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Lina20 [59]
3 years ago
5

By what factor is the intensity of sound at a rock concert louder than that of a whisper when the two intensity levels are 120 d

B and 20 dB, respectively
Physics
1 answer:
Kipish [7]3 years ago
3 0

Answer:

The intensity of sound at a rock concert is louder than that of a whisper by a factor of 1 x 10¹⁰

Explanation:

Given;

rock concert sound intensity level, β₁ = 120 dB

whisper sound intensity level, β₂ = 20 dB

The sound intensity level is given as;

\beta = 10Log(\frac{I}{I_o} )\\\\

where;

I₀ is the threshold sound intensity of hearing = 10⁻¹² W/m²

I is the sound intensity

Intensity of sound at rock concert ;

120 =  10Log(\frac{I}{10^{-12}} )\\\\12 =  Log(\frac{I}{10^{-12}} )\\\\10^{12} = \frac{I}{10^{-12}}\\\\I = 10^{12}  * 10^{-12}\\\\I = 10^0\\\\I = 1 \ W/m^2

The intensity of sound of a whisper;

20 =  10Log(\frac{I}{10^{-12}} )\\\\2 =  Log(\frac{I}{10^{-12}} )\\\\10^{2} = \frac{I}{10^{-12}}\\\\I = 10^{2}  * 10^{-12}\\\\I = 10^{-10} \ W/m^2\\\\

Determine the factor by which the intensity of sound at a rock concert louder than that of a whisper

\frac{I_{Concert}}{I_{whisper}} = \frac{1}{10^{-10}} \\\\\frac{I_{Concert}}{I_{whisper}} = 1 * 10^{10}\\\\I_{Concert} = 1 * 10^{10}*I_{whisper}

Therefore, the intensity of sound at a rock concert is louder than that of a whisper by a factor of 1 x 10¹⁰

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      The resonance frequency is  f = 18.0 kHz =  18*10^{3} Hz

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Generally the inductance of this air-core inductor is mathematically represented as

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This inductance can also be mathematically represented as

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