Sure, let’s go through the problem step-by-step.
### Part (a)
To find the intensity ratio between the hair dryer and the air conditioner, given their sound levels in decibels, we use the following formula:
[tex]\[
\beta_2 - \beta_1 = 10 \log \left(\frac{I_2}{I_1}\right)
\][/tex]
Where:
- [tex]\(\beta_2 = 70 \, \text{dB}\)[/tex] (sound intensity level of the hair dryer)
- [tex]\(\beta_1 = 50 \, \text{dB}\)[/tex] (sound intensity level of the air conditioner)
Step-by-Step Solution:
1. Calculate the difference in decibel levels.
[tex]\[
\beta_2 - \beta_1 = 70 \, \text{dB} - 50 \, \text{dB} = 20 \, \text{dB}
\][/tex]
2. Substitute [tex]\(\beta_2 - \beta_1\)[/tex] into the original formula.
[tex]\[
20 = 10 \log \left(\frac{I_2}{I_1}\right)
\][/tex]
3. Divide both sides of the equation by 10.
[tex]\[
2 = \log \left(\frac{I_2}{I_1}\right)
\][/tex]
4. To solve for [tex]\(\frac{I_2}{I_1}\)[/tex], rewrite the equation in exponential form.
[tex]\[
\frac{I_2}{I_1} = 10^2
\][/tex]
5. Calculate [tex]\(10^2\)[/tex].
[tex]\[
\frac{I_2}{I_1} = 100
\][/tex]
Hence, the sound from the hair dryer is 100 times more intense than the sound from the air conditioner.
### Part (b)
To determine the decibel rating of the teacher's voice, given that it is 80 times as intense as a 20 dB sound, we use the formula:
[tex]\[
\beta_2 = \beta_1 + 10 \log \left(\frac{I_2}{I_1}\right)
\][/tex]
Where:
- [tex]\(\frac{I_2}{I_1} = 80\)[/tex] (the teacher’s voice is 80 times as intense)
- [tex]\(\beta_1 = 20 \, \text{dB}\)[/tex]
Step-by-Step Solution:
1. Substitute the known values into the formula.
[tex]\[
\beta_2 = 20 \, \text{dB} + 10 \log(80)
\][/tex]
2. Calculate [tex]\(10 \log(80)\)[/tex].
Note: [tex]\(\log(80) \approx 1.903\)[/tex]
[tex]\[
10 \log(80) \approx 10 \times 1.903 = 19.03
\][/tex]
3. Add this result to [tex]\(\beta_1\)[/tex].
[tex]\[
\beta_2 = 20 + 19.03 = 39.03 \, \text{dB}
\][/tex]
Therefore, the decibel rating of the teacher's voice is approximately 39.03 dB.
