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Vanyuwa [196]
3 years ago
11

The diagrams at the right show the path of light as it passes from air into the three solids. What are the possibilities of the

solids in A, B and C? Explain.

Physics
1 answer:
OleMash [197]3 years ago
3 0

Answer:

when light falls on denser medium it may be reflected back or enters the medium and if it enters the medium it refracts towards the normal line of the medium.

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Which of these is a benefit of replacing a coal-burning power plant with a
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A. Reduced greenhouse gas emissions.
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Which of the statements concerning light are true? The speed of light is the same no matter what material it is traveling throug
wel

Answer:

The statements that are true concerning light are the last three statements:

  • Its propagation direction is perpendicular to both the electric field and the magnetic field.
  • It moves at a constant speed through a vacuum.
  • The speed of light in matter is less than the speed of light in a vacuum.

Explanation:

<em>Light</em> is <em>electromagnetic waves.  </em>

The properties of the electromagnetic waves were established by James Clerk Maxwell.

They included that they are the result of the oscillation of a <em>magnetic field </em>in phase with an <em>electric field</em> which are always is always <em>perpendicular</em> to each other.

Also, the electromagnetic waves propagate at right-angles to the direction of both the magnetic and the electric field,  meaning that they are a type of transverse wave.

Thus, the second statement (<em>"Its propagation direction is parallel to both the electric field and the magnetic field"</em>) is false, and the fourth statement ("Its propagation direction is perpendicular to both the electric field and the magnetic field") is true.

On the other hand, it is a postulate of the special theory of relativity that the speed of light is a constant (absolute value) in vacuum: nothing can travel faster than what light travels in vacuum. Thus, the fifth statement, <em>"It moves at a constant speed through a vacuum"</em> is true.

About the speed of light in matter, it is always less than the speed of light in vacuum. Thus, the first statement, "<em>the speed of light is the same no matter what material it is traveling through</em>", and the third statement "<em>the speed of light in matter is greater than the speed of light in a vacuum"</em> are false; while the last statement, "<em>the speed of light in matter is less than the speed of light in a vacuum</em>" is true.

The explanation on why the speed of light is less in a medium than in vacuum is related with the fact that at nanoscopic level the waves suffer polarization which means deviations from the straighi path, which makes that the net straight propagation is slower.

8 0
3 years ago
Which cells in the immune system identify pathogens and distinguish one pathogen from another?
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Answer:

T cell

Explanation:

7 0
3 years ago
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A dancer starts 3 meters from the curtain then moves 8 meters from the curtain in 15 seconds. What was his velocity?
myrzilka [38]
I am pretty sure it is either 45 or 0.2
8 0
3 years ago
A sound source A and a reflecting surface B move directly toward each other. Relative to the air, the speed of source A is 28.7
aleksandrvk [35]

(a) 1440.5 Hz

The general formula for the Doppler effect is

f'=(\frac{v+v_r}{v+v_s})f

where

f is the original frequency

f is the apparent frequency

v is the velocity of the wave

v_r is the velocity of the receiver (positive if the receiver is moving towards the source, negative otherwise)

v_s is the velocity of the source (positive if the source is moving away from the receiver, negative otherwise)

Here we have

f = 1110 Hz

v = 334 m/s

In the reflector frame (= on surface B), we have also

v_s = v_A = -28.7 m/s (surface A is the source, which is moving towards the receiver)

v_r = +62.2 m/s (surface B is the receiver, which is moving towards the source)

So, the frequency observed in the reflector frame is

f'=(\frac{334 m/s+62.2 m/s}{334 m/s-28.7 m/s})1110 Hz=1440.5 Hz

(b) 0.232 m

The wavelength of a wave is given by

\lambda=\frac{v}{f}

where

v is the speed of the wave

f is the frequency

In the reflector frame,

f = 1440.5 Hz

So the wavelength is

\lambda=\frac{334 m/s}{1440.5 Hz}=0.232 m

(c) 1481.2 Hz

Again, we can use the same formula

f'=(\frac{v+v_r}{v+v_s})f

In the source frame (= on surface A), we have

v_s = v_B = -62.2 m/s (surface B is now the source, since it reflects the wave, and it is moving towards the receiver)

v_r = +28.7 m/s (surface A is now the receiver, which is moving towards the source)

So, the frequency observed in the source frame is

f'=(\frac{334 m/s+28.7 m/s}{334 m/s-62.2 m/s})1110 Hz=1481.2 Hz

(d) 0.225 m

The wavelength of the wave is given by

\lambda=\frac{v}{f}

where in this case we have

v = 334 m/s

f = 1481.2 Hz is the apparent in the source frame

So the wavelength is

\lambda=\frac{334 m/s}{1481.2 Hz}=0.225 m

8 0
3 years ago
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