Question

the loudness level of a sound, d, in decibels, is given by the formula d = 10 log (10^{12}i), where i is the intensity of the sound, in watts per meter². decibel levels range from 0, a barely audible sound, to 160, a sound resulting in a ruptured eardrum. the sound of a certain animal can be heard 500 miles away, reaching an intensity of 7.1×10^{5} watts per meter². determine the decibel level of this sound. at close range, can the sound of this animal rupture the human eardrum?
the decibel level of this animal’s sound is approximately 189 decibels. (round to the nearest whole number as needed.)
at close range, can the sound of this animal rupture the human eardrum?
a. no, the sound cannot rupture the human eardrum.
b. yes, the sound can rupture the human eardrum.

Explanation:

Step1: Recall the decibel formula

The formula for decibel level \( D \) is \( D = 10\log(10^{12}I) \), where \( I \) is the intensity in watts per meter². Given \( I = 7.1\times 10^{6} \) watts per meter².

Step2: Substitute \( I \) into the formula

First, calculate \( 10^{12}I \): \( 10^{12}\times7.1\times 10^{6}=7.1\times 10^{18} \) (using the property of exponents \( a^m\times a^n = a^{m + n} \)).

Step3: Compute the logarithm

Now, find \( \log(7.1\times 10^{18}) \). We know that \( \log(ab)=\log(a)+\log(b) \), so \( \log(7.1\times 10^{18})=\log(7.1)+\log(10^{18})\approx0.8513 + 18=18.8513 \) (since \( \log(10^{x})=x \) and \( \log(7.1)\approx0.8513 \)).

Step4: Multiply by 10 to get decibels

Multiply the result by 10: \( D = 10\times18.8513 = 188.513\approx189 \) decibels (rounded to the nearest whole number).

Step5: Determine if it can rupture eardrum

The rupture range is up to 160 decibels. Since \( 189>160 \), the sound can rupture the human eardrum.

Answer:

The decibel level is approximately 189 decibels. For the rupture question, the answer is B. Yes, the sound can rupture the human eardrum.