ch23-p075 - (e) Now the flux is ( ) ( )( ) cos 693kg s...

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(b) Since water flows only through area wd , the flux through the larger area is still 693 kg/s. (c) Now the mass flux is ( wd /2) ρ v = (693 kg/s)/2 = 347 kg/s. (d) Since the water flows through an area ( wd /2), the flux is 347 kg/s.
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Unformatted text preview: (e) Now the flux is ( ) ( )( ) cos 693kg s cos34 575 kg s wd v = = . 75. (a) The mass flux is wd v = (3.22 m) (1.04 m) (1000 kg/m 3 ) (0.207 m/s) = 693 kg/s....
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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