In a water pistol, a piston drives water through a larger tube of area A 1 into

Question

In a water pistol, a piston drives water through a larger tube of area A1 into a smaller tube of area A2 as shown in the figure below. The radius of the large tube is 1.20 cm and that of the small tube is 1.50 mm. The smaller tube is 3.00 cm above the larger tube.

(a) If the pistol is fired horizontally at a height of 1.30 m, determine the time interval required for water to travel from the nozzle to the ground. Neglect air resistance and assume atmospheric pressure is 1.00 atm.
s

(b) If the desired range of the stream is 7.50 m, with what speed v2 must the stream leave the nozzle?
m/s

(c) At what speed v1 must the plunger be moved to achieve the desired range?
m/s

(d) What is the pressure at the nozzle?
Pa

(e) Find the pressure needed in the larger tube.
Pa

(f) Calculate the force 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.)

Explanation / Answer

Required formulas are

a) time = sqrt ( 2*height/ gravity)

T = sqrt(2*1.3/9.8) = 0.515 sec

b) velocity2 = distance/time

v2 = 7.5/0.515 = 14.56 m/s

c) velocity1 = (area2*velocity2)/(area1)

v1 = pi*(1.5*10^(-3))^2*14.56/[pi*(1.2*10^(-2))^2] = 0.227 m/s

d) the pressure at the nozzle is atmospheric pressure. then

p1 = 1.01*10^5 = 101000pa

e) p2 = p1 + (1/2)(1000)( (velocity2)^2 - (velocity1)^2)

p2 = 101000 + (1/2)*(1000)*[(14.56)^2 - (0.227)^2]

p2 = 206971 pa

f) force = (p2 - p1)*area1

force = (206971 - 101000)*(22/7)*(1.2*10^(-2))^2 = 47.96 N

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