426_Physics ProblemsTechnical Physics

426_Physics ProblemsTechnical Physics - 428 P14.50 Fluid...

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428 Fluid Mechanics P14.50 The assumption of incompressibility is surely unrealistic, but allows an estimate of the speed: Pg y v y v v v 11 1 2 22 2 2 2 2 2 5 1 2 1 2 1 0 00 0 0 2 8 70 1 2 120 2 1 00 0 287 1 013 10 347 ++ = ++= = −× = ρρ ... .. . . atm atm kg m Nm kg m ms 3 2 3 ej af P14.51 (a) h P v 3 2 1 2 = + + vg h 3 2 = If h = 100 . m , v 3 443 = m s (b) y vP v = + + ρ 1 2 0 1 2 2 2 03 2 Since vv 23 = , PP g y =− 0 FIG. P14.51 Since P 0 y P g ≤= × = 0 5 1013 10 98 10 3 . . . Pa 10 kg m m s m 33 2 e j *P14.52 Take points 1 and 2 in the air just inside and outside the window pane. P v gy P v gy 2 12 2 2 2 1 2 1 2 = PP 02 2 0 1 2 130 112 += + kg m m s 3 bg 20 81 5 . P a (a) The total force exerted by the air is outward, PA PA PA PA 1200 81 5 4 1 5 489 −=−+ = m m N ou twa rd 2 a fa f (b) PA PA v A 2 2 2 1 2 1 2 224 4 15 196 −= = = . . kg m m s m m kN outward 3 a f P14.53 In the reservoir, the gauge pressure is P = × 200 800 10 5 4 . . N 2.50 10 m Pa 2 From the equation of continuity:
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