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denis-greek [22]
3 years ago
5

What is the speed of light (in m/s) in air? (Enter your answer to at least four significant figures. Assume the speed of light i

n a vacuum is 2.997 ✕ 108 m/s.) m/s What is the speed of light (in m/s) in polystyrene? m/s
Physics
1 answer:
jenyasd209 [6]3 years ago
8 0

Answer:

The speed of light in air is 2.996x10⁸ m/s, and polystyrene is 1.873x10⁸ m/s.

Explanation:

To find the speed of light in air and in polystyrene we need to use the following equation:

c_{m} = \frac{c}{n}

Where:

c_{m}: is the speed of light in the medium

n: is the refractive index of the medium

In air:

c_{a} = \frac{c}{n_{a}} = \frac{2.997 \cdot 10^{8} m/s}{1.0003} = 2.996 \cdot 10^{8} m/s

In polystyrene:

c_{p} = \frac{c}{n_{p}} = \frac{2.997 \cdot 10^{8} m/s}{1.6} = 1.873 \cdot 10^{8} m/s  

Therefore, the speed of light in air is 2.996x10⁸ m/s, and polystyrene is 1.873x10⁸ m/s.

I hope it helps you!

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Lukalu is rappelling off a cliff. The parametric equations that describe her horizontal and vertical position as a function of t
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Answer:

2.5 s, 5 m

Explanation:

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0=-16t^2 +100\\16t^2 = 100\\t=\sqrt{\frac{100}{16}}=2.5 s

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3 years ago
Provide descriptions and examples of the following forces: friction, gravity and elastic.
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Answer:

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The friction force is the force exerted by a surface as an object moves across it or makes an effort to move across it. There are at least two types of friction force - sliding and static friction. Though it is not always the case, the friction force often opposes the motion of an object. For example, if a book slides across the surface of a desk, then the desk exerts a friction force in the opposite direction of its motion. Friction results from the two surfaces being pressed together closely, causing inter molecular attractive forces between molecules of different surfaces. As such, friction depends upon the nature of the two surfaces and upon the degree to which they are pressed together. The maximum amount of friction force that a surface can exert upon an

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3 years ago
An important diagnostic tool for heart disease is the pressure difference between blood pressure in the heart and in the aorta l
butalik [34]

Answer:

a)   f ’’ = f₀ \frac{1 + \frac{v}{c} }{1- \frac{v}{c} } , b)   Δf = 2 f₀ \frac{v}{c}

Explanation:

a) This is a Doppler effect exercise, which we must solve in two parts in the first the emitter is fixed and in the second when the sound is reflected the emitter is mobile.

Let's look for the frequency (f ’) that the mobile aorta receives, the blood is leaving the aorta or is moving towards the source

                    f ’= fo\frac{c+v}{c}

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                   f ’’ = f’ \frac{c}{ c-v}

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therefore the received frequency is

                 f ’’ = f₀   \frac{c}{c-v}

let's simplify the expression

                f ’’ = f₀ \frac{c+v}{c-v}

                f ’’ = f₀ \frac{1 + \frac{v}{c} }{1- \frac{v}{c} }

         

b) At the low speed limit v <c, we can expand the quantity

                 (1 -x)ⁿ = 1 - x + n (n-1) x² + ...

                 ( 1- \frac{v}{c} ) ^{-1} = 1 + \frac{v}{c}

 

                f ’’ = fo ( 1+ \frac{v}{c}) ( 1 + \frac{v}{c} )

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leave the linear term

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the sound difference

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               Δf = 2 f₀ \frac{v}{c}

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