QuestionApril 20, 2025

The acoustic power is changed by -3 dB. The beam intensity __ decreases by half is multiplied by 4 doubles is multiplied by 3

The acoustic power is changed by -3 dB. The beam intensity __ decreases by half is multiplied by 4 doubles is multiplied by 3
The acoustic power is changed by -3 dB. The beam intensity __
decreases by half
is multiplied by 4
doubles
is multiplied by 3

Solution
4.1(270 votes)

Answer

decreases by half Explanation 1. Relate acoustic power change to intensity A -3 \, \text{dB} change corresponds to halving the intensity. The formula for decibel change is **dB = 10 \cdot \log_{10}(\frac{I_2}{I_1})**, where I_2/I_1 = 0.5 for -3 \, \text{dB}.

Explanation

1. Relate acoustic power change to intensity<br /> A $-3 \, \text{dB}$ change corresponds to halving the intensity. The formula for decibel change is **$dB = 10 \cdot \log_{10}(\frac{I_2}{I_1})$**, where $I_2/I_1 = 0.5$ for $-3 \, \text{dB}$.
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