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sukhopar [10]
3 years ago
7

Which of the statements concerning light are true? The speed of light is the same no matter what material it is traveling throug

h. Its propagation direction is parallel to both the electric field and the magnetic field. The speed of light in matter is greater than the speed of light in a vacuum. 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.
Physics
1 answer:
wel3 years ago
8 0

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.

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ivanzaharov [21]

Answer:

1 +2=3 so thats it

Explanation:

Because i need points

4 0
3 years ago
The speed of sound in air is around 330 m/s. If a bat emits a single high-pitched ‘click’ of sound in a cave that is 25m wide, c
Bingel [31]

Answer:

0.15 s

Explanation:

From the question given above, the following data were obtained:

Speed of sound (v) = 330 m/s

Distance (x) = 25 m

Time (t) =?

The time taken for the echo of the sound to the bat can be obtained as follow:

v = 2x / t

330 = 2 × 25 / t

330 = 50 / t

Cross multiply

330 × t = 50

Divide both side by 330

t = 50 / 330

t = 0.15 s

Thus, it will take 0.15 s for the echo of the sound to the bat

4 0
2 years ago
swings a 5.5 kg cup of water in a vertical circle of radius 1.9 m. (a) What minimum speed must the cup have in this demo if the
Tanzania [10]

Answer:

4.32

Explanation:

The centripetal acceleration of any object is given as

A(cr) = v²/r, where

A(c) = the centripetal acceleration

v = the linear acceleration

r = the given radius, 1.9 m

Since we are not given directly the centripetal acceleration, we'd be using the value of acceleration due to gravity, 9.8. This means that

9.8 = v²/1.9

v² = 1.9 * 9.8

v² = 18.62

v = √18.62

v = 4.32 m/s

This means that, the minimum speed the cup must have so as not to get wet or any spill is 4.32 m/s

6 0
2 years ago
A particle (mass = 2.0 mg, charge = −6.0 μC) moves in the positive direction along the x axis with a velocity of 3.0 km/s. It en
Virty [35]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The acceleration would be  a = 0.003* 6 =0.018\  m/s^2    

Explanation:

The objective of this solution is to obtain the acceleration of the particle

       Now looking at Newton law which is mathematically represented as

                       F = ma

  Where F is the force experience by a particle

              m is the mass of the particle

              a is the acceleration of the particle

  And also this force is equivalent to magnetic force in a magnetic field which is mathematically represented as

                    F = qvB

Where q is the charge of the particle

             v is the velocity of the  charge

             B is the magnetic field the charge is under it influence

  Now equating this two formulas

                   ma = qvB

 Making a the subject we have

                  a = \frac{qvB}{m}

In the question the direction of the is in the positive x-axis which is i hence the direction would be in the i direction

      So substituting  (2.0i +3.0j+4.0k)mT = (2.0i +3.0j+4.0k)*10^{-3}T for B

                    a = \frac{q}{m} * v (2.0i +3.0j +4.0k)*10^{-3}

      Substituting       3.0 Km /s = 3.0*10^{3}\ m/s  for v  and -6.0 \muC = -6.0*10^{-6} C for q

                     a = \frac{-6.0*10^{-6}}{2.0*10^{-3}} * 3.0*10^{3} *(2i+3j+4k) *10^{-3}

                       a = 0.003 * 3i(2i+3j+4k)

                      a = 0.003 *((3*2)i \ \cdot i \ +(3*3) i \ \cdot \ j  \ + (3*4)i \ \cdot \ k)

According to vector multiplication

                                             i \cdot i = j \cdot j = k\cdot k = 1\\\\and \ i\cdot j = i\cdot k  = 0

     So

               a = 0.003* 6 =0.018\  m/s^2          

     

8 0
3 years ago
pulse train with a frequency of 1 MHz is counted using a modulo-1024 ripple-counter built with J-K flip flops. For proper operat
omeli [17]

Answer:

The maximum permissible propagation delay per flip flop stage is<u> 100 </u>n sec

Explanation:

1024 ripple counter has 10 J-K flip flops(210 = 1024).  

So the total delay will be 10×x where x is the delay of each J-K flip flops.

The period of the clock pulse is 1× 10⁻⁶ s.

Now

10x <= 10⁻⁶ s

x <= 100 ns

x= 100 ns for prpoer operation.

pulse train with a frequency of 1 MHz is counted using a modulo-1024 ripple-counter built with J-K flip flops. For proper operation of the counter, the maximum permissible propagation delay per flip flop stage is <u>100 </u>n sec.

4 0
3 years ago
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