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stepan [7]
3 years ago
13

Every day a certain amount of water evaporates from Earth’s oceans, lakes, and land surface and forms water vapor and clouds in

the atmosphere. Every day a certain amount of rain falls back to Earth. Make the reasonable assumption that, on average, the energy absorbed by the evaporation and lifting of the water is equal to the energy released by its condensation and falling back to earth. The evaporation of one mole of water requires approximately 40.6×103 J of heat. An equal amount of heat is released when a mole of water condenses.
QUESTION: The annual volume of rainfall on Earth is approximately 4.9×105 km3 and the average cloud altitude is 8.8 km above Earth’s surface. How much energy, in joules, is required every day to evaporate and lift the water?
Physics
1 answer:
pochemuha3 years ago
8 0

Answer:

0.0000000010 joules

Explanation:

Amount : 1540.3 nanojoules (nJ)

Equals : 0.0 joules (J)

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3 years ago
A ball is thrown straight down with a speed of Va2.00 from a building 40.00 meters high. How much S time passes before the ball
-Dominant- [34]

Answer:

Option C is the correct answer.

Explanation:

Considering vertical motion of ball:-

Initial velocity, u =  2 m/s

Acceleration , a = 9.81 m/s²

Displacement, s = 40 m

We have equation of motion s= ut + 0.5 at²

Substituting

   s= ut + 0.5 at²

    40 = 2 x t + 0.5 x 9.81 x t²

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   t = 2.66 s   or t = -3.06 s

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6 0
3 years ago
before dancing on a smooth wooden floor, ballet dancers sometimes sometimes put a sticky powered called rosin on their shoes sol
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3 0
2 years ago
An ideal spring hangs from the ceiling. A 2.15 kg mass is hung from the spring, stretching the spring a distance d = 0.0895 m fr
Igoryamba

Answer:

The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

Explanation:

Given that,

Mass = 2.15 kg

Distance = 0.0895 m

Amplitude = 0.0235 m

We need to calculate the spring constant

Using newton's second law

F= mg

Where, f = restoring force

kx=mg

k=\dfrac{mg}{x}

Put the value into the formula

k=\dfrac{2.15\times9.8}{0.0895}

k=235.41\ N/m

We need to calculate the kinetic energy of the mass

Using formula of kinetic energy

K.E=\dfrac{1}{2}mv^2

Here, v = A\omega

K.E=\dfrac{1}{2}m\times(A\omega)^2

Here, \omega=\sqrt{\dfrac{k}{m}}^2

K.E=\dfrac{1}{2}m\times A^2\sqrt{\dfrac{k}{m}}^2

K.E=\dfrac{1}{2}kA^2

Put the value into the formula

K.E=\dfrac{1}{2}\times235.41\times(0.0235)^2

K.E=0.06500\ J

Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

8 0
3 years ago
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