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3 years ago

If the speed of light in a sphere is 2 × 108 m/s, its index of refraction is

1 answer:

7 0

Refractive index, n = v1/v2

Where,
v1 = Speed of light in vacuum = 3*10^8 m/s
v2 = Speed of light in a medium (sphere) = 2*10^8 m/s

Therefore,
n= (3*10*8)/(2*10^8) = 1.5

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The process of wind blowing sand from one location to another is called

lesya [120]

Answer:

weathering

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2 years ago

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The amount of diffraction depends on the size of the obstacle and the wavelength of the wave.

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I believe your answer is TRUE!
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3 years ago

When light encounters a barrier with slits cut it in it, the light will bend through the slits creating a pattern like that seen

lawyer [7]

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reflection

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A physics professor demonstrates the Doppler effect by tying a 600 Hz sound generator to a 1.0-m-long rope and whirling it aroun

nalin [4]

Answer:

Therefore the the highest frequency is 620Hz and lowest frequency is 580Hz

Explanation:

Given data

Source Frequency fs=600Hz

Length r=1.0m

RPM=100 rpm

The speed of the generator is calculated as:

$v_{s}=rw\\v_{s}=r(2\pi f)$

Substitute the given values

$v_{s}=(1.0m)2\pi (\frac{100}{60}rev/s )\\v_{s}=10.47m/s$

For approaching generator the frequency is calculated as:

$f_{+}=\frac{f_{s}}{1-\frac{v_{s}}{v} }\\f_{+}=\frac{600Hz}{1-\frac{10.47m/s}{343m/s} } \\f_{+}=620Hz$

On the other hand,for the receding generator,Doppler's effect is expressed as:

$f_{-}=\frac{f_{s}}{1+\frac{v_{s}}{v} }\\f_{-}=\frac{600Hz}{1+\frac{10.47m/s}{343m/s} } \\f_{-}=580Hz$

Therefore the the highest frequency is 620Hz and lowest frequency is 580Hz

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3 years ago

Can someone solve this problem and explain to me how you got it

evablogger [386]

Answer:

question5: F=74312.5N

question6: charge at the end of antenna=0.37N

Explanation:

Coulomb's law: the magnitude of the force of attraction or repulsion due to two charges is proportional to the product of the magnitude of the charges and inversely proportional to the square of distance between the charges.

⇒ $F\alpha\frac{q1*q2}{r^{2}}$