Final physics cheat sheet PDF

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A summary of all the formulas and ways to work through physics problems....

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• Constant acceleration equations • v = vo + at • x = x0 + 1/2(vo + v)t • x = x0 + vot + 1/2at2 • x = xo + vt - 1/2at2 • V2 = v02 + 2a(x-x0)

• Acceleration + slope of velocity vs time graph • Average speed= total distance/Δt • Average velocity= Δx/Δt0 • Density = mass x volume • Displacement= Δx = xf - xi • Speed= distance/time • Velocity = slope of position vs time graph (y-y/x-x)

Volumes • Sphere 4/3 π r3 • Cylinder πr2h Areas • Cube 2lw + 2wh + 2lh • Sphere 4πr2 • Cylinder 2πr2 + 2πrh

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F = µv2 v = √F/µ λ = 2L L = λ/2 v=fλ

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Largest resultant amp possible: add Smallest resultant amp possible: subtract Speed a wave travels down string: v = √(T/µ) Tension wire must be stretched for fundamental: F = 4mlf2 Wavelength at certain frequency: λ = v/f

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Distance of sound intensity: r = √[P/(4πr)] What distance is sound barely audible: (1) I1 = I010(β1/10) (2) r0 = √(I1r12)/I0 Wavelength of amb siren from standing position: λ’ = (vsound + vamb)/f0 Frequency car driver hears amb siren: f’ = [(vsound + vcar)/(vsound + vamb)]f Wavelength person between vehicles here’s amb siren: λ’ = (vsound + vamb)/f Frequency driver going opposite way from siren: f’ = [(vsound - vcar)/(vsound + vamb)]f Tension in string: F = 4mlf2 *m in kg *l in m

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Length required if pipe open at both ends: (1) λ = vs/f (2) L0 = λ/2 Length if pipe closed at one end: Lc = λ/4 Which wave has a # wavelength: (1) λquarter = 4(L/λ) (2) nodes = (2L/λ) + 1/2 Given beat frequency, what is frequency: (1) f’ = [vsound / (vsound - vtrain)f (2) f = fb/ (f’ - 1) *vtrain in m/s *f in (1) is just variable

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I = P/(4πr2) I0 = 1 x 10-12 vs = 340 m/s fb = |f’ - f|

- Power = W/t - 1 hp = 746 Watt - Energy = Pt

• How much water could energy vaporize: m = Q/hvap • How much energy is required for change: (1) T-0°C = cicemT (2) ice-water = hwm (3) water-100°C = cwm(100) (4) water-steam = hvapm (5) steam-given T = csteamm(T-100) (6) add together

• Equilibrium temp: T = ∜(R/σe) • °F to °C 5/9(°F - 32) • °C to K °C + 273 • °C to °F 9/5(°C) + 32 • Temp both must be heated: T1 = To + (db - da)/(αada - αbdb)

Efficiency of heat engine: e = 1 - (Qcold/Qhot) Max theoretical efficiency: e = 1 - (Tcold/Thot) *T in K Given actual efficiency, how much mechanical power: P = (eQhot)/t *e as decimal COP = Thot/(Thot - Tcold) COP = [cm(T1 - T2)]/Pt *P in W (1 hp=746W) *t in s *T in K Electric power required to maintain temp: rate given (in W) Electric power required to maintain temp w/coefficient factor: (1) COP = Tin/(Tin

• What height after temp rises: (1) Li = L/2 (2) Lf = Li(1 + αΔT) (3) y = √(L2f - L2i) • Depth of layer: yoil = (h - y)/[1 - (specific gravity)] • Pressure at entrance of pipe: P1 = P2 - ρgh + 1/2ρ(v22 - v21)

• Change in entropy: ΔS = (mLv)/T *T in K *m in kg • Frequency: cycles/seconds • Period: 1/frequency

• Apparent weight of accelerated object: (1) a = = mag x given weight (in kg)

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• Speed of flow at end of siphon: v = √2gh1 • How long to siphon x water: t = 4V/πd2v *V is given; convert 1L = 10-3 m3 • Max value of h2: h2 = (Patm/ρg) - h1

Amplitude: 1/2distance Amplitude of mass-spring system: (1) K = (4π2m)/T2 (2) A = √2E/K Gravitational acceleration: g = (4n2π2l)/t2 *l in m Wavelength: c/f *if f in GHz, 1 GHz = 109 Hz v2/r

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• Absolute pressure at depth: P = Patm + ρgh • How much heat to raise temp: Q = mcΔT • Final temp: (1) mw = ρwvw (2) Tf = (mwcwT2 + mbcbT1)/ (mwcw + mbcb)

• Average KE of each atom: K = 3/2kBT *T in K

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SohCahToa Force F = ma Tension T = m(g+a) or Force for a SPRING F = -kx Kparallel = K1 + K2 Time of flight t= (2sinθ)/g (use if takes off & lands at same point T of F (different points) ttotal = (2vosinθ)/g Max height= y= [vo2(sinθ)2]/2g Range= R = [2vo2(sinθ)(cosθ)]/g Magnitude F = m(Δv/Δt) or F = ma • Time it takes for cars to become even t= Δx/ (v1 - v2)

in m3 • Gauge pressure after temp increase: (1) P1 = Pg + Patm (2) P2 = [(P1T2)/T1] - Patm

• Net energy transferred to gas by heat: (1) ΔV = Vf - Vi (2) Q = ΔV - WIAF • Work done by steam: W = nRT - n(mP/ρ) *P x 101300 Pa *ρ in g/cm3 so x (1 m/ 100 cm)3 • Change in internal energy: ΔU = nmLv - w *m in kg

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(2) magnitude =

• Young’s modulus to calculate

√(g2

+

a2)

(3) apparent weight

Minimum angular velocity needed to keep a person from slipping downward: vmin = √g/rµ Centripetal acceleration of the end of the rod: a = v2/r Force would be needed to hold an object to the rod: F= ma *m in kg Period of rotation of the rod: T = 2πr/v *r is given length Magnitude of the force that maintains circular motion acting on the puck: F = mtiedg Linear speed of the puck: v = √Fr/mpuck How fast can a car take the curve without skidding (no friction): v = √rgtanθ Minimum speed a ball must have at the top of the circle: v = √gr *r is just l How far apart are the students sitting: r2 = (Gm1m2)/Fg

mg • Bulk modulus to calculate

• Area for steel bolt is πr2 • Area for steel plate is 2πrt (t = thickness) • • • • • • •

Moments of inertia Disk I = 1/2mr2 Solid sphere I = 2/5mr2 Hollow sphere I = 2/3mr2 Long thin rod: I = 1/2ml2total Parallel axis theorum I = 1/12ml2 + m(l/2)2

• Angular speed at top after string is pulled off: (1) W = T x l (2) WF = √(2W/I) • Calculate final angular speed of system: (1) Ii = I + 2mR2 (2) IF = I + 2mr2 (3) ωF = ω(Ii /If) • Pressure exerted by each leg: P = [(m1 + m2)g]/4πr2 *r in m

- Momentum p = mv - Velocity: v = p/m

Patm = 101.3 kPa or 101300 Pa σ = 100 unless given e=1 −23 kB = 1.38066 × 10 J/K U = internal energy Natoms = 6.022 x 1023 NOT NEGATIVE!! R = 8.31

• Estimate cross sectional area: A = Fg/ρhg • Flow of speed of gas: v = flow rate/πr2 • Speed of water flowing thru constricted segment: v2 = v1(d12/d22) • Gauge pressure at 2nd diameter: P2gauge = P1gauge - 1/2ρ(v22 - v12) *make sure P1 is same unit as 2nd part

• If D=0 or F=0 or cos90, then W=0 (there is no work being done) • Work energy theorem: Wnet = ΔKE = 1/2mv2 - 1/2mv2initial • The net work done by all forces acting on a system equals its change in kinetic energy • Kinetic energy: KE = 1/2mv2 (m in kg and v in m/s) • Gravitational potential energy: ΔPEg = mgh • Potential energy of a spring/work done to stretch a spring: 1/2kx2 • FOR CONSERVATIVE FORCES KE + PE = constant. KEi + PEi = KEf + PEf • Friction is a nonconservative force, Wnet = ΔKE = Wnonc + Wcona • Value of the heavier mass: (1) find total mass m1 + m2 = 2KE/v2 (2) m1 -m2 = (mtotalv)/gt (3) m1 = [mtotal + (m1 - m2)]/2 • Value of the lighter mass: m2 = [mtotal - (m1 - m2)]/2 • If Δy = 0, then no normal or gravitational force • Work done against friction: Wfriction = mgr - KE • How much work must be done to stretch a spring given meters: (1) to get final displacement, add additional meters given to length of spring (2) W = 1/2K(x2final - x2initial) • Given power and speed, magnitude of the force: P = w/t turns into P = Fd/t turns into F = P/v • How much energy is consumed by the motor daily: (1) find power in J/s from hp using dimensional analysis. I hp=750 watts, 1 watt=1J/s (2) Use dimensional analysis to convert hours to seconds (3) W = Pt • How much it costs to run the compressor each day: (1) convert watts to kilowatts (2) multiply kW by hours (3) multiply by cents • Height above the ground is the block released: (1) find h1 using (x2/4h2) + µl (2) h = h1 - h2 *h2 is negative • Speed of the block when it leaves the track: vx = √2g(h1 - µl) • Total speed of block when hits ground: (1) vy = √-2gh2 (2) vf = √v2x + v2y • A ball is dropped on a spring & compresses it, spring force constant: K = 2mg(h + x) *m in kg x2 • • Block is pushed against a spring horizontally, when spring releases, it moves block, what is Δx: Δx = √(mv2block/K) • Speed of block at top of track: (1) find energy at B EB = 1/2mv2block (2)find Wf = -fπr (3) total energy = EB + Wf (4) v = √ 2E t total - 2ght • Centripetal acceleration of the block at top of track: ac = vt0p2/r m

• Magnitude of average force exerted on ball by the wall: |F| = -(2mvcosθ)/Δt • Average net force exerted: F = I/Δt • What is more important in determining amount of damage an object gets in a collision: total momentum changed per unit time • Speed of astronaut after throwing tank: (1) find v2 = given v (2) v1 = (-m2v2)/m1 *m2 is object thrown • Initial speed of bullet after hits block: v1 = [(m1 + m2)/ • Find velocity of object that is slower after collision: vif = [m1v1 + m2v2 - m2(v1 - v2)]/(m1 + m2) *can leave in g & cm • Angle 2 coupled cars skid: (1) find pfx = m2v2 (2) find pfy = m1m2 (3) tanθ = pfx/pfy • Find force FT provided by cable: FT = mg/sinθ

- 1 rev = 2π rad, 1 min = 60 s - Through what angle does it rotate (given angular speed in rad/ - Angular acceleration (given winitial, rev and t): [2(θ - winitial)]/t2 - Angular acceleration (given winitial, rev/min and t): (1) find wfinal by converting rev/min to rad/s

(2) (wfinal - winitial)/t

- Angular speed (given vtangential and x): w = vtan/x - Angle rotates (given winitial, angular acceleration, & t): θ = winitialt + 1/2aacc.t2

- Magnitude of Hx on pivot: Hx = 1/2 mgsinθ - Force exerted by hammer on nail: R = Fh/lcosθ - Force exerted by surface on point of contact w hammer head: (1) find f = Rsinθ - F (2) find n = Rcosθ (3) F = √f2 + n2

- Magnitude of force required to apply at a point so the board -

stays horizontal when mass is at end: (1) find x2 = (ltotal/2) - x1 (2) Fa = [(x2mboard + x3madded)/x1]g Magnitude of closer point: FB = FA + (m1 + m2)g *make sure 2nd part is in kN Force must be exerted on bicep: FB = [(x1 + x2)mg]/x2cosθ Bulk modulus to find final v: (1) Δv = - ΔPvi/B (2) vf = vi + Δv Force exerted by wedge on sphere at point: FB = mg/sinθ...