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3 years ago

What is the refractive index of a medium?

Physics

2 answers:

Travka [436]3 years ago

7 0

The refractive index is the speed of light in a vacuum over the speed of light in the medium.

The answer is B. the ratio of the velocity of light in a vacuum over the velocity of light in the medium

Verdich [7]3 years ago

4 0

Answer: Option B: the ratio of the velocity of light in a vacuum over the velocity of light in the medium, n = c/v

Explanation:

When a wave of light enters in some materials, the velocity changes depending on the material, and this is why some times when light enters in something, for example, a glass of water, the "path" of the light changes (and you can see some cool visual effects)

then, we define the refractive index of a medium as:

n = c/v

where n is the refractive index, c is the velocity of the light in the vacuum and v is the velocity of the light in the material, here you can see that n is always greater or equal than 1 ( in the case n = 1, we also have v= c)

Then, the correct option is:

option B: the ratio of the velocity of light in a vacuum over the velocity of light in the medium

You might be interested in

The total resistance of a 15-ohm, an 65-ohm and a 35-ohm resistor connected in parallel

yanalaym [24]

Answer:

I think its 9.0397 Ohms

Explanation:

take the reciprocal of all the resistances: 1/15, 1/65, 1/35

then add them: = 151/1365

then reciprocal the answer: =1365/151

And chuck it on a calculator: =9.04 Ohms

I think this is right but I'm not entirely sure. Tell me if I'm right by the way!

3 0

3 years ago

An important difference between a universal and a split-phase motor is that the split-phase motor has A. two brushes attached to

olga nikolaevna [1]

An important difference between a universal and a split-phase motor is that the split-phase motor has

A. two brushes attached to the stator.

B. a single coil formed on the rotor.

<u>C. two windings on the stator. </u>

D. an armature with a commutator.

6 0

3 years ago

A Boeing 737 airliner has a mass of 20,000 kg and the total area of both wings (top or bottom) is 100 m2. What is the pressure d

kvv77 [185]

Answer:

The correct option is A = 1960 N/m²

Explanation:

Given that,

Mass m= 20,000kg

Area A = 100m²

Pressure different between top and bottom

Assume the plane has reached a cruising altitude and is not changing elevation. Then sum the forces in the vertical direction is given as

∑Fy = Wp + FL = 0

where

Wp = is the weight of the plane, and

FL is the lift pushing up on the plane.

Let solve for FL since the mass of the plane is given:

Wp + FL = 0

FL = -Wp

FL = -mg

FL = -20,000× -9.81

FL = 196,200N

FL should be positive since it is opposing the weight of the plane.

Let Equate FL to the pressure differential multiplied by the area of the wings:

FL = (Pb −Pt)⋅A

where Pb and Pt are the static pressures on bottom and top of the wings, respectively

FL = ∆P • A

∆P = FL/A

∆P = 196,200 / 100

∆P = 1962 N/m²

∆P ≈ 1960 N/m²

The pressure difference between the top and bottom surface of each wing when the airplane is in flight at a constant altitude is approximately 1960 N/m². Option A is correct

5 0

3 years ago

Read 2 more answers

A teacher points to a microscope sitting on the desk. Which conclusion best explains

Blababa [14]

Answer:

The object is sitting

Explanation:

Therefore the object cannot be in current motion as no force has been acted upon it (exept gravitational force)

8 0

3 years ago

A motorcycle daredevil plans to ride up a 2.0-m-high, 20° ramp, sail across a 10-m-wide pool filled with hungry crocodiles, and

il63 [147K]

Answer:

He becomes a croco-dile food

Explanation:

From the question we are told that

The height is h = 2.0 m

The angle is $\theta = 20^o$

The distance is $w = 10m$

The speed is $u = 11 m/s$

The coefficient of static friction is $\mu = 0.02$

At equilibrium the forces acting on the motorcycle are mathematically represented as

$ma = mgsin \theta + F_f$

where $F_f$ is the frictional force mathematically represented as

$F_f =\mu F_x =\mu mgcos \theta$

where $F_x$ is the horizontal component of the force

substituting into the equation

$ma = mgsin \theta + \mu mg cos \theta$

$ma =mg (sin \theta + \mu cos \theta )$

making a the subject of the formula

$a = g(sin \theta = \mu cos \theta )$

substituting values

$a = 9.8 (sin(20) + (0.02 ) cos (20 ))$

$= 3.54 m/s^2$

Applying " SOHCAHTOA" rule we mathematically evaluate that length of the ramp as

$sin \theta = \frac{h}{l}$

making $l$ the subject

$l = \frac{h}{sin \theta }$

substituting values

$l = \frac{2}{sin (20)}$

$l = 5.85m$

Apply Newton equation of motion we can mathematically evaluate the final velocity at the end of the ramp as

$v^2 =u^2 + 2 (-a)l$

The negative a means it is moving against gravity

substituting values

$v^2 = (11)^2 - 2(3.54) (5.85)$

$v= \sqrt{79.582}$

$= 8.92m/s$

The initial velocity at the beginning of the pool (end of ramp) is composed of two component which is

Initial velocity along the x-axis which is mathematically evaluated as

$v_x = vcos 20^o$

substituting values

$v_x = 8.92 * cos (20)$

$= 8.38 m/s$

Initial velocity along the y-axis which is mathematically evaluated as

$v_y = vsin\theta$

substituting values

$v_y = 8.90 sin (20)$

$= 3.05 m/s$

Now the motion through the pool in the vertical direction can mathematically modeled as

$y = y_o + u_yt + \frac{1}{2} a_y t^2$

where $y_o$ is the initial height,

$u_y$ is the initial velocity in the y-axis

$a_y$ is the initial acceleration in the y axis with a constant value of ( $g = 9.8 m/s^2$ )

at the y= 0 which is when the height above ground is zero

Substituting values

$0 = 2 + (3.05)t - 0.5 (9.8)t^2$

The negative sign is because the acceleration is moving against the motion

$-(4.9)t^2 + (2.79)t + 2m = 0$

Solving using quadratic formula

$\frac{-b \pm \sqrt{b^2 -4ac} }{2a}$

substituting values

$\frac{-3.05 \pm \sqrt{(3.05)^2 - 4(-4.9) * 2} }{2 *( -4.9)}$

$t = \frac{-3.05 + 6.9}{-9.8} \ or t = \frac{-3.05 - 6.9}{-9.8}$

$t = -0.39s \ or \ t = 1.02s$

since in this case time cannot be negative

$t = 1.02s$

At this time the position the motorcycle along the x-axis is mathematically evaluated as

$x = u_x t$

x $=8.38 *1.02$

$x =8.54m$

So from this value we can see that the motorcycle would not cross the pool as the position is less that the length of the pool

7 0

3 years ago