Question

1. how does the intensity of a 55 - db sound compare to that of a 10 - db sound?

10^(loudness of sound 1 in db)-(loudness of sound 2 in db)/10
10^(55 - 10) ÷ 10

Explanation:

Step1: Identify the values

Sound 1 (55 dB) and Sound 2 (10 dB). Use the formula \(10^{\frac{(L_1 - L_2)}{10}}\) where \(L_1 = 55\), \(L_2 = 10\).

Step2: Calculate the difference in dB

Subtract \(L_2\) from \(L_1\): \(55 - 10 = 45\).

Step3: Divide by 10

\(\frac{45}{10}=4.5\).

Step4: Calculate the power of 10

\(10^{4.5}=10^{4 + 0.5}=10^4\times10^{0.5}\approx10000\times3.1623\approx31623\).

Answer:

The intensity of the 55 - dB sound is approximately 31623 times that of the 10 - dB sound.