All categories

3 years ago

Speed of light in a medium with n > 1

Physics

2 answers:

Sidana [21]3 years ago

7 0

Answer:

Explanation:

Without thinking, the answer is C.

The problem is with D. Of course it is going to depend on the transparency. If the material is a rock, then the light won't go through, and n >>>1. I don't think you are intended to think that way, but I sure wish D wasn't there. I'd answer C, but be prepared to see it come out as D.

Tcecarenko [31]3 years ago

7 0

Answer:

it and i quote you "is less than in free space

Explanation:

i cant explain

You might be interested in

An uncharged capacitor is connected to the terminals of a 4.0 V battery, and 6.0 μC flows to the positive plate. The 4.0 V batte

Ugo [173]

Answer:

1 μC extra charge will be flow here

Explanation:

Given data

battery V1 = 4.0 V

flows Q1 = 6.0 μC

replace battery V2 = 7.0 V

to find out

what happen if we replace battery

solution

we apply here principal of capacitor

that is Q directly proportional voltage

so we say Q2/Q1 = V2/ V1

put all value here

Q2/Q1 = V2/ V1

Q2/6 = 7/ 6

Q2 = 7

so we see here 7 μC will be flow

and Q = Q2 - Q1 = 7 - 6 = 1 μC

so we also say that 1 μC extra charge will be flow here

5 0

4 years ago

Which equations could be used as is, or rearranged to calculate for frequency of a wave? Check all that apply.

amm1812

-- Equations #2 and #6 are both the same equation,
and are both correct.

-- If you divide each side by 'wavelength', you get Equation #4,
which is also correct.

-- If you divide each side by 'frequency', you get Equation #3,
which is also correct.
With some work, you can rearrange this one and use it to calculate
frequency.

Summary:

-- Equations #2, #3, #4, and #6 are all correct statements,
and can be used to find frequency.

-- Equations #1 and #5 are incorrect statements.

7 0

3 years ago

Read 2 more answers

Answer the following questions for a mass that is hanging on a spring and oscillating up and down with simple harmonic motion. N

LiRa [457]

Answer:

1. equilibrium

2. bottom

3. bottom

4. nowhere

5. bottom

6. top & bottom

7. equilibrium

8. equilibrium

1. No

2. Yes

Explanation:

According to the following equation of motion for SHM:

$x(t) = A\cos(\omega t + \phi)$

where A is the amplitude, ω is the angular frequency, and ∅ is the phase angle.

Furthermore, the velocity and acceleration functions are as follows:

$y(t) = -\omega A\sin(\omega t + \phi)\\a(t) = -\omega^2 A\cos(\omega t + \phi)$

1. The acceleration is zero at the equilibrium. At the equilibrium, the net force on the object is zero. And according to Newton's Second Law, if the net force is zero, then the acceleration is zero as well.

2. The forces on the object in a vertical spring are the weight of the object and the spring force.

F = mg - kx

Since mg is constant along the motion, then the net force is maximum at the amplitude. For the special case in this question, the mass is always below the rest length of the spring. So the net force is maximum at the lower amplitude, because x is greater in magnitude at the lower amplitude. According to Newton's Second Law, acceleration is proportional to the net force, hence the acceleration is at a maximum at the bottom.

3. As explained above, the magnitude of the net force is at a maximum at the lower amplitude, that is bottom.

4. The spring force is defined by Hooke's Law: F = -kx. Since the oscillation is small enough so that the mass is always below the rest length of the spring, then x is always greater than zero, hence nowhere in the motion will the spring force becomes zero.

5. As explained above, the force of gravity is constant and the spring force is proportional to the displacement, x. Therefore, the spring force is at a maximum at the lower amplitude, that is bottom.

6. The speed is zero when the mass is instantaneously at rest, that is the amplitude.

7. The net force on the mass is zero at the equilibrium.

8. The speed is at a maximum at the equilibrium.

1. We will use the equation of motions given above. For simplicity, let's take ∅ = 0. At half its amplitude:

$\frac{A}{2} = A\cos(\omega t)\\\frac{1}{2} = \cos(\omega t)\\\omega t = \pi / 3$

Then the velocity at that point is

$v(t) = -\omega A\sin(\pi /3) = -\omega A (0.866)$

The maximum speed is where the acceleration is equal to zero:

$0 = -\omega^2 A\cos(\omega t)\\\omega t = \pi / 2\\v_{max} = -\omega A\sin(\pi /2) = -\omega A$

Comparing the maximum velocity to the velocity at A/2 yields that it is not half the maximum velocity:

$-\omega A(0.866) \neq -\omega A$

2. The maximum acceleration is at the amplitude.

$A = A\cos(\omega t)\\\omega t = 2\pi\\a_{max} = -\omega^2 A\cos(2\pi) = -\omega^2 A$

And the acceleration at A/2 is

$\frac{A}{2} = A\cos(\omega t)\\\omega t = \pi / 3\\a(t) = -\omega^2 A\cos(\pi / 3) = -\omega^2 A (0.5)$

Comparing these two results yields that the acceleration at half the amplitude is half the maximum acceleration.

5 0

3 years ago

In Ptolemy's Earth-centered model for the solar system, Venus's phase is never full as viewed from Earth because it always lies

tiny-mole [99]

<h2>Answer: in full phase</h2>

Explanation:

Galileo observed that Venus presented phases (such as those of the moon) together with a variation in size; observations that are only compatible with the fact that <u>Venus rotates around the Sun and not around Earth. </u>

This is because Venus presented <u>its smaller size when it was in full phase and the largest size when it was in the new one</u>, when it is between the Sun and the Earth.

In other words: When Venus is in its full phase it is not possible to observe it from the Earth because always in this phase, Venus, the Sun and the Earth are in conjunction (or aligned). This means <u>the Sun is between Venus and Earth</u>, in the orbit that both planets describe around this star.

Therefore, from the Earth, the Sun covers Venus.

This fact along with other discoveries were presented by Galileo to the Catholic Church (which supported the geocentric theory at that time) as a proof that completely refuted Ptolemy's geocentric system and affirmed Copernicus' heliocentric theory.

7 0

3 years ago

Describe the current atmosphere on Mars. What evidence suggests that it must have been different in the past?

alina1380 [7]

Answer:

The evidence of water that once was on mars

Explanation:

The martian atmosphere consists mostly of carbon dioxide, but is very thin, less than 1% of Earth's atmosphere. However, the strong evidence of water on Mars in the past that our missions have found means that the atmosphere must have been thicker and warmer, or water would have evaporated away very quickly.

3 0

3 years ago