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Agata [3.3K]
3 years ago
9

Jake calculates that the frequency of a wave is 230 hertz, and its wave is moving at 460 m/s. What is the wavelength of the wave

?
Physics
2 answers:
Harrizon [31]3 years ago
7 0
To find the wavelength of a wave you have to divide the speed by the frequency..it will equal 2 m...460/230=2
V=F Ω
wave speed=frequency*wavelength
Crazy boy [7]3 years ago
3 0

Explanation:

The relation between frequency, wavelength, and velocity is as follows.

                   f = \frac{v}{\lambda}

where,     f = frequency

                v = velocity

          \lambda = wavelength

It is given that frequency of the wave is 230 hertz (or 230 per second) and velocity of wave is 460 m/s. Therefore, calculate the wavelength as follows.

               f = \frac{v}{}

or,            \lambda = \frac{v}{f}

                                    = \frac{460 m/s}{230s^{-1}}

                                    = 2 m

Therefore, we can conclude that the wavelength of the wave is 2 m.

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levacccp [35]

Answer:

3.82746e+26 watts

Explanation:

There are two ways to solve this problem. One way is to use the equation

L = 4πσR²T⁴

where

L = the sun's bolometric (all-spectrum) luminous power

σ = 5.670374419e-8 W m⁻² K⁻⁴ = the Stefan-Boltzmann constant

R = 6.957e+8 meters = the sun's radius

T = 5771.8 K = the sun's effective temperature

You find that

L = 3.82746e+26 watts

The other way to solve the problem is to use the Planck integral for radiant flux.

L = 4π²R ∫(v₁,v₂) 2hv³/{c² exp[hv/(kT)]−1} dv

where

h = 6.62607015e-34 J sec

c = 299792458 m sec⁻¹

k = 1.380649e-23 J K⁻¹

v₁ = 0 = frequency band lower bound, in Hz

v₂ = ∞ = frequency band upper bound, in Hz

You find, once again, that

L = 3.82746e+26 watts

The advantage of using the Planck integral becomes clear when you want to calculate the sun's luminous power only in a specific band, rather than across the entire spectrum. For example, if we do the calculation again, except that we use

v₁ = 4.1e+14 = frequency band lower bound, in Hz

v₂ = 7.7e+14 Hz = frequency band upper bound, in Hz

restricting ourselves to the visible spectrum. We find that

L (visible) = 1.56799e+26 watts

So the fraction of the sun's luminosity that is in the visible spectrum is

L (visible) / L = 0.4096686

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Explanation:

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