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 A. The radius of the cylinder and tube are, respectively, 1.40 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 10 Pa.) Ay (a) If the pistol is fired horizontally at a height of 1.10 m, determine the time interval (in s) required for water to travel from the nozzle to the ground. Neglect air resistance. 0.57 This is a projectile motion problem. Write the kinematic equation for the y-component of the water's trajectory. Note that y = 0 when t = Te, the time of flight. s (b) If the desired range of the stream is 7.90 m, with what speed v(in m/s) must the stream leave the nozzle? 13.325 x Write the kinematic equation for the y-component of the water's trajectory. Note that x = R, the range when t = Ty the time of flight. m/s (c) At what speed V (in m/s) must the plunger be moved to achieve the desired range? 0.13825 The speed of the fluid (v, and v.,) at points 1 and 2 is related to the cross-sectional areas (A, and A.,) by the continuity equation. Knowing the radius of the cylinder and tube, you can determine their cross-sectional area. m/s (d) What is the pressure (in Pa) at the nozzle? 1.013e5 Pa (e) Find the pressure (in Pa) needed in the cylinder. 197149.76 x Write Bernoulli's equation for the cylinder and tube using subscripts 1 and 2. Use the center of the cylinder as the point of zero gravitational potential energy. Knowing P, Y can determine the desired pressure. Pa yy, and V, and V, you (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.) 67.72 X

Answer 1

from equation of motion for constant acceleration in vertical direction S, = u.,t +ża,e? -h=0+{x(-8)T h = {&T T, - VI To = vas. ; T = 0.474s from equation of motion in horizontal direction R=v2T, v2 = 1 02 = .794 = 16.67 x m/s from continuity equation volume flow rate remains same Q = Q2 Aj = Az U2 ario = xr2 02 (1.4%)(01) = (0.15) (16.67) (0.15)' (16.67) - 0.191 x mis (1.4) Prozzle = 1.013 x 10 Pa from Bernoulli's equation P. + {pv? + psh, = P2 +2pvz + pghz P1 = P2 +3pv - 2pv + pgh2 - 0 P1 = P2 + 1pc - 0?) + pgh P1 = (1.013 10 )+ (1) X (1000) X (16.672 -0.191")+(1000) X (9.8) (80) P1 = 240520.21 P1 = 2.405 x 10 Pa F = APA F =(2.405 – 1.013) X (10%)x()x(1000 F = 85.71 N