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ZanzabumX [31]
4 years ago
6

A beam of light is propagating in the x direction. The electric-field vector Group of answer choices can oscillate in any arbitr

ary direction in space. must oscillate in the z direction. must oscillate in the yz plane. must oscillate in the x direction. must have a steady component in the x direction.
Physics
1 answer:
sammy [17]4 years ago
4 0

Answer:

Option C is correct.

The electric-field vector must oscillate in the yz plane.

Explanation:

Light, in waveform, is an electromagnetic wave.

And electromagnetic waves are known to have their electric and magnetic field perpendicular to each other and also simultaneously perpendicular to the direction of propagation of the wave.

If the velocity of direction of propagation of the wave is in one direction, the electric-field vector must be in a direction we are sure is perpendicular to this direction of wave propagation and the wave's magnetic field.

Of the options provided, only option B (z-direction) and option C (yz-plane) show a direction that is indeed perpendicular to the direction of propagation of the wave (x-axis).

And truly, the electric-field vector for this wave can be in any of the two directions without breaking the laws of physics, but the electric-field vector oscillating in the yz-plane is a more general answer as it covers all the possible directions that the electric-field can oscillate in, including the one specified by option B (z-direction).

Hence, the correct answer is option C.

Hope this Helps!!!

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Answer:

<h2>138 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 69 × 2

We have the final answer as

<h3>138 N</h3>

Hope this helps you

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C) Explain the following.<br>Keftles, cooking pans and iron boxes have polished surfaces​
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Two energy transfers take place when a book hits the ground, Which type of energy transfers are those
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Answer:

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III. When you are on a rollercoaster, at what point will you have the highest amount of
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Read 2 more answers
An open-end mercury manometer is connected to a low-pressure pipeline that supplies a gas to a laboratory. Because paint was spi
Svetllana [295]

Answer:

a

P_G  = 14.03 \  psig  

b

h_m =   0.148 \  m

Explanation:

From the question we are told that

The pressure of the manometer when there is no gas flow is P_{m} =  15.5 \  psig  =  15.5 *  6894.76 =  106868.78 \ N/m^2

The level of mercury is h  =  950 \ mm  =  0.950 \  m

The drop in the mercury level at the visible arm is d =  39.0 =  0.039 \  m

Generally when there is no gas flow the pressure of the manometer is equal to the gauge pressure which is mathematically represented as

P_g  =  P_m  =  g *  \delta h  * \rho

Here \rho is the density of mercury with value \rho = 13.6 *10^{3} kg/m^3

and \delta h is the difference in the level of gas in arm one and two

So

\delta h  =  \frac{106868.78}{  13.6 *10^{3} *  9.8 }

\delta h  = 0.802 \  m

Generally the height of the mercury at the arm connected to the pipe is mathematically represented as

h_m =   0.950 -  0.802

=> h_m =   0.148 \  m

Generally from manometry principle we have that

P_G + \rho * g  * d   -  \rho *  g  * [h - (h_m + d)] = 0

Here P_G is the pressure of the gas

P_G +13.6 *10^{3} * 9.8  * 0.039    -  13.6 *10^{3}  *  9.8  * [0.950 - (0.148 + 0.039)] = 0

P_G  =  9.6724 04 *10^{4} \  N/m^2

converting to  psig

P_G  = \frac{ 9.6724 04 *10^{4} }{6894.76}

P_G  = 14.03 \  psig

6 0
3 years ago
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