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Unformatted text preview: Numerical Answers to Phys 101 Final Exam - December 2010 Question 1 not released Question 2 Vç0.637 m/s vz= 70.2 m/s vz= 0.637 m/s Pr=Po=1.013x10sPa Pz=47520Pa P:=Po=1.013x10sPa Need to consider change in pressure for water as well as mercury h=0A20m Question 3a iJ C<B=D<A iil C+D<B<A Question 3b T= 0.727 s t=0,E0=L.72J t=T, E=0.96 J Question 4 l,=3.0 m f=0.50 Hz v=1.50 m/s Qo=2.1 rad D=0.10m sin(2.09x+3.14tJ Question 5 i) 772.5 H2,772.5 Hz iÐ 1.67.5 H2,172.5 Hz iii) At position b will hear beats ivJ Constructive vJ Increase vrJ Max path difference is 4 À, will hear minima when path difference is?u/2,3)t/2 [between points c and d ) so 2 more minima with path difference 5)u/2 andT?u/z. viiJ Question 6a a) [15t2Jx10 bJ (39t5) pm Question 6b i) 2.5x10r m iiJ less @ Dept of Fhysics and .åstronomy UBC UUIú' Question 2 (10 marks) Pipe 2 The structure above consists of three connected sections of cylindrical pipes. Pipes I and 3 have a diameter of 10 cm and Pipe 2 has a diameter of 2.5 cm and is located 20 cm above Pipes I anã 3. See the table below. A mercury manometer is connected to the fìrst two pipes. Water is flowing through the thlee pipes with a flow rate of 0-005ò m3A. The fluid in the lou,er part of the manãmeter is mercùry (p: r:Ooõ kglm1. Noieìhat r:Ooõ kglm1....
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This note was uploaded on 01/15/2012 for the course PHYS 101 taught by Professor Bates during the Winter '08 term at UBC.
- Winter '08