+154 When comparing the sound measured in decibles, a normal convo between two people measured 60 dB while the threshold of pain from the loud noise is ofter concidered to be 130 dB.
The formula is β=10log(I/Io)
Io =10-12 w/m2
a.) determine how many times louder the intensity of the thrreshold of pain is compaired to the conversation
I need help solving... 60dB=10log(I/10-12)
b.) if the convo intensity was trippled louder, determine what the new decibel level would be.
+33616 With β = 10*log(l/lo) then l = lo*10(β/10)
a) lconv = lo*10(60/10) or lconv = lo106 lpain = lo*10(130/10) or lpain = lo*1013
Therefore lpain/lconv = 1013/106 = 1013-6 = 107 . That is, the intensity at the pain threshold is ten million times greater than at conversation level.
See if you can now do part b)
+33616 With β = 10*log(l/lo) then l = lo*10(β/10)
a) lconv = lo*10(60/10) or lconv = lo106 lpain = lo*10(130/10) or lpain = lo*1013
Therefore lpain/lconv = 1013/106 = 1013-6 = 107 . That is, the intensity at the pain threshold is ten million times greater than at conversation level.
See if you can now do part b)
+154 This helped so much actually thanks ! I got 65dB for the seccond answer
