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The loudness, [tex]L[/tex], measured in decibels (dB), of a sound intensity, [tex]I[/tex], measured in watts per square meter, is defined as [tex]L=10 \log \frac{I}{I_0}[/tex], where [tex]I_0=10^{-12}[/tex] is the least intense sound a human ear can hear. What is the approximate loudness of a dinner conversation with a sound intensity of [tex]10^{-7}[/tex]?

A. [tex]-58 \text{ dB}[/tex]
B. [tex]-50 \text{ dB}[/tex]
C. [tex]9 \text{ dB}[/tex]
D. [tex]50 \text{ dB}[/tex]



Answer :

GinnyAnswer
To solve for the loudness [tex]\( L \)[/tex] in decibels of a sound with intensity [tex]\( I = 10^{-7} \)[/tex] watts per square meter, we use the formula given:

[tex]\[ L = 10 \log \frac{I}{I_0} \][/tex]

where [tex]\( I_0 = 10^{-12} \)[/tex] watts per square meter is the reference intensity, the threshold of hearing.

Step-by-step solution:

1. Substitute the given values into the formula:

[tex]\[ L = 10 \log \frac{10^{-7}}{10^{-12}} \][/tex]

2. Simplify the fraction inside the logarithm:

[tex]\[ \frac{10^{-7}}{10^{-12}} = 10^{-7} \div 10^{-12} \][/tex]

Using the property of exponents [tex]\( a^{-m} / a^{-n} = a^{n-m} \)[/tex]:

[tex]\[ 10^{-7} \div 10^{-12} = 10^{(-7) - (-12)} = 10^{5} \][/tex]

Therefore:

[tex]\[ L = 10 \log (10^5) \][/tex]

3. Evaluate the logarithm:

[tex]\[ \log (10^5) = 5 \][/tex]

This is because the logarithm with base 10 of [tex]\( 10^5 \)[/tex] is just the exponent 5.

4. Multiply by 10 to find [tex]\( L \)[/tex]:

[tex]\[ L = 10 \times 5 = 50 \][/tex]

Thus, the approximate loudness of a dinner conversation with a sound intensity of [tex]\( 10^{-7} \)[/tex] watts per square meter is:

[tex]\[ L = 50 \text{ dB} \][/tex]

Therefore, the correct answer is:
[tex]\[ \boxed{50 \text{ dB}} \][/tex]

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