Unformatted text preview: 19-16 ry7*tz : lo-eergs/s- cm2 Therefore Ps : (l2go)1/2 : Q'O-e x2xl.29 x t0-3 x 3.3 v 1g+1r/z : 2.9 x 10-4 d1m/ cmz Let the distance of the furthest detectable object be D. In 70 msec the sound pulse must travel 2D (fhere and back). Therefore D:t#: :1150crt : 11.5m The minimum detectable size is on the order of the wavelength (l) of the sound. From Eq. 12-1 ^:|:*# :L7 xlo-zcm 70x10-3x3.3x104 25...
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This note was uploaded on 03/07/2011 for the course PHYS 102 taught by Professor Armenn.kocharian during the Winter '11 term at California State University Los Angeles .
- Winter '11