Physics 1 Problem Solutions 315

# Physics 1 Problem Solutions 315 - × 10 5 Pa = 101 . 325...

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Appendix 1 Introductory Physics Equation Sheet 311 v 1 f = ( m 1 m 2 ) v 1 i + 2 m 2 v 2 i m 1 + m 2 1-d Elastic Collision Expression v 2 f v 1 f = ( v 2 i v 1 i ) v rel f = v rel i 1-d Elastic Collision Expressions 14 Rotational Kinematics θ = s r ω = v r α = a r ω avg = Δ θ Δ t ω = lim Δ t 0 Δ θ Δ t α avg = Δ ω Δ t α = lim Δ t 0 Δ ω Δ t ω = αt + ω 0 θ = 1 2 αt 2 + ω 0 t θ = 1 2 ( ω 0 + ω ) t ω 2 = ω 2 0 + 2 αθ θ = 1 2 αt 2 + ωt 15 Simple Harmonic Motion P = 1 f ω = 2 πf F = kx PE = 1 2 kx 2 ω = r k m P = 2 π r m k E mech total = 1 2 mv 2 max = 1 2 kx 2 max = 1 2 mv 2 + 1 2 kx 2 x ( t ) = x max cos( ωt ) v ( t ) = ωx max sin( ωt ) a ( t ) = ω 2 x max cos( ωt ) v max = ωx max a max = ω 2 x max P = 2 π R g 16 Fluids ρ = m V P = F A P = P 0 + ρgy depth P = P 0 ρgy height ρ air = 1 . 21 kg / m 3 ρ water = 998 kg / m 3 at 20 C and 1 atm 1 atm = 1 . 01325
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Unformatted text preview: × 10 5 Pa = 101 . 325 kPa = 14 . 696 psi = 760 torr Pascal’s principle P = P ext − ρg ( y − y ext ) Δ P = Δ P ext Archimedes’s principle F buoy = m fuid dis g = ρ fuid V fuid dis g Floating condition f = V 1 V = ρ 2 − ρ obj ρ 2 − ρ 1 equation of continuity for ideal ±uid R V = Av = Constant Bernoulli’s equation P + 1 2 ρv 2 + ρgy = Constant...
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## This note was uploaded on 11/16/2011 for the course PHY 2053 taught by Professor Buchler during the Fall '06 term at University of Florida.

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