Ask Question
6 July, 05:55

Jim quadruples the distance between himself and a sound source. What is the change in decibels of the sound he hears?

A. - 6 dB

B. - 12 dB

C. + 6 dB

D. + 12 dB

+5
Answers (1)
  1. 6 July, 06:08
    0
    Quadrupling the distance reduces the sound to 1/16 of its original intensity.

    10 log (1/16) = - 12 dB (choice-B)
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “Jim quadruples the distance between himself and a sound source. What is the change in decibels of the sound he hears? A. - 6 dB B. - 12 dB ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers