Question

A village maintains a large tank with an open top, containing water for emergencies. The water can drain from the tank through a hose of diameter 6.60 cm. The hose ends with a nozzle of diameter 2.20 cm. A rubber stopper is inserted into the nozzle. The water level in the tank is kept 7.50 m above the nozzle. (a) Calculate the friction force exerted on the stopper by the nozzle. (b) The stopper is removed. What mass of water flows from the nozzle in 2.00 h

Answer 1

Answer:

(A) Frictional force will be equal to 27.92 N

(B) Mass is equal to 33120 kg

Explanation:

(A) Diameter of nozzle d = 2.20 cm

So radius $r=\frac{d}{2}=\frac{2.20}{2}=1.10cm$

eight h = 7.5 m

Density of water $\rho =1000kg/m^3$

Acceleration due to gravity $g=9.8m/sec^2$

Pressure on the rubber stopper

$P=\rho hg$

$p=1000\times 7.5\times 9.8=73500Pa$

Area of cross section $A=\pi r^2$

$A=3.14\times 0.01^2=3.799\times 10^{-4}m^2$

So force $F=PA$

$F=73500\times 3.799\times 10^{-4}=27.92N$

(B) Speed of the water through nozzle

$v=\sqrt{2gh}$

$=\sqrt{2\times 9.8\times 7.5}=12.12m/sec$

Volume of water flow

$V=vA=12.12\times 3.799\times 10^{-4}=0.0046m^3/sec$

Mass of water flow per sec

$m=\rho V=1000\times 0.0046=4.6kg$

Total time t = 2 hour = 2×3600 = 7200 sec

So total mass flow

$m=4.6\times 7200=33120kg$