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Anon25 [30]
3 years ago
7

Calculate the index of refraction for a medium in which the speed of light is 1.239 ✕ 108 m/s. (The speed of light in vacuum is

2.998 ✕ 108 m/s. Enter your answer to at least three decimal places.)
Physics
1 answer:
True [87]3 years ago
3 0

Answer:

Refractive index, n = 2.419

Explanation:

It is given that,

Speed of light in the medium, v=1.239\times 10^8\ m/s

Speed of light in the vacuum, c=2.998\times 10^8\ m/s

Let n is the index of refraction for a medium. We know that the refractive index for any medium is equal to the ratio of speed of light in vacuum to the speed of light in medium. It can be written as :

n=\dfrac{c}{v}

n=\dfrac{2.998\times 10^8\ m/s}{1.239\times 10^8\ m/s}

n = 2.419

So, the index of refraction for a medium is 2.419. Hence, this is the required solution.

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What is the magnitude of the force acting on a spring with a spring constant of 275 N/m that is stretched 14.3 cm?
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A force of 500 N acts horizontally on a 10,000 g body. What is its horizontal acceleration
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momentum A proton interacts electrically with a neutral HCl molecule located at the origin. At a certain time t, the proton’s po
arlik [135]

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\ m/s

Explanation:

F = Force =

m = Mass of proton = 1.7\times 10^{-27\ kg

t = Time taken = 2\times 10^{-14}\ s

Acceleration is given by

a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{}{1.7\times 10^{-27}}\\\Rightarrow a=\ m/s^2

v=u+at\\\Rightarrow v=+\times 2\times 10^{-14}\\\Rightarrow v=+\times 2\times 10^{-14}\\\Rightarrow v=+\\\Rightarrow v=\ m/s

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6 0
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The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles
belka [17]

Answer:

F_a=5.67\times 10^{-5}\ N

<u />F_b=3.49\times 10^{-5}\ N

F_c=9.16\times 10^{-5}\ N

Explanation:

Given:

  • mass of particle A, m_a=363\ kg
  • mass of particle B, m_b=517\ kg
  • mass of particle C, m_c=154\ kg
  • All the three particles lie on a straight line.
  • Distance between particle A and B, x_{ab}=0.5\ m
  • Distance between particle B and C, x_{bc}=0.25\ m

Since the gravitational force is attractive in nature it will add up when enacted from the same direction.

<u>Force on particle A due to particles B & C:</u>

F_a=G. \frac{m_a.m_b}{x_{ab}^2} +G. \frac{m_a.m_c}{(x_{ab}+x_{bc})^2}

F_a=6.67\times 10^{-11}\times (\frac{363\times 517}{0.5^2}+\frac{363\times 154}{(0.5+0.25)^2})

F_a=5.67\times 10^{-5}\ N

<u>Force on particle C due to particles B & A:</u>

<u />F_c=G.\frac{m_c.m_b}{x_{bc}^2} +G.\frac{m_c.m_a}{(x_{ab}+x_{bc})^2}<u />

F_c=6.67\times 10^{-11}\times (\frac{154\times 517}{0.25^2}+\frac{154\times 363}{(0.25+0.5)^2} )

F_c=9.16\times 10^{-5}\ N

<u>Force on particle B due to particles C & A:</u>

<u />F_b=G.\frac{m_b.m_c}{x_{bc}^2} -G.\frac{m_b.m_a}{x_{ab}^2}<u />

<u />F_b=6.67\times 10^{-11}\times (\frac{517\times 154}{0.25^2}-\frac{517\times 363}{0.5^2}  )<u />

<u />F_b=3.49\times 10^{-5}\ N<u />

3 0
3 years ago
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