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Unformatted text preview: alexander (jra2623) oldhomework 26 Turner (92510) 1 This printout should have 11 questions. Multiplechoice 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 crosssectional area of 4 . 22398 10 5 m 2 . The crosssectional area of the needle is 1 . 51631 10 8 m 2 . In the absence of a force on the plunger, the pressure everywhere is 1 . 0 atm. A force of magnitude 2 . 72993 N is exerted on the plunger, making medicine squirt from the needle. 2 . 72993 N v 2 P 2 A 2 A 1 P 1 v 1 Find the medicines 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: 11 . 3692 m / s. Explanation: Let : A 1 = 4 . 22398 10 5 m 2 , A 2 = 1 . 51631 10 8 m 2 , F = 2 . 72993 N , and = water = 1 . 00 10 3 kg / m 3 . P 2 = P Applying Bernoullis equation, with y = 0 along the horizontal axis of the syringe and needle, P 1 = P 2 + 1 2 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 v 2 2 = F A 1 v 2 = radicalBigg 2 F A 1 = radicalBigg 2 (2 . 72993 N) (4 . 22398 10 5 m 2 ) (1000 kg / m 3 ) = 11 . 3692 m / s . 002 (part 1 of 2) 10.0 points A liquid of density 1140 kg / m 3 flows with speed 1 . 33 m / s into a pipe of diameter 0 . 12 m . The diameter of the pipe decreases to 0 . 05 m at its exit end. The exit end of the pipe is 6 . 28 m lower than the entrance of the pipe, and the pressure at the exit of the pipe is 1 . 4 atm . P 1 1 . 33 m / s . 12 m P 2 v 2 . 05 m 1 . 4 atm 6 . 28 m What is the velocity v 2 of the liquid flowing out of the exit end of the pipe? Assume the viscosity of the fluid is negligible and the fluid is incompressible. The acceleration of gravity is 9 . 8 m / s 2 and P atm = 1 . 013 10 5 Pa. Correct answer: 7 . 6608 m / s. Explanation: Let : v 1 = 1 . 33 m / s , r 1 = 0 . 06 m , and r 2 = 0 . 025 m . From the continuity equation, the volume of liquid entering the pipe during a certain time alexander (jra2623) oldhomework 26 Turner (92510) 2 interval must equal the volume of liquid leav ing the pipe during that time interval: A 1 v 1 = A 2 v 2 v 2 = A 1 A 2 v 1 = r 2 1 r 2 2 v 1 = parenleftbigg r 1 r 2 parenrightbigg 2 v 1 = parenleftbigg . 06 m . 025 m parenrightbigg 2 (1 . 33 m / s) = 7 . 6608 m / s ....
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 Summer '08
 Kaplunovsky
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