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Blizzard [7]
3 years ago
13

The speed of light in a vacuum is approximately 2.99 × 108 m/s, and the speed of light through a piece of glass is approximately

1.97 × 108 m/s.
What is the index of refraction for the piece of glass?

a) 0.66
b) 1.52
c) 1.02 × 10^8
d) 4.96 × 10^8
Physics
2 answers:
Vlad1618 [11]3 years ago
7 0

Answer:

The index of refraction for the piece of glass is 1.52

Explanation:

Given that,

Speed of light in vacuum, c=2.99\times 10^8\ m/s

Speed of light through a piece of glass is, v=1.97\times 10^8\ m/s

The ratio of speed of light in vacuum to the speed of light in a medium is called refractive index i.e.

n=\dfrac{c}{v}

n=\dfrac{2.99\times 10^8}{1.97\times 10^8}

n = 1.52

Hence, the index of refraction for the piece of glass is 1.52

exis [7]3 years ago
6 0

Option (b) is correct.The index of refraction for the glass is 1.52

Explanation:

velocity of light in vacuum= C= 2.99 x 10⁸m/s

Velocity of light in glass = V= 1.97 x 10⁸m/s

The refractive index is given by n=\frac{c}{v}

n= 2.99 x 10⁸/1.97 x 10⁸m/s

n= 1.52

Thus the refractive index of glass is 1.52

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Answer:

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mg = m \frac{v^2}{r}

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g = 9.8 m/s^2 is the acceleration of gravity

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Solving for v,

v=\sqrt{gr}=\sqrt{(9.8 m/s^2)(0.824 m)}=2.84 m/s

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3 years ago
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A 0.750 kg block is attached to a spring with spring constant 13.0 N/m . While the block is sitting at rest, a student hits it w
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To solve this problem we will apply the concepts related to energy conservation. From this conservation we will find the magnitude of the amplitude. Later for the second part, we will need to find the period, from which it will be possible to obtain the speed of the body.

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KE = PE

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Rearranging to find the Amplitude we have,

A = \sqrt{\frac{mv^2}{k}}

Replacing,

A = \sqrt{\frac{(0.750)(31*10^{-2})^2}{13}}

A = 0.0744m

(B) For this part we will begin by applying the concept of Period, this in order to find the speed defined in the mass-spring systems.

The Period is defined as

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Replacing,

T = 2\pi \sqrt{\frac{0.750}{13}}

T= 1.509s

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v = \frac{2\pi}{T} * \sqrt{A^2-x^2}

v = \frac{2\pi}{T} * \sqrt{A^2-0.75A^2}

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7 0
3 years ago
A solenoid with an inductance of 8 mH is connected in series with a resistance of 5 Ω and an EMF forming a series RL circuit. A
monitta

Answer:

induced EMF = 240 V

and by the lenz's law  direction of induced EMF is opposite to the applied EMF

Explanation:

given data

inductance = 8 mH

resistance = 5 Ω

current = 4.0 A

time t = 0

current grow = 4.0 A to 10.0 A

to find out

value and the direction of the induced EMF

solution

we get here induced EMF of induction is express as

E = - L \frac{dI}{dt}    ...................1

so E = - L \frac{I2 - I1}{dt}

put here value we get

E = - 8 × 10^{-3} \frac{10 - 4}{0.2*10^{-3}}

E = -40 ×  6

E = -240

take magnitude

induced EMF = 240 V

and by the lenz's law we get direction of induced EMF is opposite to the applied EMF

5 0
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Murrr4er [49]

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