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Unformatted text preview: alexander (jra2623) – homework 04 – 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 Consider the units for the gas constant [ R ] = m 2 · kg s 2 · K , air pressure [ P ] = kg m · s 2 , tem perature [ T ] = K, molar mass [ μ ] = kg, and volume density [ ρ ] = kg / m 3 . Assuming the speed of sound depends only on R , P , T and μ , find the correct relation ship. 1. v ∝ radicalbigg μ T R 2. v ∝ radicalbigg T R P 3. v ∝ radicalbig R μ T 4. v ∝ radicalbigg μ T R P 5. v ∝ radicalBigg T R μ 6. v ∝ radicalbigg P T R 7. v ∝ radicalBigg R T μ correct 8. v ∝ radicalBigg P T μ P 9. v ∝ √ R P T 10. v ∝ radicalbigg R T P Explanation: Under our assumption the most general ex pression for the speed of sound will be v = P a μ b R c T d . [ v ] = [ P ] a [ μ ] b [ R ] c [ T ] d m s = parenleftbigg kg m · s 2 parenrightbigg a (kg) b parenleftbigg m 2 · kg s 2 · K parenrightbigg c (K) d m 1 s 1 = (m) a +2 c (s) 2 a 2 c × (kg) a + b + c (K) c + d . Equating the exponents, a + 2 c = 1 2 a 2 c = 1 a + b + c = 0 c + d = 0 . The solution of the system is a = 0, b = 1 2 , c = 1 2 , d = 1 2 , so the final expression for v is v = P μ 1 / 2 R 1 / 2 T 1 / 2 = radicalBigg R T μ = radicalBigg [ R ] [ T ] [ μ ] = radicaltp radicalvertex radicalvertex radicalvertex radicalbt m 2 · kg s 2 · K K kg = m s ....
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This note was uploaded on 10/20/2011 for the course PHY 302K taught by Professor Kaplunovsky during the Summer '08 term at University of Texas.
 Summer '08
 Kaplunovsky
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