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oksano4ka [1.4K]
3 years ago
6

An 11,000-watt radio station transmits at 880 kHz. Calculate the energy of a single photon at the transmitted frequency.

Physics
1 answer:
prisoha [69]3 years ago
8 0

The energy of a single photon at the transmitted frequency is \bold{5834.4 \times 10^{-31} \mathrm{J}}

Answer: Option b

<u>Solution:</u>

Energy of photon is given as E=\frac{h c}{\lambda}

Where  c is the velocity of Light \left(3 \times 10^{8} m / s\right)

h is planck's constant  \left(6.63 \times 10^{-34}\right)

λ is the wavelength of photon  

Energy of photon can be rewritten as E=h \times f

Where f is the frequency of photon  

Frequency of photon is obtained by dividing velocity of light by wavelength of photon.

f=\frac{c}{\lambda}

E=6.63 \times 10^{-34} \times 880 \times 10^{3}=5834.4 \times 10^{-31} \mathrm{J}

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Answer:

(a) I=8.84*10^{-9}W/m^{2}

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The source is 1.00 μW

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The intensity at a certain distance from a point source that emits sound wave is given as:

I=\frac{P_{s} }{4\pi r^{2} }\\I=\frac{1*10^{-6}W  }{4\pi (3m)^{2} } \\I=8.84*10^{-9}W/m^{2}

For Part (b)

The Sound level is given by

β=(10dB)×log(I/I₀)

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I₀=1×10⁻¹²W/m²

Substitute the given values to find Sound level

So

\beta =(10dB)log(\frac{8.84*10^{-9}W/m^{2}  }{1*10^{-12}W/m^{2}} )\\\beta =39.46dB

4 0
3 years ago
A student has a small piece of steel.
Simora [160]

Answer:

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use appropriate apparatus and methods to measure volume and mass and use that to investigate density

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5.444\times 10^{-4}

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P_A=m_Av_A, P_B=m_Bv_B\\\\\therefore m_Av_A,=m_Bv_B

Let the energy released during the decay be Q:

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