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3 years ago

7

The energy per unit volume in an electromagnetic wave is

1 answer:

andre [41]3 years ago

3 0

Answer: sum of the electric field energy density and magnetic field energy density.

Explanation:

The total energy stored in an electromagnetic wave is equal to the sum of energy stored in the electric and magnetic fields. In that case, the energy stored per unit volume, or energy density of the electromagnetic wave, is the sum of the electric field energy density and magnetic field energy density.

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Electric power is to be generated by installing a hydraulic turbine-generator at a site 120 m below the free surface of a large

iogann1982 [59]

Answer:

P = 2234400 Watt

Explanation:

Power generation is given as rate of work done

Here the turbine generator is located at 120 m below the free surface

So here rate of work done is given as rate of potential energy

$Power = \frac{dw}{dt}$

$Power = \frac{mgh}{t}$

so we have

$Power = (\frac{dm}{dt})gh$

now we have

$Power = 1900(9.8)(120) = 2234400 Watt$

so the power generation potential will be 2234400 Watt

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3 years ago

How resistance, current, and voltage behave in a series circuit.?

mylen [45]

b.the covalent circuit is held throw magnetic waves

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4 years ago

What is the main function of an earthworm organ

Annette [7]

Answer:

brain

Explanation:

brain and blood

7 0

3 years ago

An unknown object is placed inside of a spherical container and dropped from an airplane. When empty, the spherical container ha

hichkok12 [17]

Answer:

The mass of unknown object is 8.62Kg

Explanation:

To develop this problem it is necessary to apply the equations related to the Drag force and the Force of Gravity.

For the given point, that is, the moment at which the terminal velocity is reached, the two forces equalize, that is,

$F_D =F_g$

By definition we know that the Drag force is defined as

$F_D= \frac{1}{2} C_d \rho A V^2$

Where,

$C_d =$ Drag coefficient

$\rho =$ Density

A =Cross-sectional Area

V = Velocity

In the other hand we have,

$F_g = (m_1 +m_2) g$

Where,

$m_1 =$ Mass of sphere

$m_2 =$ Mass of unknown object

Equating the two equations we have to

$(m_1 +m_2) g=\frac{1}{2} C_d \rho A V^2$

Re-arrange for m_2,

$m_2 = \frac{1}{2g} C_d \rho A V^2 -m_1$

Our values are given by,

$C_d = 0.5$

$\rho = 1.22Kg/m^3$

$V = 66.7m/s$

$m_1 = 3Kg$

$d= 32.7*10^{-2}m$

$r = 16.35*10^{-2}m$

Replacing in the equation we have,

$m_2 = \frac{1}{2(9.8)}(0.5) (1.22) (\pi*(16.35*10^{-2})^2)*66.7^2 -3$

$m_2 = 8.62Kg$

<em>Therefore the mass of unknown object is 8.62Kg</em>

4 0

3 years ago

Light is reflected from a crystal of table salt with an index of refraction of 1.544. An analyser is placed to intercept the ref

Vitek1552 [10]

Answer:

hola me llamo bruno y tu?

Explanation:

pero yo soy de mexico

5 0

3 years ago