The loudness, *L*, measured in decibels (Db), of a sound intensity, *I*, measured in watts per square meter, is defined as , where  and is the least intense sound a human ear can hear. Brandon is trying to take a nap, and he can barely hear his neighbor mowing the lawn. The sound intensity level that Brandon can hear is 10-10. Ahmad, Brandon’s neighbor that lives across the street, is mowing the lawn, and the sound intensity level of the mower is 10-4. How does Brandon’s sound intensity level compare to Ahmad’s mower? 


Brandon’s sound intensity is  the level of Ahmad’s mower.

Brandon’s sound intensity is   the level of Ahmad’s mower.

Brandon’s sound intensity is 20 times the level of Ahmad’s mower.

Brandon’s sound intensity is 80 times the level of Ahmad’s mower.

The formula for loudness, \( L \), in decibels (dB), is given by:

\[
L = 10 \log\left(\frac{I}{I_0}\right)
\]

where \( I \) is the sound intensity in watts per square meter, and \( I_0 \) is the reference intensity (the least intense sound a human ear can hear), which is \( I_0 = 10^{-12} \, \text{W/m}^2 \).

### Brandon's sound intensity:
Brandon can hear a sound intensity level of \( I_B = 10^{-10} \, \text{W/m}^2 \). Using the formula:

\[
L_B = 10 \log\left(\frac{10^{-10}}{10^{-12}}\right)
= 10 \log\left(10^2\right)
= 10 \times 2 = 20 \, \text{dB}
\]

### Ahmad's mower sound intensity:
Ahmad's mower produces a sound intensity level of \( I_A = 10^{-4} \, \text{W/m}^2 \). Using the formula:

\[
L_A = 10 \log\left(\frac{10^{-4}}{10^{-12}}\right)
= 10 \log\left(10^8\right)
= 10 \times 8 = 80 \, \text{dB}
\]

### Comparing sound intensities:
The comparison between Brandon's sound intensity (\( I_B = 10^{-10} \)) and Ahmad's mower sound intensity (\( I_A = 10^{-4} \)) is a ratio:

\[
\frac{I_A}{I_B} = \frac{10^{-4}}{10^{-10}} = 10^6
\]

So, Ahmad's mower sound intensity is **1,000,000 times** more intense than what Brandon hears. Therefore, **Brandon’s sound intensity is \(\frac{1}{1,000,000}\)** the level of Ahmad’s mower.

None of the answer options directly provide this precise comparison. However, if we assume the question intends to refer to a logarithmic comparison, we see that Brandon’s sound intensity is significantly smaller than Ahmad’s, not 20 or 80 times larger.

Let me know if you need more details!

Here are 5 related questions:
1. How does decibel scale work in terms of sound intensity ratios?
2. How would the comparison change if Brandon heard a sound intensity of \( 10^{-8} \)?
3. What is the formula to convert sound intensity to decibels?
4. How much louder is a 90 dB sound compared to a 50 dB sound?
5. What is the threshold of pain in terms of sound intensity level?

**Tip:** The decibel scale is logarithmic, so an increase of 10 dB represents a tenfold increase in intensity!

The loudness, L, measured in decibels (dB), of a sound intensity, I, measured in watts per square meter, is defined as L = 10 log(I/I_0), where I_0 is the least intense sound a human ear can hear. Brandon is trying to take a nap, and he can barely hear his neighbor mowing the lawn. The sound intensity level that Brandon can hear is 10^-10. Ahmad, Brandon’s neighbor that lives across the street, is mowing the lawn, and the sound intensity level of the mower is 10^-4. How does Brandon’s sound intensity level compare to Ahmad’s mower?

Explore how to compare sound intensity levels using the decibel scale. In this problem, Brandon hears a sound intensity of 10^-10 W/m^2 while Ahmad's lawn mower produces an intensity of 10^-4 W/m^2. Learn how to use the decibel formula to calculate and compare sound intensities. Suitable for Grades 9-12.

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