Problem_Set_#14

(1/2gt 1 2 = vt 2 )*2 = gt 1 2 = 2vt 2 d = .5g* t 1 2 d = 330* t 2 t 1 +t 2 =1.5 d = 300(1.5 - t 1 ) 300(1.5 - t 1 ) = .5g* t 1 2 --> 300(1.5 - t 1 ) = .5*9.8* t 1 2 t = 1.46sec d = .5*9.8* t 1 *2 = 10.44m or 11m (b) If the depth of the well were doubled, the time required to hear the splash would less than 3.0 seconds; this is because the fall time increases. 36. The distance to a point source is decreased by a factor of three.   (a)   By what multiplicative factor does the intensity increase?   (b)   By what additive amount does the intensity level increase? Solution: (a) Solve by calculating the ratio of intensities. I new /I old = (P/4πr new 2 )/(P/4πr old 2 ) = (r old /r new ) = (r old /1/3r old ) = 9 (b) Solve for intensity by substituting the new intensity level to find the change. β = 10log(I new /I 0 ) = 10log(9I old /I 0 ) --> 10log(I old /I 0 )+10log(9) = β old +10log(9) β new - β old = 10log(9) = +9.54dB