final-solutions

final-solutions - Physics 262, Final Exam, May 20, 2002,...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Physics 262, Final Exam, May 20, 2002, Dr. Baden Try to do as many problems as you can, and show your work clearly. There are a lot of problems here, more than you might have guessed, the reason for this is to try to cover as much material as possible to try to measure what you’ve learned. I don’t expect everyone to finish every problem, so please do the problems you are more sure of first, then go back to the others. Credit will not be given for answers with no work shown. Partial credit will be given. There are 170 points total you can get on the exam. Problem 1 (10 points). Water flows thru a fire hose of diameter 5.4cm at a rate of 0.105 m 3 /s . The fire hose ends in a nozzle that has a diameter of 2.20 cm . Calculate the speed that water exits the nozzle. The conservation of flow equation is Av = constant=0.105m 3 /s . The area of a hose with diameter D is given by 4 2 D A π = . The continuity equation is then v D Av 4 2 = = 0.105m3/s. Solving for the final velocity gives s m D v f / 2 . 276 4 105 . 0 2 = = Problem 2 (10 points). An astronaut on the Moon wishes to measure the local value of g (as in “F=mg”) on the moon (g moon ) by timing pulses traveling down a wire that has a large mass suspended from it. Assume that the wire has a mass of 2.20 g and a length of 1.10 m , and that a 7.50 kg mass is suspended from it. A pulse requires 0.015 s to traverse the length of the wire. Calculate g moon from these data, neglecting the mass of the wire. The velocity of waves on a string of mass m , length L , and under tension T is given by L m T v = . In this problem, the tension comes from gravity due to the weight M=1.5kg , so T=Mg so we have L m Mg v = . If the wave takes 6.1s to traverse the string of length 4.1m, then the velocity is v=1.1m/0.015s=73.3m/s . Solving for g gives 2 2 2 2 2 / 43 . 1 1 . 1 5 . 7 0022 . 0 / 3 . 73 s m m kg kg s m ML m v g = = = Problem 3 (10 points). A point source emits sound waves with an average power output of 1.3 W . a) find the intensity and sound level (in dB) 4.50 m from the source. b) find the distance at which the sound level is 95 dB . a) the intensity is defined as power/area, and if the source is a point source then the waves are emitted as spherical waves, and have an area given by A=4pr 2 at radius r. The intensity at r=4.5m is then given by 2 2 / 0051 . 0 5 . 4 4 3 . 1 m W W A P I = = = . The sound level in dB is given by 0 log 10 I I = β where I 0 is the threshold intensity of 10 -12 W/m 2 , so at 4.5m the sound level is dB 1 . 97 10 0051 . 0 log 10 12 = = - .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
b) You use the same formula for sound level, but now you have to solve for the intensity where the level is 95dB: 2 3 2 5 . 2 5 . 9 12 10 0 / 10 16 . 3 / 10 10 10 10 m W x m W I I - - - = = = = β and solving for area using the formula A P I = gives 2 3 4 5 . 411 10 16 . 3 / 3 . 1 r x I P A π = = = = - . Solving for the radius gives r =5.7m
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 11/16/2010 for the course PHYSICS 2012 taught by Professor Mikey during the Spring '10 term at Temple.

Page1 / 6

final-solutions - Physics 262, Final Exam, May 20, 2002,...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online