5 pl | (8e23p75)Water in an irrigation ditch of width w = 3.92 m and depth d1.24

Question

5 pl | (8e23p75)Water in an irrigation ditch of width w = 3.92 m and depth d1.24 m flows with a speed of 0.247 m/s. The mass flux of the flowing water through an imaginary surface is the product of the wate's density (1000 k/ d its volue ux through that surface. Find the mass flux through the following imaginary surfaces: (a) a surface of area wd, entirely in the water, perpendicular to the flow; (in kg/s) 27 BO 1.20 × 103 DO 2. 12 × 101 EO 282 × 103 GO 500 x 10 HO 6.65 103 CO 1.60× 103 FO 3.76 × 103 10. AO 9.03 x 102

Explanation / Answer

the mass flux through surface area is

w*d*rho*v = (3.92)(1.24)(1000)(0.247)

= 1.2*10^3

option (B) is correct

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