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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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Suppose Galileo dropped a lead ball (100 kilograms) and a glass ball (1 kilogram) from the Leaning Tower of Pisa. Which one hit
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They will both hit the ground at the same time

Explanation:

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What happens when you want to move the boat forward? You pull the oars toward yourself.Explain why you do this.
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Answer:

You pull on the oars. By the third law, the oars push back on your hands, but that’s irrelevant to the motion of the boat. The other end of each oar (the blade) pushes against the water. By the third law, the water pushes back on the oars, pushing the boat forward.

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2 years ago
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If a glass rod rubbed with Silk piece is taken to a ball pen rubeed with wool the ball pen_____​
WARRIOR [948]

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The glass rod remains charged due to electrification by the silk

3 0
3 years ago
A 52 N sled is pulled across a cement sidewalk at constant speed. A horizontal force of 36 N is exerted. What is the coefficient
Andre45 [30]

Answer:

μ = 0.692

Explanation:

In order to solve this problem, we must make a free body diagram and include the respective forces acting on the body. Similarly, deduce the respective equations according to the conditions of the problem and the directions of the forces.

Attached is an image with the respective forces:

A summation of forces on the Y-axis is performed equal to zero, in order to determine the normal force N. this summation is equal to zero since there is no movement on the Y-axis.

Since the body moves at a constant speed, there is no acceleration so the sum of forces on the X-axis must be equal to zero.

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5 0
3 years ago
In this problem, you will practice applying this formula to several situations involving angular acceleration. In all of these s
riadik2000 [5.3K]

Answer:

Part a)

\alpha = \frac{2(m_1 - m_2)g}{(m_1 + m_2)L}

Part b)

\alpha = \frac{6(m1 - m_2)g}{3(m_1 + m_2)L + m_{bar}L}

Explanation:

As we know that the see saw bar is massless so here torque due to two masses is given as

\tau = I\alpha

here we will have

\tau = (m_1g - m_2g)(\frac{L}{2})

now we will have inertia of two masses given as

I = (m_1 + m_2)(\frac{L}{2})^2

now we have

I = (m_1 + m_2)\frac{L^2}{4}

now the angular acceleration is given as

\alpha = \frac{\tau}{I}

so we have

\alpha = \frac{2(m_1 - m_2)g}{(m_1 + m_2)L}

Part b)

Now if the rod is not massles then we will have total inertia given as

I = (m_1 + m_2)(\frac{L}{2})^2 + \frac{m_{bar}L^2}{12}

so we will have

I = (m_1 + m_2)\frac{L^2}{4} + \frac{m_{bar}L^2}{12}

now the acceleration is given as

\alpha = \frac{\tau}{I}

\alpha = \frac{6(m1 - m_2)g}{3(m_1 + m_2)L + m_{bar}L}

7 0
3 years ago
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