The plunger at the end of syringe has a cross-sectional area of 2.00 cm^2, and t

Question

The plunger at the end of syringe has a cross-sectional area of 2.00 cm^2, and the narrow end has a cross-sectional area of 1.00 mm^2. The syringe contains 30 cc's of saline solution. The plunger is initially compressed by a spring which is squished 1.50 cm from it's equilibrium position. The narrow end is blocked by a cap that exerts force of 5.00 mN to prevent the solution from leaking out. Assume the syringe is held horizontally. What is the gauge pressure at the wide end of the syringe? a) 1.00 times 10^3 Pa b) 2.00 times 10^3 Pa c) 3.00 times 10^3 Pa d) 4.00 times 10^3 Pa e) 5.00 times 10^3 Pa How much force does the spring exert on the plunger? a) 1.00 N b) 1.00 times 10^3 N c) 1.00 kN d) Both b and c are correct. e) 10.0 kN What is the spring constant? a) 3.33 N/m b) 6.67 times 10^1 N/m c) 6.67 times 10^2 N/m d) 6.67 times 10^3 N/m e) 6.67 times 10^4 N/M If the cap pops off, how fast is the solution leaving the nozzle if when plunger is moving into the syringe at 2.00 cm/s? a) 200 cm/s b) 400 cm/s c) 600 cm/s d) 800 cm/s e) 1000 cm/s

Explanation / Answer

By Pascal law, pressure at plunger-fluid contact point 1, is equal to pressure at fluid-cap contact point 2.

So, force being exerted on plunger , by spring whose net displacement is .015m = F=k*0.015

Force being exerted by the cap=0.005N

Pressure being exerted by the cap=Force being exerted by the cap/area=0.005/10^(-6) =5000Pa

Pressure at 1 & 2 are same, so, 5000=0.015k / 2*10(-4)

So k = 66.67N/m

Spring force on plunger=F=k*x=66.67*0.015 = 1N

Gage pressure at point 1= (atmospheric pressure+ spring pressure)-atmospheric pressure = spring's pressure = spring force/ area = 1 / 2*10^(-4) = 5000N/m^2

moving at 0.02m/s , change volume exiting is = 0.02*2*10(-4) m^3/s = 4*10^-6  m^3/s

Equating this at other end,

4*10^-6  m^3/s =speed*area= v*10^(-6)

So, speed = 4m/s = 400cm/s

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