Answer:
10 Times (greater)
Explanation:
The given parameters are;
The number of decibels of the first sound, 35-db
The number of decibels of the second sound, 25 db
We have;
[tex]\beta (dB) = 10 \cdot log_{10} \left (\dfrac{I}{I_0 \right)}[/tex]
Where;
I₀ = 10⁻¹² W/m² = The lowest perceivable sound
Therefore, we have;
[tex]I = I_0 \times 10^{\left (\dfrac{\beta }{10} \right) }[/tex]
Substituting the known values, gives;
When β = 35-db, we get;
[tex]I_{35} = 10^{-12} \times 10^{\left (\dfrac{35 }{10} \right) } = 10^{-12} \times 10^{3.5} = 10^{-8.5}[/tex]
When β = 25-db, we get;
[tex]I_{25} = 10^{-12} \times 10^{\left (\dfrac{25 }{10} \right) } = 10^{-12} \times 10^{2.5} = 10^{-9.5}[/tex]
Therefore, we get the number of times the intensity of a 35-db sound is compared to a 25-db sound, which is, I₃₅/I₂₅ is given as follows;
[tex]\dfrac{I_{35}}{I_{25}} = \dfrac{10^{-8.5}}{10^{-9.5}} = 10[/tex]
Therefore, the intensity of a 35-db sound is 10 times greater than the intensity of a 25-db sound
