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Ratling [72]
3 years ago
11

Snell's Law: Light goes from material having a refractive index of n 1 into a material with refractive index n 2. If the refract

ed light is bent away from the normal, what can you conclude about the indices of refraction
Physics
1 answer:
Luda [366]3 years ago
4 0

Answer:

a) the light is close to normal therefore the reference incidence of medium 1 is less than medium n2 where the ray is transmitted.

b) The ray is far from normal in this case the refractive index of medium 1 is greater than index of medium 2

Explanation:

The expression for the angle of refraction is

          n₁ sin θ₁ = n₂ sin θ₂

refractive index n₁ is for incident light and n₂ is for transmitted light.

We have two cases

a) the light is close to normal therefore the reference incidence of medium 1 is less than medium n2 where the ray is transmitted.

b) The ray is far from normal in this case the refractive index of medium 1 is greater than index of medium 2

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A simulation shows whether changing various weather conditions increases
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Answer:I don’t know

Explanation:

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2 years ago
A double-slit experiment is set up using red light (λ = 706 nm). A first order bright fringe is seen at a given location on a sc
Elanso [62]

Answer:

λ = 470.66 nm

Explanation:

for bright fringey_m = \frac{m\lambda D}{d}

D= distance between slit and screen

d= distance between the slits

for first order bright fringe m = 1,

        y_1 = \frac{1\lambda D}{d}

         y_1 = {706*D}{d}

for dark fringe,we have

y_m = {(m + 1/2)\lambda D}{d}    

Now to get the dark fringes at the same location we should have;

(706)D/d = (m + 1/2)λD/d    

put m = 1

(1 + 1/2)λ = (706)

λ = 470.66 nm

6 0
4 years ago
Which answer choice provides the best set of labels for Wave A and Wave B?
hodyreva [135]
A has less energy and lower frequency, while B has greater energy and higher frequency.
3 0
4 years ago
Read 2 more answers
DONT ANSWER WITH A LINK PLEASE I NEED AN ANSWER FROM SOMEONE!
rjkz [21]
At 4 m/s?

How do the two kinetic energies compare to one another? QUADRUPLES !

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3 years ago
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A reasonable estimate of the moment of inertia of an ice skater spinning with her arms at her sides can be made by modeling most
Oxana [17]

Answer:

A)  I_{total} = 1.44 kg m², B) moment of inertia must increase

Explanation:

The moment of inertia is defined by

     I = ∫ r² dm

For figures with symmetry it is tabulated, in the case of a cylinder the moment of inertia with respect to a vertical axis is

      I = ½ m R²

A very useful theorem is the parallel axis theorem that states that the moment of inertia with respect to another axis parallel to the center of mass is

    I = I_{cm} + m D²

Let's apply these equations to our case

The moment of inertia is a scalar quantity, so we can add the moment of inertia of the body and both arms

      I_{total}=I_{body} + 2 I_{arm}

       I_{body} = ½ M R²

The total mass is 64 kg, 1/8 corresponds to the arms and the rest to the body

       M = 7/8 m total

       M = 7/8 64

       M = 56 kg

The mass of the arms is

      m’= 1/8 m total

      m’= 1/8 64

      m’= 8 kg

As it has two arms the mass of each arm is half

     m = ½ m ’

     m = 4 kg

The arms are very thin, we will approximate them as a particle

    I_{arm} = M D²

Let's write the equation

     I_{total} = ½ M R² + 2 (m D²)

Let's calculate

    I_{total} = ½ 56 0.20² + 2 4 0.20²

    I_{total} = 1.12 + 0.32

    I_{total} = 1.44 kg m²

b) if you separate the arms from the body, the distance D increases quadratically, so the moment of inertia must increase

6 0
4 years ago
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