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mrs_skeptik [129]
2 years ago
10

The wavelengths of visible light vary from about 300 nm to 700 nm. What is the range of frequencies of visible light in a vacuum

Physics
1 answer:
exis [7]2 years ago
6 0

The range of frequencies  of visible light in a vacuum is mathematically given as

Fmin=4.19*10^14Hz  to Fmax=1*10^15Hz

<h3>What is the range of frequencies of visible light in a vacuum?</h3>

Question Parameters:

The wavelengths of visible light vary from about 300 nm to 700 nm.

Generally, the equation for the frequency   is mathematically given as

F=C/\lambda

Therefore

For Fmax

Fmax=\frac{300*10^8}{3*10^9}

Fmax=1*10^15Hz

Where

Fmin=\frac{3*10^8}{700*10^9}

Fmin=4.19*10^14Hz

For more information on Wave

brainly.com/question/3004869

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(b) The particle displacement y of air molecules due to a sound wave is given by y 4m and w = = 0.008 cos wt sin kz. Where k - m
serious [3.7K]

The distance between two consecutive nodes and the amplitude after 0.56s are m/2 and 1.75×10^(-4) m respectively.

<h3>What's the distance between consecutive nodes of the displacement of air molecules?</h3>
  • Wavelength is the distance between two consecutive nodes or toughs or crests or anti-nodes.
  • So, distance between consecutive nodes = wavelength = 2π÷k

= 2π/(4π÷m)

= m/2

<h3>What's the amplitude after 0.56s of the displacement of air molecules?</h3>

Displacement after 0.56 s = 0.008×cos(50π×0.56s)

=1.75×10^(-4) m

Thus, we can conclude that the distance between consecutive nodes and displacement after 0.56 s are m/2 and 1.75×10^(-4) m respectively.

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: The particle displacement y of air molecules due to a sound wave is given by y=0.008coswtsinkz where k=4π÷m and w=50π rads/s.

Calculate:

I) the distance between 2 consecutive nodes

ii) the amplitude after 0.565s

Learn more about the wavelength here:

brainly.com/question/10750459

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