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The intensity, or loudness, of a sound can be measured in decibels (dB) according to the equation [tex]\( I(dB) = 10 \log \left(\frac{I}{I_0}\right) \)[/tex], where [tex]\( I \)[/tex] is the intensity of a given sound and [tex]\( I_0 \)[/tex] is the threshold of hearing intensity.

What is the intensity, in decibels, [tex]\( I(dB) \)[/tex], when [tex]\( I = 10^2 I_0 \)[/tex]?

A. 8
B. 9
C. 19
D. 80



Answer :

GinnyAnswer
To solve this problem, we'll use the formula for calculating the intensity of sound in decibels:

[tex]\[ I(dB) = 10 \log_{10}\left(\frac{I}{I_0}\right) \][/tex]

Given:
- [tex]\( I = 10^2 \cdot I_0 \)[/tex]
- [tex]\( I_0 \)[/tex] is the threshold of hearing intensity.

Step-by-step, we can substitute the given values into the formula:

1. Substitute [tex]\( I \)[/tex] with [tex]\( 10^2 \cdot I_0 \)[/tex]:

[tex]\[ I(dB) = 10 \log_{10}\left(\frac{10^2 \cdot I_0}{I_0}\right) \][/tex]

2. Simplify the fraction inside the logarithm:

[tex]\[ I(dB) = 10 \log_{10}(10^2) \][/tex]

3. Recognize that the [tex]\( I_0 \)[/tex] terms cancel each other out:

[tex]\[ I(dB) = 10 \log_{10}(10^2) \][/tex]

4. Apply the properties of logarithms. The logarithm of a power can be written as the exponent times the logarithm of the base:

[tex]\[ \log_{10}(10^2) = 2 \][/tex]

5. Multiply the result by 10:

[tex]\[ I(dB) = 10 \cdot 2 \][/tex]

6. Simplify the multiplication:

[tex]\[ I(dB) = 20 \][/tex]

Therefore, the intensity in decibels is [tex]\( 20 \, dB \)[/tex].

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