Answer:
[tex]\large \boxed{\text{110 db}}[/tex]
Step-by-step explanation:
The intensity β of a sound wave in decibels is given by the formula
[tex]\beta = 10 \log \left (\dfrac{I}{I_{0}} \right)[/tex]
and I₀ = the reference intensity (10⁻¹² W/m²)
Data:
I = 10⁻⁴ W·in⁻²
Calculations:
1. Convert watts per square inch to watts per square metre
[tex]\text{Sound level} = \dfrac{10^{-4}\text{ W}}{\text{1 in}^{2}} \times \left(\dfrac{\text{39.37 in}}{\text{1 m}}\right )^{2} = 1.5 \times 10^{-1} \text{ W$\cdot$m}^{-2}[/tex]
2. Convert the intensity to decibels.
[tex]\beta = 10 \log \left (\dfrac{1.5 \times 10^{-1}}{1\times 10^{-12}} \right) = 10\log(1.5 \times 10^{11}) = 10 \times 11 = \textbf{110 db}\\\text{The sound intensity of the firecracker is $\large \boxed{\textbf{110 db}}$}[/tex]
Answer:
80 db
Step-by-step explanation:
D=10log(I10−12)
Substitute in the intensity level, I, and then simplify to find
DD=10log(10−410−12)=10log(108).
Since log108=8, simplify to find
DD=10⋅8=80.
The decibel level is 80dB.
