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Aloiza [94]
3 years ago
10

Atmospheric conditions near a mountain range are such that a cloud at an altitude of 2.00 km contains 3.20 ✕ 10^7 kg of water va

por. How long (in years) would it take for a 2.96 kW pump to raise this amount of water (at a constant speed) from Earth's surface to the altitude of the cloud?
Physics
1 answer:
kykrilka [37]3 years ago
3 0

Answer:

time required is 6.72 years

Explanation:

Given data

mass m = 3.20 ✕ 10^7 kg

height h = 2.00 km = 2 × 10^3 m

power p = 2.96 kW  =2.96 × 10^3 J/s

to find out

time period

solution

we know work is mass × gravity force × height

and power is work / time

so we say that power =  mass  gravity force × height / time

now put all value and find time period

power =  mass × gravity force × height / time

2.96 × 10^3 =  3.20 ✕ 10^7  × 9.81× 2 × 10^3  / time

time =  62.784 × 10^10 / 2.96 × 10^3

time = 21.21081081 × 10^7 sec

time = 58.91891892 × 10^3 hours

time = 6.72 years

so time required is 6.72 years

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Light of wavelength 560 nm passes through a slit of width 0. 170 mm. (a) the width of the central maximum on a screen is 8. 00 m
faltersainse [42]

The distance between slit and the screen is 1.214m.

To find the answer, we have to know about the width of the central maximum.

<h3>How to find the distance between slit and the screen?</h3>
  • It is given that, wavelength 560 nm passes through a slit of width 0. 170 mm, and the width of the central maximum on a screen is 8. 00 mm.
  • We have the expression for slit width w as,

                           w=\frac{2*wavelength*d}{a}

where, d is the distance between slit and the screen, and a is the slit width.

  • Thus, distance between slit and the screen is,

                           d=\frac{w*a}{2*wavelength} =\frac{8*10^{-3}*0.17*10^{-3}}{560*10^{-9}*2} \\\\d=1.214m

Thus, we can conclude that, the distance between slit and the screen is 1.214m.

Learn more about the width of the central maximum here:

brainly.com/question/13088191

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