Midterm 4

# Midterm 4 - midterm 04 KELLERMANN MARC Due Dec 6 2006 11:00...

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midterm 04 – KELLERMANN, MARC – Due: Dec 6 2006, 11:00 pm 1 Gravity ~ F 21 = - G m 1 m 2 r 2 12 ˆ r 12 , for r R , g ( r ) = G M r 2 G = 6 . 67259 × 10 - 11 N m 2 /kg 2 R earth = 6370 km, M earth = 5 . 98 × 10 24 kg Circular orbit: a c = v 2 r = ω 2 r = 2 π T · 2 r = g ( r ) U = - G m M r , E = U + K = - G m M 2 r F = - d U dr = - m G M r 2 = - m v 2 r Kepler’s Laws of planetary motion: i ) elliptical orbit, r = r 0 1 - ² cos θ r 1 = r 0 1+ ² , r 2 = r 0 1 - ² ii ) L = r m Δ r Δ t -→ Δ A Δ t = 1 2 r Δ r Δ t = L 2 m = const. iii ) G M a 2 = 2 π a T · 2 1 a , a = r 1 + r 2 2 , T 2 = 4 π 2 G M · r 3 Escape kinetic energy: E = K + U ( R ) = 0 Fluid mechanics Pascal: P = F 1 A 1 = F 2 A 2 , 1 atm = 1 . 013 × 10 5 N/m 2 Archimedes: B = M g , Pascal=N/m 2 P = P atm + ρ g h , with P = F A and ρ = m V F = R P dA -→ ρ g ‘ R h 0 ( h - y ) dy Continuity equation: A v = constant Bernoulli: P + 1 2 ρ v 2 + ρ g y = const, P 0 Oscillation motion f = 1 T , ω = 2 π T S H M: a = d 2 x dt 2 = - ω 2 x , α = d 2 θ dt 2 = - ω 2 θ x = x max cos( ω t + δ ), x max = A v = - v max sin( ω t + δ ), v max = ω A a = - a max cos( ω t + δ ) = - ω 2 x , a max = ω 2 A E = K + U = K max = 1 2 m ( ω A ) 2 = U max = 1 2 k A 2 Spring: m a = - k x Simple pendulum: m a θ = m α ‘ = - m g sin θ Physical pendulum: τ = I α = - m g d sin θ Torsion pendulum: τ = I α = - κ θ Wave motion Traveling waves: y = f ( x - v t ), y = f ( x + v t ) In the positive x direction: y = A sin( k x - ω t - φ ) T = 1 f , ω = 2 π T , k = 2 π λ , v = ω k = λ T Along a string: v = q F μ Reflection of wave: fixed end: phase inversion open end: same phase General: Δ E = Δ K + Δ U = Δ K max P = Δ E Δ t = 1 2 Δ m Δ t ( ωA ) 2 Waves: Δ m Δ t = Δ m Δ x · Δ x Δ t = Δ m Δ x · v P = 1 2 μ v ( ω A ) 2 , with μ = Δ m Δ x Circular: Δ m Δ t = Δ m Δ A · Δ A Δ r · Δ r dt = Δ m Δ A · 2 π r v Spherical: Δ m Δ t = Δ m Δ V · 4 π r 2 v Sound v = q B ρ , s = s max cos( k x - ω t - φ ) Δ P = - B Δ V V = - B ∂s ∂x Δ P max = B κ s max = ρ v ω s max Piston: Δ m Δ t = Δ m Δ V · A Δ x Δ t = ρ A v Intensity: I = P A = 1 2 ρ v ( ω s max ) 2 Intensity level: β = 10 log 10 I I 0 , I 0 = 10 - 12 W/m 2 Plane waves: ψ ( x, t ) = c sin( k x - ω t ) Circular waves: ψ ( r, t ) = c r sin( k r - ω t ) Spherical: ψ ( r, t ) = c r sin( k r - ω t ) Doppler effect: λ = v T , f 0 = 1 T , f 0 = v 0 λ 0 Here v 0 = v sound ± v observer , is wave speed relative to moving observer and λ 0 = ( v sound ± v source ) /f 0 , detected wave length established by moving source of frequency f 0 . f received = f reflected Shock waves: Mach Number= v source v sound = 1 sin θ Superposition of waves Phase difference: sin( k x - ωt ) + sin( k x - ω t - φ ) Standing waves: sin( k x - ω t ) + sin( k x + ω t ) Beats: sin( kx - ω 1 t ) + sin( k x - ω 2 t ) Fundamental modes: Sketch wave patterns String: λ 2 = , Rod clamped middle: λ 2 = , Open-open pipe: λ 2 = , Open-closed pipe: λ 4 = Temperature and heat Conversions: F = 9 5 C + 32 , K = C + 273 . 15 Constant volume gas thermometer: T = a P + b Thermal expansion: α = 1 d ‘ dT , β = 1 V d V dT Δ = α ‘ Δ T , Δ A = 2 α A Δ T , Δ V = 3 α V Δ T Ideal gas law: P V = n R T = N k T R = 8 . 314510 J / mol / K = 0 . 0821 L atm / mol / K k = 1 . 38 × 10 - 23 J / K, N A = 6 . 02 × 10 23 , 1 cal=4.19 J Calorimetry: Δ Q = c m Δ T, Δ Q = L Δ m First law: Δ U = Δ Q - Δ W , W = R P dV Conduction: H = Δ Q Δ t = - k A Δ T Δ , Δ T i = - H A i k i Stefan’s law: P = σ A e T 4 , σ

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