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natita [175]
3 years ago
9

Yellow light is incident upon a medium from air at an angle of 45.0°. The light bends toward the normal and refracts at 28.0°. W

hat is the index of refraction of yellow light in this medium? DOK 2
a. 1.61
b. 0.66
c. 1.51
d. 1.00
Physics
2 answers:
Sav [38]3 years ago
6 0

i would say C. This link should ex plain it well https://www.britannica.com/science/refractive-index

Flauer [41]3 years ago
3 0

Answer:

c. 1.51

Explanation:

i = angle of incidence made by yellow light at air-medium interface = 45.0°

r = angle of refraction made by yellow light at air-medium interface = 28.0°

n₁ = index of refraction of air = 1

n₂ = index of refraction of medium = ?

Using Snell's law air-medium interface

n₁ Sini = n₂ Sinr

1 Sin45 = n₂ Sin28

n₂ = 1.51

So the correct choice is

c. 1.51

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3 0
2 years ago
What is the momentum of a photon having the same total energy as an electron with a kinetic energy of 100 keV?
statuscvo [17]

Answer:

The momentum of the photon is 1.707 x 10⁻²² kg.m/s

Explanation:

Given;

kinetic of electron, K.E = 100 keV = 100,000 eV = 100,000  x 1.6 x 10⁻¹⁹ J = 1.6 x 10⁻¹⁴ J

Kinetic energy is given as;

K.E = ¹/₂mv²

where;

v is speed of the electron

K.E = \frac{1}{2}mv^2\\\\mv^2 = 2K.E \\\\v^2 = \frac{2K.E}{m} \\\\v = \sqrt{\frac{2K.E}{m}} \\\\but \ momentum ,P = mv\\\\(v)m = (\sqrt{\frac{2K.E}{m}})m\\\\P_{photon} =  (\sqrt{\frac{2K.E}{m_e}})m_e\\\\P_{photon} =  (\sqrt{\frac{2\times 1.6\times 10^{-14}}{9.11\times10^{-31}}})(9.11\times 10^{-31})\\\\P_{photon} = 1.707 \times 10^{-22} \ kg.m/s

Therefore, the momentum of the photon is 1.707 x 10⁻²² kg.m/s

6 0
3 years ago
A hollow spherical shell has mass 8.20 kg and radius 0.220 m. It is initially at rest and then rotates about a stationary axis t
Likurg_2 [28]

Answer:

8.91 J

Explanation:

mass, m = 8.20 kg

radius, r = 0.22 m

Moment of inertia of the shell, I = 2/3 mr^2

                                                    = 2/3 x 8.2 x 0.22 x 0.22 = 0.265 kgm^2

n = 6 revolutions

Angular displacement, θ = 6 x 2 x π = 37.68 rad

angular acceleration, α = 0.890 rad/s^2

initial angular velocity, ωo = 0 rad/s

Let the final angular velocity is ω.

Use third equation of motion

ω² = ωo² + 2αθ

ω² = 0 + 2 x 0.890 x 37.68

ω = 8.2 rad/s

Kinetic energy,

K = \frac{1}{2}I\omega ^{2}

K = 0.5 x 0.265 x 8.2 x 8.2

K = 8.91 J

6 0
3 years ago
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Kazeer [188]
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8 0
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Answer:

a. FTh = 30 N

b. Fw = 30 N

c. a = 200 m/s2

Explanation:

See full explanation in the picture. Please rate as brainliest

5 0
3 years ago
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