Question

What is the decibel level of the noise from a firecracker with intensity 10^-4 watts per square inch? Use a logarithmic model

Answer 1

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.

Answer 2

Answer:

$\large \boxed{\text{110 db}}$

Step-by-step explanation:

The intensity β of a sound wave in decibels is given by the formula

$\beta = 10 \log \left (\dfrac{I}{I_{0}} \right)$

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

$\text{Sound level} = \dfrac{10^{-4}\text{ W}}{\text{1 in}^{2}} \times \left(\dfrac{\text{39.37 in}}{\text{1 m}}\right )^{2} = 1.5 \times 10^{-1} \text{ W \cdot m}^{-2}$

2. Convert the intensity to decibels.

$\beta = 10 \log \left (\dfrac{1.5 \times 10^{-1}}{1\times 10^{-12}} \right) = 10\log(1.5 \times 10^{11}) = 10 \times 11 = \textbf{110 db}\\\text{The sound intensity of the firecracker is \large \boxed{\textbf{110 db}} }$