Final answer:
The speed at which the water exits the nozzle is 19.4 m/s.
Step-by-step explanation:
To calculate the speed at which the water exits the nozzle, we can use the principle of conservation of mass. The formula for flow rate is given by Q = A×v, where Q is the flow rate, A is the cross-sectional area of the nozzle, and v is the velocity of the water. Rearranging the formula, we have v = Q/A. Given that the water is supplied to the hose at a rate of 0.493 m/s, we need to find the cross-sectional area of the nozzle.
The diameter of the nozzle is given as 5.67 mm. We can convert this to meters by dividing by 1000, which gives us a diameter of 0.00567 m. The cross-sectional area of the nozzle can be calculated using the formula A = πr2, where r is the radius of the nozzle. The radius is half the diameter, so r = 0.00567/2 = 0.002835 m. Substituting these values into the formula for velocity, we get v = 0.493/(π × 0.0028352) = 0.493/0.00002536 = 19.4 m/s.