Question

The construction of a water pistol is shown in the figure below. The cylinder with cross-sectional area A, is filled with water and when the piston is pushed (by pulling the trigger), water is forced out the tube with cross-sectional area Ag. The radius of the cylinder and tube are, respectively, 1.30 cm and 1.50 mm, and the center of the tube is a height h = 3.00 cm above the center of the cylinder. (Assume atmospheric pressure is 1.013 x 105 Pa.) A TS 1 (a) If the pistol is fired horizontally at a height of 1.30 m, determine the time interval (in s) required for water to travel from the nozzle to the ground. Neglect air resistance. (b) If the desired range of the stream is 7.70 m, with what speed v, in m/s) must the stream leave the nozzle? - m/s (c) At what speed v, (in m/s) must the plunger be moved to achieve the desired range? m/s (d) What is the pressure (in Pa) at the nozzle? (e) Find the pressure (in Pa) needed in the cylinder. Pa (f) Calculate the force (in N) that must be exerted on the trigger to achieve the desired range. (The force that must be exerted is due to pressure over and above atmospheric pressure. Enter the magnitude.)

Answer 1

a) The time taken to reach the ground is 2h t = 2(1.30m) 9.81m/s 0.514816 0.515s -(b) The nozzle speed is, d t 7.70m 0.515s =14.95145631 =15.0m/s (c) Using continuity equation 4 Av2 T(1.3x10 m= r(1.5x 103 m)' (15.0 m/s) 0.199704142m/s = 0.20m/s (d) The pressure at the nozzle is equal to the atmospheric pressure That is P=101.3 kPa 101300pa (e) The pressure required at the larger tube can be calculated by using Bemoulli's equation. 1 pvZPg2 Ppv+pgPm+ atn 1.013x10 Pa 1000 kg/m3)(15 m/s) 1 Р+ (1000 kg/m3)(0.20 m/s)* = +(1000 kg/m2 ) (9.8 m/s2 )(0.03 m) P+20 214094 P 214074 2.14x10 pa (f) The force is, f (P-Pa)A = (2.14x10 Pa -1.013x10 Pa)(1.3x 10 m)2 ann = 59.83571616 = 59.8N