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stepan [7]
2 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:
pochemuha2 years ago
8 0

Answer:

0.0000000010 joules

Explanation:

Amount : 1540.3 nanojoules (nJ)

Equals : 0.0 joules (J)

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An ant moves towards the plane mirror with speed of 2 m/s & the mirror is moved towards the ant with the same speed. What is
Stells [14]
  • Speed of ant-V_a=2m/s
  • Speed of mirror =v_b=2m/s

We know

\boxed{\sf Relative\:velocity(V_{AB})=V_A-V_B}

\\ \sf\longmapsto V_{AB}=2-2

\\ \sf\longmapsto V_{AB}=0m/s

6 0
2 years ago
Passing an electric current through a certain substance produces oxygen and sulfur. This substance cannot be a(n)
Nezavi [6.7K]

Answer:

An Element

Explanation:

Such substance cannot be an element because an element cannot be chemically disintegrated (i.e it cannot be disintegrated via chemical reaction).

4 0
2 years ago
Two uniform, solid cylinders of radius R and total mass M are connected along their common axis by a short, light rod and rest o
sveta [45]

Explanation:

A) To prove the motion of the center of mass of the cylinders is simple harmonic:

System diagram for given situation is shown in attached Fig. 1

We can prove the motion of the center of mass of the cylinders is simple harmonic if

a_{x} = -\omega^{2}  x

where aₓ is acceleration when attached cylinders move in horizontal direction:

<h3>PROOF:</h3>

rotational inertia for cylinders  is given as:

                                  I=\frac{1}{2}MR^{2} -----(1)

Newton's second law for angular motion is:

                                             ∑τ = Iα ------(2)

For linear motion in horizontal direction it is:

                                             ∑Fₓ = Maₓ ------ (3)

By definition of torque:

                                               τ  = RF --------(4)        

Put (4) and (1) in (2)

                                       RF=\frac{1}{2}MR^{2}\alpha

                                       RF=\frac{1}{2}MR^{2}\alpha

from Fig 3 it can be seen that fs is force by which the cylinders roll without slipping as they oscillate

So above equation becomes

                                   f_{s}=\frac{1}{2}MR\alpha------ (5)

As angular acceleration is related to linear by:

                                          a= R\alpha

Eq (5) becomes

                                    f_{s}=\frac{1}{2}Ma_{x}---- (6)

aₓ shows displacement in horizontal direction

From (3)

                                              ∑Fₓ = Maₓ

Fₓ is sum of fs and restoring force that spring exerts:

                                  \sum F_{x} = f_{s} - kx ----(7)

Put (7) in (3)

                                  f_{s} - kx  = Ma_{x}[/tex] -----(8)

Using (6) in (8)

                               \frac{1}{2}Ma_{x} - kx =Ma_{x}

                                     a_{x} = \frac{2k}{3M} x --- (9)

For spring mass system

                                  a= -\omega^{2} x ----- (10)

Equating (9) and (10)

                                  \omega^{2} = \frac{2k}{3M}

\omega = \sqrt{ \frac{2k}{3M}}

then (9) becomes

                                a_{x} = - \omega^{2}x

(The minus sign says that x and  aₓ  have opposite directions as shown in fig 3)

This proves that the motion of the center of mass of the cylinders is simple harmonic.

<h3 /><h3>B) Time Period</h3>

Time period is related to angular frequency as:

                                   T=\frac{2\pi }{\omega}

                                  T = 2\pi \sqrt{\frac{3M}{2k}

                           

 

5 0
2 years ago
3-m-high large tank is initially filled with water. The tank water surface is open to the atmosphere, and a sharp-edged 10-cm-di
irinina [24]

Answer:

The initial velocity of the water from the tank is 5.42 m/s

Explanation:

By applying Bernoulli equation between  point 1 and 2

\dfrac{P_1}{\rho g}+\dfrac{V_1^2}{2g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{V_2^2}{2g}+Z_2+h_L

At the point 1

P₁=0  ( Gauge pressure)

V₁= 0 m/s

Z₁=3 m

At point 2

P₂=0  ( Gauge pressure)

Z₂= 0 m/s

h_L=1.5\ m

Now by putting the values

\dfrac{P_1}{\rho g}+\dfrac{V_1^2}{2g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{V_2^2}{2g}+Z_2+h_L

Z_1-h_L=\dfrac{V_2^2}{2g}

3-1.5=\dfrac{V_2^2}{2\times 9.81}

V_2=\sqrt{2\times 1.5\times 9.81}\ m/s

V₂= 5.42 m/s

The initial velocity of the water from the tank is 5.42 m/s

3 0
3 years ago
Two long, parallel, current-carrying wires lie in an xy-plane. The first wire lies on the line y = 0.340 m and carries a current
sladkih [1.3K]

Answer:

The y-value of the line in the xy-plane where the total magnetic field is zero  U = 0.1355 \ m

Explanation:

From the question we are told that

    The distance of wire one from two along the y-axis is    y = 0.340 m

   The current on the first wire is  I_1 =  (27.5i) A

    The force per unit length on each wire is  Z =  295 \mu N/m = 295*10^{-6}  N/m

Generally the force per unit length is mathematically represented as

         Z = \frac{F}{l}  =  \frac{\mu_o I_1I_2}{2\pi y}

=>      \frac{\mu_o I_1I_2}{2\pi y}  =  295

Where  \mu_o is the permeability of free space with a constant value of  \mu_o  =  4\pi *10^{-7} \ N/A2

substituting values

       \frac{ 4\pi *10^{-7} 27.5 * I_2}{2\pi * 0.340}  =  295 *10^{-6}

=>    I_2 =  18.23 \ A

Let U  denote the  line in the xy-plane where the total magnetic field is zero

So  

      So the force per unit length of  wire 2  from  line  U is equal to the force per unit length of wire 1  from  line  (y - U)      

   So  

         \frac{\mu_o  I_2  }{2 \pi U} =  \frac{\mu_o  I_1  }{2 \pi(y -  U) }

substituting values

          \frac{  18.23  }{ U} =  \frac{ 27.5 }{(0.34 -  U) }

         6.198 -18.23U = 27.5U

          6.198=45.73U

          U = 0.1355 \ m              

5 0
2 years ago
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