Question

A garden hose having an internal diameter of 0.740 in. ( 1.8796 cm ) is connected to a lawn sprinkler that consists merely of an enclosure with 36 holes, each 0.055 in. ( 0.1397 cm ) in diameter. If the water in the hose has a speed of 4.00 ft/s ( 121.920 cm/s ), at what speed does it leave the sprinkler holes?

Answer 1

Answer:

It will leave the sprinkler at speed of $v_2=613.87m/sec$

Step-by-step explanation:

We have given internal diameter of the garden hose $d_1=0.740in=1.8796cm$

So radius $r_1=\frac{d_1}{2}=\frac{1.8796}{2}=0.9398cm$

So area $A_1=\pi r_1^2=3.14\times 0.9398^2=2.7733cm^2$

Water in the hose has a speed of 4 ft/sec

So $v_1=4ft/sec=121.92cm/sec(As\ 1ft/sec\ =30.48cm/sec)$

Number of holes n = 36

Diameter of each hole $d_2=0.1397cm$

So radius $r_2=0.0698cm$

So area $A_2=\pi r^2=3.14\times 0.0698^2=0.0153cm^2$

From continuity equation

$A_1v_1=nA_2v_2$

$2.7733\times 121.92=36\times 0.0153\times v_2$

$v_2=613.87m/sec$