Answer:
(a).The speed of the water in the nozzle is 3.014 m/s.
(b). The pressure in the nozzle is 1.86 atm.
Explanation:
Given that,
Nozzle diameter = 0.25 in = 0.00635 m
Hose pipe diameter = 0.64 in = 0.016256 m
Pressure = 1.9 atm =192518 Pa
(a). We need to calculate the speed of the water in the nozzle
Flow Speed at the inlet pipe will be given by using Continuity Equation



Where, A = area of pipe


Put the value into the formula


The speed of the water in the nozzle is 3.014 m/s.
(b). We need to calculate the pressure in the nozzle
Using Bernoulli's Theorem,

Where, 


Put the value into the formula



Hence, (a).The speed of the water in the nozzle is 3.014 m/s.
(b). The pressure in the nozzle is 1.86 atm.