# hw42 - hinojosa(jlh3938 – homework 42 – Turner...

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Unformatted text preview: hinojosa (jlh3938) – homework 42 – Turner – (58185) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The hypodermic syringe contains a medicine with the same density of water. The barrel of the syringe has a cross-sectional area of 2 . 87055 × 10 − 5 m 2 . The cross-sectional area of the needle is 2 . 47867 × 10 − 8 m 2 . In the absence of a force on the plunger, the pressure everywhere is 1 . 0 atm. A force of magnitude 1 . 05374 N is exerted on the plunger, making medicine squirt from the needle. 1 . 05374 N v 2 P 2 A 2 A 1 P 1 v 1 Find the medicine’s flow speed through the needle. Assume that the pressure in the nee- dle remains at atmospheric pressure, that the syringe is horizontal, and that the speed of the emerging fluid is the same as the speed of the fluid in the needle. Correct answer: 8 . 56839 m / s. Explanation: Let : A 1 = 2 . 87055 × 10 − 5 m 2 , A 2 = 2 . 47867 × 10 − 8 m 2 , F = 1 . 05374 N , and ρ = ρ water = 1 . 00 × 10 3 kg / m 3 . P 2 = P Applying Bernoulli’s equation, with y = 0 along the horizontal axis of the syringe and needle, P 1 = P 2 + 1 2 ρ v 2 2 , since the syringe is horizontal ( h 1 = h 2 ) and v 1 is negligible in comparison with the fluid speed inside the needle ( v 1 ≪ v 2 ). Choose y = 0 along the horizontal axis of the plunger + needle. Then P 1 − P 2 = 1 2 ρ v 2 2 = F A 1 v 2 = radicalBigg 2 F A 1 ρ = radicalBigg 2 (1 . 05374 N) (2 . 87055 × 10 − 5 m 2 ) (1000 kg / m 3 ) = 8 . 56839 m / s . 002 (part 1 of 2) 10.0 points Note: P atm = 1 . 013 × 10 5 Pa. The viscos- ity of the fluid is negligible and the fluid is incompressible. A liquid of density 1273 kg / m 3 flows with speed 1 . 76 m / s into a pipe of diameter 0 . 11 m. The diameter of the pipe decreases to 0 . 05 m at its exit end. The exit end of the pipe is 5 . 26 m lower than the entrance of the pipe, and the pressure at the exit of the pipe is 1 . 4 atm. The acceleration of gravity is 9 . 8 m / s 2 . P 1 1 . 76 m / s . 11 m P 2 v 2 . 05 m 1 . 4 atm 5 . 26 m What is the velocity v 2 of the liquid flowing out of the exit end of the pipe? Correct answer: 8 . 5184 m / s. Explanation: Let : v 1 = 1 . 76 m / s , d 1 = 0 . 11 m , and d 2 = 0 . 05 m . Basic Concepts The continuity equation for incompressible fluids is A v = constant . hinojosa (jlh3938) – homework 42 – Turner – (58185) 2 Bernoulli’s equation for fluid flow along streamlines is P + ρ g h + 1 2 ρ v 2 = constant . Solution: The continuity equation tells us that the volume of liquid entering the pipe during a certain time interval must equal the volume of liquid leaving the pipe during that time interval, so A 1 v 1 = A 2 v 2 ....
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## This note was uploaded on 03/26/2010 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas.

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hw42 - hinojosa(jlh3938 – homework 42 – Turner...

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