

37.5 J

Explanation:

With work done equation: W=Fs

W=75*0.50=37.5 J

or use mgh=(75)(0.5) which is the same

The surface pressure of the atmosphere is about 14.7 psi (pounds per square inch). How many pounds per square yard does that amount to

Answers

Answer:

14.7 psi is equal to 19051.2 pounds per square yard.

Explanation:

Dimensionally speaking, a square yard equals 1296 square inches. Therefore, we need to multiply the atmospheric pressure by 1296 to obtain its equivalent in pounds per square yard. That is:

$p = 14.7\,(lbf)/(in^(2))* 1296\,(in^(2))/(yd^(2))$

$p = 19051.2\,(lbf)/(yd^(2))$

14.7 psi is equal to 19051.2 pounds per square yard.

A flat circular coil having a diameter of 25 cm is to produce a magnetic field at its center of magnitude, 1.0 mT. If the coil has 100 turns how much current must pass through the coil?