HomePlumbing TipsWater emerges from an inverted conical vessel at a rate of cubic centimeters per minute at the same time?
Posted in Plumbing Tips on 20th July 2013

water runs out of an inverted conical tank at a rate of cubic centimeters per min at the same time?
Water emerges from an inverted conical vessel with a rate of 8600.00 cm ³ per minute at the same time that water is pumped into the vessel at a constant rate. The tank has height of 7,000 meters and the diameter at the top is 5.5000 meters. If the water level of 25,000 centimeters per minute increases as the height of the water column is 2.000 m, find the rate is pumped to the water in the tank in cubic centimeters per minute Best Answer (s):.

reply by Lawrence
Given … R = D / 2 = 5.5 m / cm 2 = 550/2 = 275 cm H = 7 m = 700 cmh = 2m = 200 CMDh / dt = 25 cm / MinDV / dt (out) = 8600 cc / min —————— Apply similar triangles … r / R = h / Hr/275 = h/700dannr = 11/28 hThe tank volume is defined by … V = Ï€ r ² h / 3To the dependent variable, r eliminate, substitute r = 11/28 hV = (Ï€ / 3) (11 / 28 h) ² (h) V = (Ï€ / 3) (11/28) ² (h) ³ V = (121Ï€ / 2352) (h) ³ Take derivative dV / dh findendV / dh = (363Ï€ / 2352) (h ² ) Since dV / dt = (dV / dh) (dh / dt), danndV / dt (tank) = (363Ï€ / 2352) (h ²) (dh / dt) dV / dt (tank) = (363Ï€ / 2352) (200 cm ² ²) (25 cm / min), dV / dt (tank) = (363Ï€ / 2352) (40000 cm ²) (25 cm / min), dV / dt (tank) = (363Ï€ / 2352) (cm 1000000 / min ³) dV / dt (tank) = 484,863 cm / min ³, since dV / dt (tank) = dV / dt (in) – dV / dt (out), danndV / dt (in) = dV / dt (Tank) + dV / dt (out) dV / dt (in) = 484 863 cm ³ / min + 8600 cc / MinDV / dt (in) = 493 463 cm ³ / min

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