Physics Mathematics

100 Best Physics Formulas Every Student Should Know

The 100 physics formulas students use most, from kinematics and energy to circuits, optics and modern physics, each with symbols, units and a worked example.

This is a working reference for the physics formulas that appear again and again in high school and first-year university courses: motion, forces, energy, rotation, gravity, fluids, waves, heat, electricity, magnetism, optics and modern physics.

Each of the 100 entries gives the formula, what every symbol means with its SI unit, when to use it, and a short worked example. Use it as a revision checklist. If you cannot say when a formula applies, or you cannot redo its example without looking, that is the one to practise.

How this was checked. Formulas follow the free OpenStax textbooks University Physics (Volumes 1–3) and College Physics 2e. Constants are the 2022 CODATA values published by NIST. Every worked example was recomputed in Python, and results are rounded to a sensible number of significant figures. Examples use g=9.8 m/s2g = 9.8\ \text{m/s}^2 unless stated otherwise. Full references are at the end.

Constants used in this article

Constant Symbol Value
Speed of light in vacuum cc 2.998×1082.998 \times 10^8 m/s (exact: 299,792,458 m/s)
Gravitational constant GG 6.674×10−116.674 \times 10^{-11} N·m²/kg²
Planck constant hh 6.626×10−346.626 \times 10^{-34} J·s (exact)
Elementary charge ee 1.602×10−191.602 \times 10^{-19} C (exact)
Electron mass mem_e 9.109×10−319.109 \times 10^{-31} kg
Boltzmann constant kBk_B 1.381×10−231.381 \times 10^{-23} J/K (exact)
Molar gas constant RR 8.3148.314 J/(mol·K)
Avogadro constant NAN_A 6.022×10236.022 \times 10^{23} mol⁻¹ (exact)
Stefan–Boltzmann constant σ\sigma 5.670×10−85.670 \times 10^{-8} W/(m²·K⁴)
Vacuum permittivity ε0\varepsilon_0 8.854×10−128.854 \times 10^{-12} F/m
Vacuum permeability μ0\mu_0 1.257×10−61.257 \times 10^{-6} N/A²
Coulomb constant k=1/(4πε0)k = 1/(4\pi\varepsilon_0) 8.988×1098.988 \times 10^9 N·m²/C²
Standard gravity gg 9.806659.80665 m/s² (exact, by definition); about 9.8 m/s² in examples

Values from the NIST CODATA 2022 tables. "Exact" means the value is fixed by the definition of the SI units since 2019.

Kinematics

The equations of motion for constant acceleration, plus projectiles (OpenStax University Physics Volume 1, chapters 3–4). In formulas 3 to 6, uu is the initial velocity, vv the final velocity, aa the acceleration, tt the time and ss the displacement.

1. Average speed and velocity

vˉ=ΔxΔt\bar{v} = \frac{\Delta x}{\Delta t}

Symbols: Δx\Delta x is displacement (m) for velocity, or total distance for speed; Δt\Delta t is time taken (s).

Use it when: you need an average over a whole trip, not the value at one instant.

Example: A drive of 120 km in 1.5 h has an average speed of 80 km/h, or 80/3.6=22.280 / 3.6 = 22.2 m/s.

2. Acceleration

a=ΔvΔta = \frac{\Delta v}{\Delta t}

Symbols: Δv\Delta v is change in velocity (m/s), Δt\Delta t is time (s), aa is in m/s².

Use it when: velocity changes at a steady rate, or you want the average acceleration.

Example: A car going from rest to 27 m/s in 9.0 s has a=27/9.0=3.0a = 27 / 9.0 = 3.0 m/s².

3. Velocity–time equation

v=u+atv = u + at

Use it when: acceleration is constant and you know (or want) the time, but not the displacement.

Example: A ball rolling at 5 m/s that accelerates at 2 m/s² for 4 s reaches v=5+2×4=13v = 5 + 2 \times 4 = 13 m/s.

4. Displacement–time equation

s=ut+12at2s = ut + \tfrac{1}{2}at^2

Use it when: you know the starting velocity, the acceleration and the time, and want the displacement.

Example: A stone dropped from rest falls s=12(9.8)(32)=44.1s = \tfrac{1}{2}(9.8)(3^2) = 44.1 m in 3 s.

5. Velocity–displacement equation

v2=u2+2asv^2 = u^2 + 2as

Use it when: time is unknown and not needed, such as braking-distance problems.

Example: A car at 25 m/s braking at 5 m/s² stops when 0=252−2(5)s0 = 25^2 - 2(5)s, so s=62.5s = 62.5 m.

6. Displacement from average velocity

s=(u+v)2 ts = \frac{(u + v)}{2}\,t

Use it when: acceleration is constant and you know both velocities and the time, but not the acceleration.

Example: A train speeding up from 10 m/s to 30 m/s in 20 s travels s=10+302×20=400s = \frac{10 + 30}{2} \times 20 = 400 m.

7. Range of a projectile

R=v02sin⁡2θgR = \frac{v_0^2 \sin 2\theta}{g}

Symbols: v0v_0 is launch speed (m/s), θ\theta is launch angle above horizontal, RR is horizontal range (m).

Use it when: the projectile lands at the same height it was launched from and air resistance is negligible. Range is greatest at 45°.

Example: At 20 m/s and 45°, R=202×sin⁡90°/9.8=40.8R = 20^2 \times \sin 90° / 9.8 = 40.8 m.

8. Time of flight

T=2v0sin⁡θgT = \frac{2 v_0 \sin\theta}{g}

Use it when: the projectile lands at its launch height.

Example: At 20 m/s and 30°, T=2×20×0.5/9.8=2.04T = 2 \times 20 \times 0.5 / 9.8 = 2.04 s.

9. Maximum height of a projectile

H=v02sin⁡2θ2gH = \frac{v_0^2 \sin^2\theta}{2g}

Use it when: you need the peak height, where the vertical velocity is zero.

Example: At 20 m/s and 30°, H=400×0.25/19.6=5.10H = 400 \times 0.25 / 19.6 = 5.10 m.

Forces and momentum

Newton's laws and the forces they act through (OpenStax University Physics Volume 1, chapters 5, 6 and 9).

10. Newton's second law

Fnet=maF_{\text{net}} = ma

Symbols: FnetF_{\text{net}} is net force (N), mm is mass (kg), aa is acceleration (m/s²). 1 N = 1 kg·m/s².

Use it when: relating the total force on an object to its acceleration. Always use the net force.

Example: A 1,200 kg car accelerating at 2.5 m/s² needs a net force of 1200×2.5=30001200 \times 2.5 = 3000 N.

11. Weight

W=mgW = mg

Use it when: finding the gravitational force on a mass. Mass stays the same everywhere; weight depends on gg.

Example: A 60 kg person weighs 60×9.8=58860 \times 9.8 = 588 N on Earth and 60×1.62=97.260 \times 1.62 = 97.2 N on the Moon.

12. Friction

fs≤μsN,fk=μkNf_s \le \mu_s N, \qquad f_k = \mu_k N

Symbols: μs\mu_s and μk\mu_k are the coefficients of static and kinetic friction (no units), NN is the normal force (N).

Use it when: surfaces are sliding (fkf_k) or about to slide (maximum fsf_s).

Example: Sliding a 50 kg box across a level floor with μk=0.30\mu_k = 0.30 needs a force of 0.30×50×9.8=1470.30 \times 50 \times 9.8 = 147 N to keep it moving at constant speed.

13. Hooke's law

F=−kxF = -kx

Symbols: kk is the spring constant (N/m), xx is extension or compression from the natural length (m). The minus sign means the force opposes the displacement.

Use it when: a spring or elastic material is within its elastic limit.

Example: Stretching a spring with k=200k = 200 N/m by 5.0 cm needs 200×0.050=10200 \times 0.050 = 10 N.

14. Centripetal force

Fc=mv2r,ac=v2rF_c = \frac{mv^2}{r}, \qquad a_c = \frac{v^2}{r}

Use it when: something moves in a circle. FcF_c is not a new force; it is the net inward force supplied by tension, friction, gravity or a normal force.

Example: A 1,000 kg car taking a 50 m radius curve at 15 m/s needs 1000×152/50=45001000 \times 15^2 / 50 = 4500 N of friction from the road.

15. Forces on an inclined plane

F∥=mgsin⁡θ,N=mgcos⁡θF_{\parallel} = mg\sin\theta, \qquad N = mg\cos\theta

Use it when: an object rests or slides on a slope at angle θ\theta, with no other forces perpendicular to the slope.

Example: For a 10 kg crate on a 30° slope, the force down the slope is 98sin⁡30°=4998 \sin 30° = 49 N and the normal force is 98cos⁡30°=84.998\cos 30° = 84.9 N.

16. Pressure

P=FAP = \frac{F}{A}

Symbols: PP in pascals (1 Pa = 1 N/m²), FF is force perpendicular to the surface (N), AA is area (m²).

Use it when: a force is spread over an area.

Example: A 600 N person standing on 0.040 m² of shoe sole exerts 15,00015,000 Pa. Balanced on a 1 cm² heel (10−410^{-4} m²), the pressure would be 6.0×1066.0 \times 10^6 Pa.

17. Momentum

p=mvp = mv

Use it when: analysing collisions, explosions and stopping forces. Momentum is a vector, in kg·m/s.

Example: A 0.145 kg baseball at 40 m/s has p=5.8p = 5.8 kg·m/s.

18. Impulse–momentum theorem

J=F Δt=ΔpJ = F\,\Delta t = \Delta p

Use it when: a force acts for a short time. Stretching the time reduces the average force for the same change in momentum.

Example: Catching that 5.8 kg·m/s baseball over 0.05 s takes an average force of 5.8/0.05=1165.8 / 0.05 = 116 N. Stopping it in 0.005 s would take 1,160 N, which is why you pull your hands back when catching.

19. Conservation of momentum

m1u1+m2u2=m1v1+m2v2m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

Use it when: no net external force acts during a collision or explosion. It applies to elastic and inelastic collisions.

Example: A 1,000 kg car at 20 m/s hits a stationary 1,500 kg car and they lock together: v=1000×202500=8.0v = \frac{1000 \times 20}{2500} = 8.0 m/s.

Work, energy and power

Energy accounting is often the fastest route to a mechanics answer (OpenStax University Physics Volume 1, chapters 7–8).

20. Work done by a constant force

W=Fdcos⁡θW = Fd\cos\theta

Symbols: FF is force (N), dd is displacement (m), θ\theta is the angle between them. Work is in joules (J).

Use it when: a constant force moves an object. A force perpendicular to the motion does no work.

Example: Pulling a suitcase 20 m with 50 N at 30° above the horizontal does 50×20×cos⁡30°=86650 \times 20 \times \cos 30° = 866 J of work.

21. Kinetic energy

KE=12mv2KE = \tfrac{1}{2}mv^2

Use it when: finding the energy of motion. Doubling the speed quadruples the kinetic energy.

Example: A 1,000 kg car at 20 m/s has 12(1000)(202)=200,000\tfrac{1}{2}(1000)(20^2) = 200{,}000 J, or 200 kJ.

22. Gravitational potential energy (near Earth)

PE=mghPE = mgh

Use it when: height changes are small compared with Earth's radius. For orbits, use formula 41.

Example: A 70 kg person climbing 3.0 m of stairs gains 70×9.8×3.0=205870 \times 9.8 \times 3.0 = 2058 J, about 2.1 kJ.

23. Elastic potential energy

PEspring=12kx2PE_{\text{spring}} = \tfrac{1}{2}kx^2

Use it when: energy is stored in a stretched or compressed spring.

Example: A spring with k=400k = 400 N/m compressed by 0.10 m stores 12(400)(0.102)=2.0\tfrac{1}{2}(400)(0.10^2) = 2.0 J.

24. Work–energy theorem

Wnet=ΔKE=12mv2−12mu2W_{\text{net}} = \Delta KE = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mu^2

Use it when: you know the net work and want a speed, or the other way round, without needing time.

Example: Slowing a 1,000 kg car from 20 m/s to 10 m/s takes W=50,000−200,000=−150,000W = 50{,}000 - 200{,}000 = -150{,}000 J. The negative sign means the brakes remove energy.

25. Power

P=Wt=ΔEtP = \frac{W}{t} = \frac{\Delta E}{t}

Use it when: you need the rate of energy transfer. Power is in watts (1 W = 1 J/s).

Example: Climbing those 3.0 m of stairs (2,058 J) in 4.0 s takes 2058/4.0=5152058 / 4.0 = 515 W.

26. Power from force and velocity

P=FvP = Fv

Use it when: a steady force keeps an object moving at constant speed.

Example: A cyclist pushing against 20 N of drag at 10 m/s produces 20×10=20020 \times 10 = 200 W.

27. Efficiency

η=useful energy (or power) outtotal energy (or power) in×100%\eta = \frac{\text{useful energy (or power) out}}{\text{total energy (or power) in}} \times 100\%

Use it when: comparing machines. Efficiency is always less than 100% for real devices.

Example: A motor taking 400 W of electrical power and delivering 300 W of mechanical power is 75% efficient.

Rotation

Each linear formula has a rotational partner: mass becomes moment of inertia, force becomes torque (OpenStax University Physics Volume 1, chapters 10–12). Angles must be in radians.

28. Angular velocity and linear speed

ω=ΔθΔt,v=ωr\omega = \frac{\Delta\theta}{\Delta t}, \qquad v = \omega r

Symbols: ω\omega in rad/s, rr is radius (m), vv is the speed of a point at radius rr (m/s).

Use it when: converting between spinning rate and the speed of a point on the rotating object. One revolution is 2π2\pi rad.

Example: A wheel turning at 600 revolutions per minute has ω=600×2π/60=62.8\omega = 600 \times 2\pi / 60 = 62.8 rad/s. A point on its 0.35 m rim moves at 62.8×0.35=22.062.8 \times 0.35 = 22.0 m/s.

29. Angular acceleration

α=ΔωΔt,at=αr\alpha = \frac{\Delta\omega}{\Delta t}, \qquad a_t = \alpha r

Use it when: a spin rate changes. ata_t is the tangential acceleration of a point at radius rr.

Example: A fan reaching 62.8 rad/s from rest in 4.0 s has α=15.7\alpha = 15.7 rad/s². A point 0.30 m from the axis has at=4.71a_t = 4.71 m/s².

30. Torque

τ=rFsin⁡θ\tau = rF\sin\theta

Symbols: rr is distance from the pivot to where the force acts (m), θ\theta is the angle between rr and FF. Torque is in N·m.

Use it when: working out the turning effect of a force.

Example: A 40 N push at right angles on a spanner 0.25 m long gives τ=0.25×40=10\tau = 0.25 \times 40 = 10 N·m.

31. Newton's second law for rotation

τnet=Iα\tau_{\text{net}} = I\alpha

Use it when: relating net torque to angular acceleration about a fixed axis.

Example: A net torque of 10 N·m on a wheel with I=0.50I = 0.50 kg·m² gives α=20\alpha = 20 rad/s².

32. Moment of inertia

I=∑miri2,Isolid disc=12MR2,Ihoop=MR2I = \sum m_i r_i^2, \qquad I_{\text{solid disc}} = \tfrac{1}{2}MR^2, \qquad I_{\text{hoop}} = MR^2

Use it when: finding how hard an object is to spin. Other common shapes: solid sphere 25MR2\tfrac{2}{5}MR^2, rod about its centre 112ML2\tfrac{1}{12}ML^2.

Example: A 2.0 kg solid disc of radius 0.20 m has I=12(2.0)(0.202)=0.040I = \tfrac{1}{2}(2.0)(0.20^2) = 0.040 kg·m².

33. Rotational kinetic energy

KErot=12Iω2KE_{\text{rot}} = \tfrac{1}{2}I\omega^2

Use it when: an object spins. A rolling object has both 12mv2\tfrac{1}{2}mv^2 and 12Iω2\tfrac{1}{2}I\omega^2.

Example: The 0.040 kg·m² disc spinning at 50 rad/s stores 12(0.040)(502)=50\tfrac{1}{2}(0.040)(50^2) = 50 J.

34. Angular momentum and its conservation

L=Iω,I1ω1=I2ω2L = I\omega, \qquad I_1\omega_1 = I_2\omega_2

Use it when: no external torque acts, so LL stays constant.

Example: A skater spinning at 2.0 rad/s with I=3.0I = 3.0 kg·m² pulls in their arms to I=1.0I = 1.0 kg·m² and spins at ω2=3.0×2.0/1.0=6.0\omega_2 = 3.0 \times 2.0 / 1.0 = 6.0 rad/s.

35. Conditions for equilibrium

∑F=0,∑τ=0\sum F = 0, \qquad \sum \tau = 0

Use it when: an object is at rest and not rotating, such as beams, ladders, shelves and seesaws.

Example: A 30 kg child sits 2.0 m from a seesaw's pivot. A 40 kg child balances them at x=30×2.0/40=1.5x = 30 \times 2.0 / 40 = 1.5 m on the other side.

Gravitation and orbits

Newton's law of gravitation and the motion of satellites (OpenStax University Physics Volume 1, chapter 13). Examples use Earth's mass ME=5.972×1024M_E = 5.972 \times 10^{24} kg and mean radius RE=6.371×106R_E = 6.371 \times 10^6 m.

36. Newton's law of universal gravitation

F=Gm1m2r2F = \frac{G m_1 m_2}{r^2}

Symbols: rr is the distance between the centres of mass (m).

Use it when: finding the attraction between any two masses.

Example: Earth and the Moon (7.35×10227.35 \times 10^{22} kg, 3.84×1083.84 \times 10^8 m apart) attract each other with about 1.98×10201.98 \times 10^{20} N.

37. Gravitational field strength

g=GMR2g = \frac{GM}{R^2}

Use it when: finding surface gravity on a planet or gg at a height above it (use r=R+hr = R + h).

Example: For Earth, g=6.674×10−11×5.972×1024/(6.371×106)2=9.82g = 6.674 \times 10^{-11} \times 5.972 \times 10^{24} / (6.371 \times 10^6)^2 = 9.82 m/s². The measured value varies from about 9.78 to 9.83 m/s² with latitude and altitude.

38. Orbital speed (circular orbit)

v=GMrv = \sqrt{\frac{GM}{r}}

Use it when: a satellite is in a circular orbit of radius rr measured from the planet's centre.

Example: At 400 km altitude, r=6.771×106r = 6.771 \times 10^6 m and v=7.67v = 7.67 km/s, about the speed of the International Space Station.

39. Kepler's third law

T2=4π2GM r3T^2 = \frac{4\pi^2}{GM}\,r^3

Use it when: relating an orbit's period to its radius (or semi-major axis for an ellipse).

Example: For the 400 km orbit above, T=2πr3/GM=5545T = 2\pi\sqrt{r^3/GM} = 5545 s, about 92 minutes.

40. Escape velocity

vesc=2GMRv_{\text{esc}} = \sqrt{\frac{2GM}{R}}

Use it when: finding the minimum launch speed to leave a planet with no further thrust, ignoring air resistance.

Example: For Earth, vesc=11.2v_{\text{esc}} = 11.2 km/s, exactly 2\sqrt{2} times the orbital speed at the same radius.

41. Gravitational potential energy (general)

U=−GMmrU = -\frac{GMm}{r}

Use it when: heights are large compared with the planet's radius. U=0U = 0 at infinite distance, so bound objects have negative energy.

Example: Raising 1 kg from Earth's surface to 400 km takes GM(1/RE−1/r)=3.70×106GM(1/R_E - 1/r) = 3.70 \times 10^6 J. The approximation mghmgh would give 3.93×1063.93 \times 10^6 J, about 6% too much, because gg falls with height.

Fluids

Pressure, buoyancy and flow (OpenStax University Physics Volume 1, chapter 14). Fresh water has ρ≈1000\rho \approx 1000 kg/m³; air at room conditions about 1.2 kg/m³.

42. Density

ρ=mV\rho = \frac{m}{V}

Use it when: comparing materials or predicting whether something floats.

Example: A 500 g stone with a volume of 200 cm³ has ρ=2.5\rho = 2.5 g/cm³, or 2,500 kg/m³.

43. Pressure at depth

P=P0+ρghP = P_0 + \rho g h

Symbols: P0P_0 is pressure at the surface (Pa), hh is depth (m).

Use it when: finding pressure in a liquid at rest.

Example: At 10 m in fresh water, the water adds 1000×9.8×10=981000 \times 9.8 \times 10 = 98 kPa. With atmospheric pressure (101 kPa), the total is about 199 kPa, roughly double the surface pressure.

44. Archimedes' principle

Fb=ρfluid Vdisplaced gF_b = \rho_{\text{fluid}}\, V_{\text{displaced}}\, g

Use it when: finding the upward force on a submerged or floating object. An object floats if it can displace its own weight of fluid.

Example: A fully submerged 0.0020 m³ block in water feels a buoyant force of 1000×0.0020×9.8=19.61000 \times 0.0020 \times 9.8 = 19.6 N.

45. Pascal's principle (hydraulics)

F1A1=F2A2\frac{F_1}{A_1} = \frac{F_2}{A_2}

Use it when: a confined fluid transmits pressure between two pistons.

Example: 100 N on a 4.0 cm² piston produces 100×(400/4.0)=10,000100 \times (400 / 4.0) = 10{,}000 N on a 400 cm² piston, enough to lift about 1,000 kg.

46. Continuity equation

A1v1=A2v2A_1 v_1 = A_2 v_2

Use it when: an incompressible fluid flows through a pipe that changes width.

Example: Water at 1.0 m/s in a 2.0 cm-diameter hose speeds up to 16 m/s in a 0.50 cm nozzle, because the area is 16 times smaller.

47. Bernoulli's equation

P+12ρv2+ρgh=constantP + \tfrac{1}{2}\rho v^2 + \rho g h = \text{constant}

Use it when: steady, non-viscous, incompressible flow along a streamline. Faster flow at the same height means lower pressure.

Example: Water speeding up from 2.0 m/s to 4.0 m/s in a horizontal pipe drops in pressure by 12(1000)(42−22)=6000\tfrac{1}{2}(1000)(4^2 - 2^2) = 6000 Pa.

48. Drag equation

FD=12ρv2CDAF_D = \tfrac{1}{2}\rho v^2 C_D A

Symbols: CDC_D is the drag coefficient (no units), AA is frontal area (m²).

Use it when: estimating air or water resistance at everyday speeds, where drag grows with v2v^2.

Example: A cyclist with CDA=0.30C_D A = 0.30 m² at 10 m/s in air of density 1.2 kg/m³ feels 12(1.2)(102)(0.30)=18\tfrac{1}{2}(1.2)(10^2)(0.30) = 18 N of drag and must produce 18×10=18018 \times 10 = 180 W just to overcome it.

Oscillations, waves and sound

Oscillators, wave speed, sound levels and the Doppler effect (OpenStax University Physics Volume 1, chapters 15–17).

49. Simple harmonic motion

x(t)=Acos⁡(ωt+ϕ)x(t) = A\cos(\omega t + \phi)

Symbols: AA is amplitude (m), ω\omega is angular frequency (rad/s), ϕ\phi is the phase constant.

Use it when: an object oscillates under a restoring force proportional to displacement.

Example: With A=0.10A = 0.10 m, f=2.0f = 2.0 Hz (ω=4π\omega = 4\pi rad/s) and ϕ=0\phi = 0, the position at t=0.10t = 0.10 s is 0.10cos⁡(0.4π)=0.0310.10\cos(0.4\pi) = 0.031 m.

50. Period of a mass on a spring

T=2πmkT = 2\pi\sqrt{\frac{m}{k}}

Use it when: a mass oscillates on an ideal spring. The period does not depend on amplitude.

Example: A 0.50 kg mass on a 200 N/m spring has T=2π0.0025=0.314T = 2\pi\sqrt{0.0025} = 0.314 s.

51. Period of a simple pendulum

T=2πLgT = 2\pi\sqrt{\frac{L}{g}}

Use it when: a pendulum swings through small angles (below about 15°). The period does not depend on mass.

Example: A 1.00 m pendulum with g=9.81g = 9.81 m/s² has T=2.01T = 2.01 s.

52. Frequency, period and angular frequency

f=1T,ω=2πff = \frac{1}{T}, \qquad \omega = 2\pi f

Use it when: converting between how often something repeats and how long each cycle takes. Frequency is in hertz (Hz).

Example: A 50 Hz mains supply has a period of 0.020 s and ω=314\omega = 314 rad/s.

53. Wave equation

v=fλv = f\lambda

Use it when: relating speed, frequency and wavelength for any wave.

Example: The note A at 440 Hz in air (343 m/s) has a wavelength of 343/440=0.780343 / 440 = 0.780 m.

54. Wave speed on a string

v=Tμv = \sqrt{\frac{T}{\mu}}

Symbols: TT is tension (N), μ\mu is mass per unit length (kg/m).

Use it when: tuning string instruments or working with standing waves on strings.

Example: A string under 60 N with μ=0.00060\mu = 0.00060 kg/m carries waves at v=100,000=316v = \sqrt{100{,}000} = 316 m/s.

55. Speed of sound in air

v≈331+0.6 TC m/sv \approx 331 + 0.6\,T_C \ \text{m/s}

Symbols: TCT_C is air temperature in °C.

Use it when: you need the speed of sound at an everyday temperature. It is an approximation valid near room temperature.

Example: At 20 °C, v≈343v \approx 343 m/s; at 30 °C, about 349 m/s.

56. Sound intensity level

β=10log⁡10 ⁣(II0) dB,I0=10−12 W/m2\beta = 10\log_{10}\!\left(\frac{I}{I_0}\right)\ \text{dB}, \qquad I_0 = 10^{-12}\ \text{W/m}^2

Use it when: converting sound intensity (W/m²) to decibels. Every 10 dB increase is a factor of 10 in intensity.

Example: I=10−5I = 10^{-5} W/m² gives β=10log⁡10(107)=70\beta = 10\log_{10}(10^7) = 70 dB, roughly busy traffic.

57. Doppler effect for sound

f′=f v±vov∓vsf' = f\,\frac{v \pm v_o}{v \mp v_s}

Symbols: vv is the speed of sound, vov_o the observer's speed, vsv_s the source's speed. Use the upper signs when they move towards each other.

Use it when: source or listener moves relative to the air.

Example: A 700 Hz siren approaching a stationary listener at 30 m/s sounds like 700×343/313=767700 \times 343 / 313 = 767 Hz, and 700×343/373=644700 \times 343 / 373 = 644 Hz as it moves away.

58. Standing waves on a string

fn=nv2L,n=1,2,3,…f_n = \frac{nv}{2L}, \qquad n = 1, 2, 3, \dots

Use it when: a string is fixed at both ends. n=1n = 1 is the fundamental; the others are harmonics.

Example: The 316 m/s string from formula 54, 0.65 m long, has a fundamental of 316/1.3=243316 / 1.3 = 243 Hz and a second harmonic of 486 Hz.

59. Beat frequency

fbeat=∣f1−f2∣f_{\text{beat}} = |f_1 - f_2|

Use it when: two close frequencies sound together and the loudness throbs.

Example: Tuning forks at 440 Hz and 444 Hz produce 4 beats per second. Musicians tune by adjusting until the beats disappear.

Heat and thermodynamics

Temperature, heat transfer, gases and engines (OpenStax University Physics Volume 2, chapters 1–4). Gas-law temperatures must be in kelvins.

60. Temperature conversions

TK=TC+273.15,TF=95TC+32T_K = T_C + 273.15, \qquad T_F = \tfrac{9}{5}T_C + 32

Use it when: switching scales. Gas laws, radiation and engine efficiency always use kelvins.

Example: 37 °C is 310.15 K and 98.6 °F.

61. Linear thermal expansion

ΔL=αL0ΔT\Delta L = \alpha L_0 \Delta T

Symbols: α\alpha is the coefficient of linear expansion (1/K), L0L_0 is original length (m).

Use it when: sizing gaps in bridges, rails and pipes.

Example: A 100 m steel bridge (α≈12×10−6\alpha \approx 12 \times 10^{-6} /K) warming by 40 K grows by 12×10−6×100×40=0.04812 \times 10^{-6} \times 100 \times 40 = 0.048 m, nearly 5 cm.

62. Specific heat

Q=mcΔTQ = mc\Delta T

Symbols: QQ is heat (J), cc is specific heat capacity (J/(kg·K)); for water c≈4186c \approx 4186 J/(kg·K).

Use it when: temperature changes without a change of state.

Example: Heating 1.0 kg of water from 20 °C to 100 °C takes 4186×80=3354186 \times 80 = 335 kJ. A 2.0 kW kettle would need at least 167 s, a bit under 3 minutes, ignoring losses.

63. Latent heat

Q=mLQ = mL

Symbols: LL is specific latent heat (J/kg): for water, fusion 3.34×1053.34 \times 10^5 J/kg and vaporisation 2.26×1062.26 \times 10^6 J/kg.

Use it when: a substance melts, freezes, boils or condenses at constant temperature.

Example: Melting 0.50 kg of ice at 0 °C takes 0.50×3.34×105=1.67×1050.50 \times 3.34 \times 10^5 = 1.67 \times 10^5 J.

64. Ideal gas law

PV=nRTPV = nRT

Symbols: PP in Pa, VV in m³, nn in moles, TT in K.

Use it when: gases are at low to moderate pressure and not close to condensing.

Example: 1 mol of gas at 273.15 K and 101,325 Pa occupies V=nRT/P=0.0224V = nRT/P = 0.0224 m³, or 22.4 L.

65. Combined gas law

P1V1T1=P2V2T2\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

Use it when: a fixed amount of gas changes state. Use absolute pressure, not gauge pressure.

Example: A tyre at 300 kPa absolute and 293 K (20 °C) cools to 263 K (−10 °C) at roughly constant volume: P2=300×263/293=269P_2 = 300 \times 263 / 293 = 269 kPa.

66. Kinetic theory of gases

KE‾=32kBT,vrms=3kBTm\overline{KE} = \tfrac{3}{2}k_B T, \qquad v_{\text{rms}} = \sqrt{\frac{3k_B T}{m}}

Symbols: mm is the mass of one molecule (kg).

Use it when: linking temperature to the motion of gas molecules.

Example: At 300 K, each molecule has an average translational kinetic energy of 6.21×10−216.21 \times 10^{-21} J. For nitrogen (m=4.65×10−26m = 4.65 \times 10^{-26} kg), vrms≈517v_{\text{rms}} \approx 517 m/s.

67. First law of thermodynamics

ΔU=Q−W\Delta U = Q - W

Symbols: ΔU\Delta U is change in internal energy, QQ is heat added to the system, WW is work done by the system (all in J). Some textbooks define WW as work done on the system and write ΔU=Q+W\Delta U = Q + W; check which convention your course uses.

Use it when: tracking energy in gases, engines and refrigerators.

Example: A gas absorbs 500 J of heat and does 200 J of work pushing a piston, so ΔU=300\Delta U = 300 J.

68. Heat conduction

P=kA ΔTdP = \frac{kA\,\Delta T}{d}

Symbols: PP is heat flow rate (W), kk is thermal conductivity (W/(m·K)), AA is area (m²), dd is thickness (m).

Use it when: heat flows steadily through a slab such as a wall or a window.

Example: Heat through 10 m² of insulation with k=0.040k = 0.040 W/(m·K), 0.10 m thick, with a 20 K temperature difference: P=0.040×10×20/0.10=80P = 0.040 \times 10 \times 20 / 0.10 = 80 W.

69. Stefan–Boltzmann law

P=εσAT4P = \varepsilon\sigma A T^4

Symbols: ε\varepsilon is emissivity (0 to 1), AA is surface area (m²), TT is surface temperature (K).

Use it when: calculating thermal radiation. The net loss to surroundings at TsT_s is εσA(T4−Ts4)\varepsilon\sigma A(T^4 - T_s^4).

Example: Skin at 306 K, area 1.8 m², ε=0.97\varepsilon = 0.97, radiates about 870 W, but in a 293 K room it also absorbs radiation, so the net loss is about 140 W. Clothing reduces this further.

70. Carnot efficiency

ηmax⁡=1−TCTH\eta_{\max} = 1 - \frac{T_C}{T_H}

Symbols: THT_H and TCT_C are the hot and cold reservoir temperatures in kelvins.

Use it when: finding the maximum possible efficiency of any heat engine between two temperatures. Real engines do worse.

Example: An engine working between 800 K and 300 K can be at most 1−300/800=62.5%1 - 300/800 = 62.5\% efficient.

Electricity

Charge, fields, circuits and capacitors (OpenStax University Physics Volume 2, chapters 5–10).

71. Coulomb's law

F=k∣q1q2∣r2F = k\frac{|q_1 q_2|}{r^2}

Use it when: finding the force between two point charges. Like charges repel; unlike charges attract.

Example: Two 1.0 µC charges 0.10 m apart push each other with 8.99×109×10−12/0.010=0.8998.99 \times 10^9 \times 10^{-12} / 0.010 = 0.899 N.

72. Electric field of a point charge

E=Fq=k∣Q∣r2E = \frac{F}{q} = k\frac{|Q|}{r^2}

Symbols: EE in N/C (equivalently V/m).

Use it when: describing the force per unit charge around a charge.

Example: At 0.10 m from a 1.0 nC charge, E=8.99×109×10−9/0.010=899E = 8.99 \times 10^9 \times 10^{-9} / 0.010 = 899 N/C.

73. Uniform electric field

E=VdE = \frac{V}{d}

Use it when: a voltage VV is applied across parallel plates a distance dd apart.

Example: 12 V across plates 4.0 mm apart gives E=12/0.0040=3000E = 12 / 0.0040 = 3000 V/m.

74. Work done moving a charge

W=qVW = qV

Use it when: a charge moves through a potential difference. This defines the electronvolt: 1 eV = 1.602×10−191.602 \times 10^{-19} J.

Example: Moving 1 C through a 12 V battery transfers 12 J. An electron accelerated through 1 V gains 1 eV.

75. Electric current

I=QtI = \frac{Q}{t}

Symbols: II in amperes (1 A = 1 C/s).

Use it when: relating charge flow to time, such as battery capacity.

Example: A battery rated 3,000 mAh can deliver 3.0 A for one hour, a total charge of 3.0×3600=10,8003.0 \times 3600 = 10{,}800 C.

76. Ohm's law

V=IRV = IR

Use it when: a component has constant resistance (an ohmic conductor at steady temperature).

Example: 9.0 V across a 450 Ω resistor drives I=9.0/450=0.020I = 9.0 / 450 = 0.020 A, or 20 mA.

77. Resistance and resistivity

R=ρLAR = \frac{\rho L}{A}

Symbols: ρ\rho is resistivity (Ω·m; copper ≈1.68×10−8\approx 1.68 \times 10^{-8} Ω·m), LL is length (m), AA is cross-sectional area (m²).

Use it when: choosing wire sizes or explaining why long, thin wires resist more.

Example: 10 m of 1.5 mm² copper wire has R=1.68×10−8×10/(1.5×10−6)=0.112R = 1.68 \times 10^{-8} \times 10 / (1.5 \times 10^{-6}) = 0.112 Ω.

78. Electrical power

P=VI=I2R=V2RP = VI = I^2R = \frac{V^2}{R}

Use it when: finding the rate a component uses energy. Pick the form that uses the quantities you know.

Example: A 2,000 W kettle on 230 V draws I=2000/230=8.7I = 2000 / 230 = 8.7 A and has a resistance of 2302/2000=26.5230^2 / 2000 = 26.5 Ω.

79. Resistors in series

Rtotal=R1+R2+R3+⋯R_{\text{total}} = R_1 + R_2 + R_3 + \cdots

Use it when: components are connected end to end, with one path for current.

Example: 100 Ω, 220 Ω and 330 Ω in series make 650 Ω.

80. Resistors in parallel

1Rtotal=1R1+1R2+⋯\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots

Use it when: components connect across the same two points. The total is always less than the smallest resistor.

Example: 100 Ω and 220 Ω in parallel give 1/(0.01+0.004545)=68.81/(0.01 + 0.004545) = 68.8 Ω. Two equal resistors in parallel give half of one.

81. Capacitance

C=QV,Cparallel plate=ε0AdC = \frac{Q}{V}, \qquad C_{\text{parallel plate}} = \frac{\varepsilon_0 A}{d}

Symbols: CC in farads (F), AA is plate area (m²), dd is separation (m), for plates separated by vacuum or air.

Use it when: relating stored charge to voltage.

Example: Two 0.010 m² plates 1.0 mm apart in air have C=8.85×10−12×0.010/0.0010=88.5C = 8.85 \times 10^{-12} \times 0.010 / 0.0010 = 88.5 pF.

82. Energy stored in a capacitor

U=12CV2U = \tfrac{1}{2}CV^2

Use it when: finding how much energy a charged capacitor holds.

Example: A 100 µF capacitor charged to 300 V, as in a camera flash, stores 12(100×10−6)(3002)=4.5\tfrac{1}{2}(100 \times 10^{-6})(300^2) = 4.5 J.

83. RC time constant

τ=RC,VC(t)=V0(1−e−t/τ)\tau = RC, \qquad V_C(t) = V_0\left(1 - e^{-t/\tau}\right)

Use it when: a capacitor charges or discharges through a resistor.

Example: With R=10R = 10 kΩ and C=100C = 100 µF, τ=1.0\tau = 1.0 s. After 1 s the capacitor reaches 63.2% of the supply voltage, and after 5τ5\tau it is over 99% charged.

84. Electrical energy

E=PtE = Pt

Use it when: working out energy use and electricity bills. With PP in kW and tt in hours, EE is in kWh (1 kWh = 3.6×1063.6 \times 10^6 J).

Example: A 2.0 kW kettle running for 3 minutes (0.05 h) uses 0.10 kWh. Multiply by your tariff per kWh to get the cost.

Magnetism and induction

Magnetic forces, fields from currents, and Faraday's law (OpenStax University Physics Volume 2, chapters 11–14).

85. Magnetic force on a moving charge

F=qvBsin⁡θF = qvB\sin\theta

Symbols: BB is magnetic field (tesla, T), θ\theta is the angle between vv and BB.

Use it when: charges move through a magnetic field, as in mass spectrometers and particle accelerators.

Example: A proton at 1.0×1061.0 \times 10^6 m/s moving perpendicular to a 0.50 T field feels 1.602×10−19×106×0.50=8.0×10−141.602 \times 10^{-19} \times 10^6 \times 0.50 = 8.0 \times 10^{-14} N.

86. Magnetic force on a current-carrying wire

F=BILsin⁡θF = BIL\sin\theta

Use it when: designing motors and loudspeakers, or predicting forces on wires in a field.

Example: A 0.20 m wire carrying 5.0 A at right angles to a 0.10 T field feels 0.10×5.0×0.20=0.100.10 \times 5.0 \times 0.20 = 0.10 N.

87. Magnetic field of a long straight wire

B=μ0I2πrB = \frac{\mu_0 I}{2\pi r}

Use it when: finding the field at distance rr from a long straight current.

Example: 10 A at 5.0 cm gives B=4.0×10−5B = 4.0 \times 10^{-5} T, or 40 µT, similar to Earth's field. That is why a compass near a heavy DC cable is unreliable.

88. Magnetic field inside a solenoid

B=μ0nIB = \mu_0 n I

Symbols: nn is turns per metre of length.

Use it when: a long coil produces a nearly uniform field inside, as in electromagnets and MRI magnets.

Example: 1,000 turns/m carrying 2.0 A gives B=1.257×10−6×1000×2.0=2.5B = 1.257 \times 10^{-6} \times 1000 \times 2.0 = 2.5 mT.

89. Magnetic flux

Φ=BAcos⁡θ\Phi = BA\cos\theta

Symbols: Φ\Phi in webers (Wb), θ\theta is the angle between BB and the normal to the area.

Use it when: setting up induction problems.

Example: A 0.20 T field through a 0.010 m² loop tilted at 60° gives Φ=0.20×0.010×0.5=1.0×10−3\Phi = 0.20 \times 0.010 \times 0.5 = 1.0 \times 10^{-3} Wb.

90. Faraday's law of induction

ε=−NΔΦΔt\varepsilon = -N\frac{\Delta\Phi}{\Delta t}

Symbols: ε\varepsilon is induced voltage (V), NN is number of turns. The minus sign is Lenz's law: the induced current opposes the change.

Use it when: a changing magnetic flux through a coil generates a voltage, as in generators, transformers and induction hobs.

Example: A 200-turn coil whose flux changes by 1.0×10−31.0 \times 10^{-3} Wb in 0.050 s produces 200×0.001/0.050=4.0200 \times 0.001 / 0.050 = 4.0 V.

91. Ideal transformer

VsVp=NsNp\frac{V_s}{V_p} = \frac{N_s}{N_p}

Use it when: stepping AC voltage up or down. In an ideal transformer, power in equals power out, so current changes the opposite way.

Example: A transformer with 2,000 primary turns and 100 secondary turns steps 230 V down to 230×100/2000=11.5230 \times 100 / 2000 = 11.5 V.

Optics

Refraction, lenses and interference (OpenStax University Physics Volume 3, chapters 1–3).

92. Snell's law and refractive index

n1sin⁡θ1=n2sin⁡θ2,n=cvn_1\sin\theta_1 = n_2\sin\theta_2, \qquad n = \frac{c}{v}

Symbols: nn is refractive index (air ≈1.00\approx 1.00, water ≈1.33\approx 1.33, glass ≈1.5\approx 1.5), angles are measured from the normal.

Use it when: light crosses a boundary between materials.

Example: Light entering water from air at 30° bends to θ2=arcsin⁡(0.5/1.33)=22.1°\theta_2 = \arcsin(0.5 / 1.33) = 22.1°.

93. Critical angle

sin⁡θc=n2n1,n1>n2\sin\theta_c = \frac{n_2}{n_1}, \qquad n_1 > n_2

Use it when: light travels from a denser to a less dense medium. Beyond θc\theta_c, all the light reflects (total internal reflection).

Example: For glass (n=1.5n = 1.5) to air, θc=41.8°\theta_c = 41.8°; for water to air, 48.8°48.8°.

94. Thin lens equation and magnification

1f=1do+1di,m=−dido,P=1f\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}, \qquad m = -\frac{d_i}{d_o}, \qquad P = \frac{1}{f}

Symbols: ff is focal length, dod_o object distance, did_i image distance, mm magnification. Lens power PP is in dioptres when ff is in metres. Sign conventions vary between textbooks; this uses the common one where real images have positive did_i.

Use it when: finding where a lens forms an image and how big it is.

Example: An object 30 cm from a 10 cm converging lens forms a real image at di=15d_i = 15 cm with m=−0.5m = -0.5: inverted and half size. A lens with f=0.10f = 0.10 m has a power of +10 D; a −2.5 D spectacle lens has f=−0.40f = -0.40 m.

95. Double-slit interference

dsin⁡θ=mλ,Δy≈λLdd\sin\theta = m\lambda, \qquad \Delta y \approx \frac{\lambda L}{d}

Symbols: dd is slit separation, mm is the order of the bright fringe, LL is distance to the screen, Δy\Delta y is fringe spacing (small-angle approximation).

Use it when: measuring wavelengths with slits or diffraction gratings.

Example: Red laser light (650 nm) through slits 0.10 mm apart onto a screen 2.0 m away gives fringes 650×10−9×2.0/10−4=13650 \times 10^{-9} \times 2.0 / 10^{-4} = 13 mm apart.

Modern physics

Photons, matter waves, mass-energy and radioactive decay (OpenStax University Physics Volume 3, chapters 5–10).

96. Photon energy

E=hf=hcλE = hf = \frac{hc}{\lambda}

Use it when: relating light's frequency or wavelength to the energy of each photon.

Example: Green light at 500 nm carries E=hc/λ=3.97×10−19E = hc/\lambda = 3.97 \times 10^{-19} J, or 2.48 eV, per photon. A handy shortcut: E (eV)≈1240/λ (nm)E\,(\text{eV}) \approx 1240 / \lambda\,(\text{nm}).

97. Photoelectric equation

KEmax⁡=hf−ϕKE_{\max} = hf - \phi

Symbols: ϕ\phi is the work function of the metal (J or eV), the minimum energy to release an electron.

Use it when: light ejects electrons from a surface. If hf<ϕhf < \phi, no electrons are emitted however bright the light.

Example: 400 nm light (3.10 eV photons) on a metal with ϕ=2.3\phi = 2.3 eV ejects electrons with up to 0.80 eV of kinetic energy.

98. De Broglie wavelength

λ=hp=hmv\lambda = \frac{h}{p} = \frac{h}{mv}

Use it when: finding the wavelength of a moving particle (for speeds well below cc).

Example: An electron at 1.0×1061.0 \times 10^6 m/s has λ=6.626×10−34/(9.109×10−31×106)=0.73\lambda = 6.626 \times 10^{-34} / (9.109 \times 10^{-31} \times 10^6) = 0.73 nm, comparable to the spacing of atoms in a crystal, which is why electrons diffract off crystals.

99. Mass–energy equivalence

E=mc2E = mc^2

Use it when: relating rest mass to energy, for example in nuclear reactions, where the products have slightly less mass than the reactants.

Example: Converting 1.0 g of mass entirely to energy would release 10−3×(3.0×108)2=9.0×101310^{-3} \times (3.0 \times 10^8)^2 = 9.0 \times 10^{13} J, about 25 GWh.

100. Radioactive decay and half-life

N=N0(12)t/T1/2=N0e−λt,λ=ln⁡2T1/2N = N_0\left(\tfrac{1}{2}\right)^{t/T_{1/2}} = N_0 e^{-\lambda t}, \qquad \lambda = \frac{\ln 2}{T_{1/2}}

Symbols: N0N_0 is the starting number of nuclei, T1/2T_{1/2} the half-life, λ\lambda the decay constant.

Use it when: predicting how much of a radioactive sample remains, or dating a sample from what remains.

Example: Carbon-14 has a half-life of about 5,730 years. After 11,460 years (two half-lives), 25% remains. Its decay constant is ln⁡2/5730=1.21×10−4\ln 2 / 5730 = 1.21 \times 10^{-4} per year.

How to use physics formulas well

Knowing a formula is only half the job. These habits prevent most lost marks:

  • Convert to SI units first. Kilometres per hour, grams, centimetres and minutes cause more wrong answers than hard physics. Convert to metres, kilograms and seconds before substituting.
  • Check the conditions. The kinematics equations need constant acceleration. The projectile range formula needs equal launch and landing heights. The pendulum formula needs small angles. Ohm's law needs constant resistance.
  • Use kelvins for gases and radiation. PV=nRTPV = nRT, P=εσAT4P = \varepsilon\sigma AT^4 and Carnot efficiency are wrong with Celsius values.
  • Use radians for rotation. v=ωrv = \omega r and at=αra_t = \alpha r only work with angles in radians.
  • Check units at the end. If a speed comes out in kg·m, something went wrong. Dimensional analysis catches many algebra slips.
  • Estimate the size. A car's braking distance should be metres, not kilometres. A household current should be a few amps, not thousands.
  • Watch sign conventions. The first law of thermodynamics, the thin lens equation and the Doppler formula all have competing conventions. Use the one your course uses, consistently.

Next steps

References