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Brrunno [24]
2 years ago
12

Describe what happens to the particles of water when it boils.

Physics
1 answer:
Nezavi [6.7K]2 years ago
8 0

Answer:

The liquid turns to a gas.

Explanation:

If a liquid is heated the particles are given more energy and move faster and faster expanding the liquid. Particles in the middle of the liquid form bubbles of gas in the liquid.

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at certain times the demand for electric energy is low and electric energy is used to pump water to a reservoir 45 m above the g
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The mass of water that must be raised is 5.25\cdot 10^7 kg

Explanation:

Since the process is 70% efficiency, the power in output to the turbine can be written as

P_{out} = 0.70 P_{in}

where P_{in} is the power in input.

The power in input can be written as

P_{in} = \frac{W}{t}

where

W is the work done in lifting the water

t = 3 h = 10,800 s is the time elapsed

The work done in lifting the water is given by

W=mgh

where

m is the mass of water

g=9.8 m/s^2 is the acceleration of gravity

h = 45 m is the height at which the water is lifted

Combining the three equations together, we get:

P_{out} = 0.70 \frac{mgh}{t}

Where

P_{out} = 150 MW = 150\cdot 10^6 W

And solving for m, we find:

m=\frac{Pt}{0.70gh}=\frac{(1.50\cdot 10^6)(10800)}{(0.70)(9.8)(45)}=5.25\cdot 10^7 kg

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A sinusoidal electromagnetic wave is propagating in vacuum. At a given point P and at a particular time, the electric field is i
olasank [31]

Answer:

a

The direction of the wave propagation is the negative  z -axis

b

The amplitude of  electric and magnetic field are  A_E= 3.35*10^5 V/m ,

A_M= 1.12 *10^{-3} T respectively

Explanation:

According to right hand rule, your finger (direction of electric field) would be pointing in the positive  x-axis  i.e towards your right let your palms be face toward the direction of the magnetic field i.e negative y-axis  (toward the ground ) Then anywhere your thumb stretched out is facing is the direction of propagation of the wave here in this case is the negative  z -axis

             The Intensity of the wave is mathematically represented as

                          I = \frac{1}{2} c \epsilon _O E_{rms}^2

Given that I = 7.43 \frac{kW}{cm^2} =  7.43 \frac{*10^3}{*10^-{4} }= 7.43*10^7 \frac{W}{m^2}

Making E_{rms} the subject we have

                   E_{rms} = \sqrt{\frac{I}{0.5*c*\epsilon_o} }

Substituting values as given on the question

                E_{rms} = \sqrt{\frac{7.43 *10^7[\frac{W}{m^2} ]}{0.5 * 3.08*10^8 *8.85*10^{-12}} }

                          = 2.37*10^5 \ V/m

The amplitude of the electric field is mathematically represented as

                  A_E = \sqrt{2} * E_{rms}

                         = \sqrt{2} * 2.37*10^5

                        A_E= 3.35*10^5 V/m

The amplitude of the magnetic field is mathematically represented as

                       A_M = \frac{A_E}{c}

Substituting value

                      A_M = \frac{3.35 *10^5}{3.0*10^8}

                             A_M= 1.12 *10^{-3} T

7 0
3 years ago
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