Dr. Thomas plays a tuba such that the sound level is 53 dB at a distance of 5 m

Question

Dr. Thomas plays a tuba such that the sound level is 53 dB at a distance of 5 m from himself. Dr.Ryder plays another tuba such that sound level is 57 dB at a distance of 5 m from him NOTESIMAGES DISCUSS UNITS STATS HELP PartDescription Answer Chk History What is the sound level at 5 m when both Dr. Thomas and Dr. Ryder play their tubas together? C: 58.455 (+0.2 dB) Try 1 - CORRECT! A: 58.455 7.69 points added to your score Hints : 1,0 | # tries: 1 show Details Dr.Stewart now joins Drs. Thomas and Ryder in playing tubas. At what sound level does Dr. Stewart need A: 57.47 to play his tuba so that the combined sound level at 5 m is 61 dB? (t0.2 dB) Try 1 - CORRECT! 7.69 points added to your score C: 57.468 Hints: 0,0 | #tries: 1 Show Details Try 1 - your answer of 134.32 m is not At what distance is the sound level from the three distinguished tuba players 34 dB? (include units with answer) correct 7.54 pts, 98% 296 try penalty Hints: 0,0 C. Format Check WARNING: 5 SF given, 3 SF expected # tries: 1 Show Details

Explanation / Answer

(A) Sound level = 10 log(I / I0)

53 = 10 log(I1 / 10^-12)

I1 = 1.995 x 10^-7 W/m^2

and 57 = 10 log(I2 / 10^-12)

I2 = 5.012 x 10^-7 W/m^2

I' = I1 + I2 = 7 x 10^-7 W/m^2

Sound level = 10 log(7 x 10^-7 / 10^-12)

= 58.455 dB

(B) 61 = 10 log(I / 10^-12)

I = 1.259 x 10^-6 W/m^2

I = I1 + I2 + I3

I3 = I - I1 - I2 = 5.582 x 10^-7 W/m^2

SL3 = 10 log(I3 / 10^-12) = 57.468 dB  

(C) for 34 dB, 34 = 10 log(I" / 10^-12)

I" = 2.512 x 10^-9 W/m^2

I r^2 = constant

(1.259 x 10^-6 ) (5^2) = (2.512 x 10^-9) r^2

r = 111.94 m Or 112 m

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