1532kformula2k5 - x = A e-bt 2 m cos ωt φ where ω = q ω...

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Mechanics Linear Motion: x = x 0 + 1 2 ( v x 0 + v x ) t , x = x 0 + v x 0 t + 1 2 a x t 2 , v x = v x 0 + a x t , v x 2 = v x 0 2 + 2 a x ( x - x 0 ) Forces: F = ma , F = d dt p , Friction: F = μN , Spring: F = - kx , Damping: F = - bv Bouyant: F B = ρV g W = r f r i F · dr , W = F · Δ r , K = 1 2 mv 2 , Δ U gravity = mgy , Δ U spring = 1 2 kx 2 , P = dW dt , P = F · v . Thermodynamics Thermal Expansion: Δ L = αL Δ T Stress and Strain: F A = Y Δ L L Ideal Gas Law: PV = nRT , K av = 3 2 kT , 1 2 kT for each degree of freedom. Thermal Conductivity: I = Δ Q Δ t = kA Δ T Δ x Black Body Radiation: P = eσAT 4 Internal Energy: U = nC V T First Law: dQ = dU + dW for an ideal gas dW = PdV . For isothermal expansion W = nRT ln( V f /V i ). For adiabatic expansion PV γ = C where C is a constant and γ = C p /C v . Work for adiabatic expansion W = V 2 V 1 PdV = C V 2 V 1 dV V γ = C 1 - γ ( V 2 1 - γ - V 1 1 - γ ) Q = mc Δ T , Q = mL C P = C V + R , C V = f 2 R where f =degrees of freedom. Oscillations ω = 2 πf , T = 1 f , x = A cos( ωt + φ ), ω 2 = k m . Damped Oscillations:
Image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: x = A e-bt 2 m cos( ωt + φ ), where ω = q ω 2-( b 2 m ) 2 , Q = 2 π E Δ E Waves v = q T μ , k = 2 π λ , v = λf , P = 1 2 μω 2 A 2 v , y = A sin( kx ∓ ωt + φ ) p = ρωvs v = q γRT M , I = P av 4 πr 2 , β = 10 dB log( I I ) Doppler Effect f = f (1 ± v D v ) (1 ∓ v s v ) Beats Δ f = f 2-f 1 Interference φ = k Δ x Standing Waves f m = mv 2 L m = 1 , 2 , 3 , . . . f m = mv 4 L m = 1 , 3 , 5 , . . ....
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern