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natta225 [31]
3 years ago
14

If the speed of light in a vaccum is c, the speed of light in a medium like glass with an index of refraction of 1.5 is : (a) 3c

/2 (b) 3c (c) 2c/3 (d) 9c/4 (e) 4c/9 Please explain in detail why it is the answer you have chosen
Physics
1 answer:
maw [93]3 years ago
7 0

Answer:

The speed of light in the medium is \dfrac{2c}{3}

(c) correct option.

Explanation:

Given that,

Speed of light in vacuum = c

Refraction index = 1.5

We need to calculate the value of speed of light in the medium

The refractive index is equal to the speed of light in vacuum divide by the speed of light in medium.

Using formula of refractive index

\mu = \dfrac{c}{v}

v=\dfrac{c}{\mu}

Where, c = speed of light in vacuum

v = speed of light in medium

Put the value into the formula

v=\dfrac{c}{1.5}

v=\dfrac{2c}{3}

Hence, The speed of light in the medium is \dfrac{2c}{3}

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A spinning disc rotating at 130 rev/min slows and stops 31 s later. how many revolutions did the disc make during this time?
gayaneshka [121]
F = 130 revs/min = 130/60 revs/s = 13/6 revs/s
t = 31s
wi = 2πf = 2π × 13/6 = 13π/3 rads/s
wf = 0 rads/s = wi + at
a = -wi/t = -13π/3 × 1/31 = -13π/93 rads/s²
wf² - wi² = 2a∅
-169π²/9 rads²/s² = 2 × -13π/93 rads/s² × ∅
∅ = 1209π/18 rads
n = ∅/2π = (1209π/18)/(2π) = 1209/36 ≈ 33.5833 revolutions.
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Find equivalent resistance between A and B​
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Explanation:

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Turn the ignition switch to start and release the key immediately or you could destroy the______________.
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Starter

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Turn the ignition switch to start and release the key immediately or you could destroy the starter.

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5 0
3 years ago
Why is an element considered a pure substance?
anastassius [24]

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3 0
3 years ago
A stone with a mass of 0.100kg rests on a frictionless, horizontal surface. A bullet of mass 2.50g traveling horizontally at 500
jolli1 [7]

Answer:

Explanation:

Given that:

mass of stone (M) = 0.100 kg

mass of bullet (m) = 2.50 g = 2.5  ×10 ⁻³ kg

initial velocity of stone (u_{stone}) = 0 m/s

Initial velocity of bullet (u_{bullet}) = (500 m/s)i

Speed of the bullet after collision (v_{bullet}) = (300 m/s) j

Suppose we represent (v_{stone}) to be the velocity of the stone after the truck, then:

From linear momentum, the law of conservation can be applied which is expressed as:

m*u_{bullet} + M*{u_{stone}}= mv_{bullet}+Mv_{stone}

(2.50*10^{-3} \ kg) (500)i+0 = (2.50*10^{-3} \ kg)(300 \ m/s)j + (0.100 \ kg)v_{stone}

(2.50*10^{-3} \ kg) (500)i- (2.50*10^{-3} \ kg)(300 \ m/s)j=  (0.100 \ kg)v_{stone}

v_{stone}= (1.25\  kg.m/s)i-(0.75\ kg m/s)j

v_{stone}= (12.5\  m/s)i-(7.5\ m/s)j

∴

The magnitude now is:

v_{stone}=\sqrt{ (12.5\  m/s)^2-(7.5\ m/s)^2}

\mathbf{v_{stone}= 14.6 \ m/s}

Using the tangent of an angle to determine the direction of the velocity after the struck;

Let θ represent the direction:

\theta = tan^{-1} (\dfrac{-7.5}{12.5})

\mathbf{\theta = 30.96^0 \ below \ the \ horizontal\ level}

5 0
3 years ago
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