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2 years ago

15

A particular material has an index of refraction of 1.25. What percent of the speed of light in a vacuum is the speed of light i

n the material?

2 answers:

Len [333]2 years ago

7 0

Answer:

The percentage of speed of light in vacuum to the speed of light in the said material is 80%

Explanation:

The common values of refractive index are between 1 and 2 since nothing can travel faster than the speed of light, therefore, no material has a refractive index lower than 1.

According to the formula $n=\frac{c}{v}$

where $n$ is the index of refraction

$c$ is the speed of light in vacuum

and $v$ is the speed of light in the material, it can be seen that n and v are inversely proportional which means greater the refractive index lower is the speed of light.

Since we know that speed of light in vacuum is 300,000 km/s using the formula we get,

$v=\frac{c}n}$

$v=\frac{300000}{1.25}=240,000 km/s$

for finding percentage,

$=\frac{240000}{300000}*100 = 80$ %

beks73 [17]2 years ago

3 0

Answer:

80% (Eighty percent)

Explanation:

The material has a refractive index (n) of 1.25

Speed of light in a vacuum (c) is 2.99792458 x 10⁸ m/s

We can find the speed of light in the material (v) using the relationship

n = c/v, similarly

v = c/n

therefore v = 2.99792458 x 10⁸ m/s ÷ (1.25) = 239 833 966 m/s

v = 239 833 966 m/s

Therefore the percentage of the speed of light in a vacuum that is the speed of light in the material can be calculated as

(v/c) × 100 = (1/n) × 100 = (1/1.25) × 100 = 0.8 × 100 = 80%

Therefore speed of light in the material (v) is eighty percent of the speed of light in the vacuum (c)

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Alenkasestr [34]

Answer:

-13.18°C

Explanation:

To develop the problem it is necessary to consider the concepts related to the thermal conduction rate.

Its definition is given by the function

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Where,

Q = The amount of heat transferred

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k = Thermal conductivity constant

A = Cross-sectional area

$\Delta T =$ The difference in temperature between one side of the material and the other

d= thickness of the material

The problem says that there is a loss of heat twice that of the initial state, that is

$Q_2 = 2*Q_1$

Replacing,

$kA\frac{\Delta T_m}{x} = 2*kA\frac{\Delta T}{x}$

$\frac{\Delta T}{x}=2*\frac{\Delta T}{x}$

$\frac{T_i-T_o}{x} = 2\frac{T_1-T_2}{x}$

$\frac{28.9-T_o}{x} = 2\frac{28.9-7.86}{x}$

Solvinf for $T_o$ ,

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Roman55 [17]

Let $\theta$ be the direction the swimmer must swim relative to east. Then her velocity relative to the water is

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The current has velocity vector (relative to the Earth)

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The swimmer's resultant velocity (her velocity relative to the Earth) is then

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Mice21 [21]

Answer:

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1 year ago

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