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At a distance of 5.0 m from a point soun...

At a distance of 5.0 m from a point sound source, the sound intensity level is 110 dB. At what distance is the intensity level 95 dB?

A

5.0 m

B

14 m

C

7.1 m

D

28 m

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The correct Answer is:
To solve the problem of finding the distance at which the sound intensity level is 95 dB, given that at a distance of 5.0 m from a point sound source the intensity level is 110 dB, we can follow these steps: ### Step 1: Understand the relationship between intensity and decibels The sound intensity level in decibels (dB) is given by the formula: \[ L = 10 \log_{10} \left( \frac{I}{I_0} \right) \] where: - \(L\) is the sound intensity level in dB, - \(I\) is the intensity of the sound, - \(I_0\) is the reference intensity (threshold of hearing, typically \(10^{-12} \, \text{W/m}^2\)). ### Step 2: Set up the equations for the two distances At a distance of 5.0 m, the intensity level is 110 dB: \[ 110 = 10 \log_{10} \left( \frac{I_1}{I_0} \right) \] At a distance \(x\) (which we need to find), the intensity level is 95 dB: \[ 95 = 10 \log_{10} \left( \frac{I_2}{I_0} \right) \] ### Step 3: Solve for intensities \(I_1\) and \(I_2\) From the first equation: \[ \log_{10} \left( \frac{I_1}{I_0} \right) = 11 \implies \frac{I_1}{I_0} = 10^{11} \implies I_1 = 10^{11} I_0 \] From the second equation: \[ \log_{10} \left( \frac{I_2}{I_0} \right) = 9.5 \implies \frac{I_2}{I_0} = 10^{9.5} \implies I_2 = 10^{9.5} I_0 \] ### Step 4: Relate the intensities to distances The intensity of sound from a point source is inversely proportional to the square of the distance from the source: \[ I_1 \propto \frac{1}{r_1^2} \quad \text{and} \quad I_2 \propto \frac{1}{r_2^2} \] Thus, we can write: \[ \frac{I_1}{I_2} = \frac{r_2^2}{r_1^2} \] Substituting the values we found: \[ \frac{10^{11} I_0}{10^{9.5} I_0} = \frac{r_2^2}{(5)^2} \] This simplifies to: \[ 10^{1.5} = \frac{r_2^2}{25} \] ### Step 5: Solve for \(r_2\) Rearranging gives: \[ r_2^2 = 25 \cdot 10^{1.5} = 25 \cdot 31.6228 \approx 790.57 \] Taking the square root: \[ r_2 \approx \sqrt{790.57} \approx 28.14 \, \text{m} \] ### Conclusion The distance at which the intensity level is 95 dB is approximately **28 m**. ---
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