Answer :
To find the intensity in decibels for each value of [tex]\( I \)[/tex], we use the formula for calculating the intensity level in decibels:
[tex]\[ I(\text{dB}) = 10 \cdot \log_{10} \left( \frac{I}{I_0} \right) \][/tex]
where [tex]\( I \)[/tex] is the intensity of the sound, and [tex]\( I_0 \)[/tex] is the reference intensity.
Let's calculate each value step by step:
### Normal Conversation:
Given:
[tex]\[ I = 10^6 I_0 \][/tex]
Using the formula:
[tex]\[ I(\text{dB}) = 10 \cdot \log_{10}(10^6) = 10 \cdot 6 = 60 \text{ dB} \][/tex]
So, for a normal conversation, the intensity in decibels is:
[tex]\[ I(\text{dB}) = 60 \text{ dB} \][/tex]
### Power Saw at 3 feet:
Given:
[tex]\[ I = 10^{11} I_0 \][/tex]
Using the formula:
[tex]\[ I(\text{dB}) = 10 \cdot \log_{10}(10^{11}) = 10 \cdot 11 = 110 \text{ dB} \][/tex]
So, for a power saw at 3 feet, the intensity in decibels is:
[tex]\[ I(\text{dB}) = 110 \text{ dB} \][/tex]
### Jet Engine at 100 feet:
Given:
[tex]\[ I = 10^{18} I_0 \][/tex]
Using the formula:
[tex]\[ I(\text{dB}) = 10 \cdot \log_{10}(10^{18}) = 10 \cdot 18 = 180 \text{ dB} \][/tex]
So, for a jet engine at 100 feet, the intensity in decibels is:
[tex]\[ I(\text{dB}) = 180 \text{ dB} \][/tex]
Therefore, we have the following results:
- Normal conversation: [tex]\( I(\text{dB}) = 60 \text{ dB} \)[/tex]
- Power saw at 3 feet: [tex]\( I(\text{dB}) = 110 \text{ dB} \)[/tex]
- Jet engine at 100 feet: [tex]\( I(\text{dB}) = 180 \text{ dB} \)[/tex]
These values represent the intensity levels in decibels for each corresponding sound intensity.
[tex]\[ I(\text{dB}) = 10 \cdot \log_{10} \left( \frac{I}{I_0} \right) \][/tex]
where [tex]\( I \)[/tex] is the intensity of the sound, and [tex]\( I_0 \)[/tex] is the reference intensity.
Let's calculate each value step by step:
### Normal Conversation:
Given:
[tex]\[ I = 10^6 I_0 \][/tex]
Using the formula:
[tex]\[ I(\text{dB}) = 10 \cdot \log_{10}(10^6) = 10 \cdot 6 = 60 \text{ dB} \][/tex]
So, for a normal conversation, the intensity in decibels is:
[tex]\[ I(\text{dB}) = 60 \text{ dB} \][/tex]
### Power Saw at 3 feet:
Given:
[tex]\[ I = 10^{11} I_0 \][/tex]
Using the formula:
[tex]\[ I(\text{dB}) = 10 \cdot \log_{10}(10^{11}) = 10 \cdot 11 = 110 \text{ dB} \][/tex]
So, for a power saw at 3 feet, the intensity in decibels is:
[tex]\[ I(\text{dB}) = 110 \text{ dB} \][/tex]
### Jet Engine at 100 feet:
Given:
[tex]\[ I = 10^{18} I_0 \][/tex]
Using the formula:
[tex]\[ I(\text{dB}) = 10 \cdot \log_{10}(10^{18}) = 10 \cdot 18 = 180 \text{ dB} \][/tex]
So, for a jet engine at 100 feet, the intensity in decibels is:
[tex]\[ I(\text{dB}) = 180 \text{ dB} \][/tex]
Therefore, we have the following results:
- Normal conversation: [tex]\( I(\text{dB}) = 60 \text{ dB} \)[/tex]
- Power saw at 3 feet: [tex]\( I(\text{dB}) = 110 \text{ dB} \)[/tex]
- Jet engine at 100 feet: [tex]\( I(\text{dB}) = 180 \text{ dB} \)[/tex]
These values represent the intensity levels in decibels for each corresponding sound intensity.
