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wlad13 [49]
2 years ago
10

A ray of light crosses a boundary between two transparent materials. The medium the ray enters has a larger index of refraction.

Which of the following statements are true?
a) The wavelength of the light increases as it transitions between materials
b) The speed of the light remains constant as it transitions between materials.
c) The frequency of the light remains constant as it transitions between materials.
d) The wavelength of the light decreases as it enters into the medium with the greater index of refraction.
e) The wavelength of the light remains constant as it transitions between materials.
f) The speed of the light decreases as it enters into the medium with the greater index of refraction.
g) The frequency of the light decreases as it enters into the medium with the greater index of refraction.
h) The frequency of the light increases as it enters into the medium with the greater index of refraction.
i) The speed of the light increases as it enters the medium with the greater index of refraction.
Physics
1 answer:
Liula [17]2 years ago
7 0

The wavelength of the light decreases as it enters into the medium with the greater index of refraction. The wavelength of the light remains constant as it transitions between materials.

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Answer:

(a) 89 m/s

(b) 11000 N

Explanation:

Note that answers are given to 2 significant figures which is what we have in the values in the question.

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v=\dfrac{2\times 620 \text{ m}}{14\text{ s}} = \dfrac{1240\text{ m}}{14\text{ s}}=88.571428\ldots \text{ m/s}= 89\text{ m/s}

(b) The tension, T, is given by

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Hence,

T = \mu\cdot v^{2}

To determine \mu, we need to know the mass of the cable. We use the density formula:

\rho = \dfrac{m}{V}

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m=\rho\cdot V

If the length is denoted by l, then

\mu = \dfrac{m}{l} = \dfrac{\rho\cdot V}{l}

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V = \pi \dfrac{d^{2}}{4} l

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T = \dfrac{8050\times\pi\times0.015^2}{4} \times 88.57^2

T = 11159.4186\ldots \text{ N} = 11000 \text{ N}

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