Formula sheet: Mass and Energy II Student’s name: Topic 1: Fluids at rest Density π= Specific weight π π πΎ= Density of fresh water π πΎπ ππ»2 0 = 1 = 1000 3 3 ππ π Hydrostatic Pressure (Manometric Pressure) ππ = ππβ Pascal’s principle π1 = π2 πΉ1 πΉ2 = π΄1 π΄2 Mechanical Pressure π π π= πΉ π΄ πΎ = ππ Atmosferic Pressure Absolute Pressure πππ‘π = 1 ππ‘π = 760 πππ»π πππ‘π = 101 325 ππ π = ππ + πππ‘π Floating force (Buoyant force) π΅ = ππ (ππ - Displaced fluid weight) Tension Force (Resulting force) πΉ = π΅ − π0 (π0 - Weight of the object) π΅ = πππ Aparent weight Height of a liquid inside a capillary tube 2π β= πππ π΄π€ = π0 − π΅ Topic 2: Fluids dynamic Volume Flow rate π= π π‘ π = π΄π£ Continuity equation π1 = π2 π΄1 π£1 = π΄2 π£2 Bernoulli’s Theorem Torricelli’s Theorem 1 1 π1 + ππβ1 + ππ£1 2 = π2 + ππβ2 + ππ£2 2 2 2 π£ = √2πβ Tema 3: Vibración Frecuency Period π= π π‘ π= π‘ π π= 1 π π= 1 π Angular frequency π= Angular displacement π π‘ π = ππ‘ π = 2ππ π= 2π π‘ Position (Linear displacement) Velocity π£ = −π΄π β π ππ(ππ‘) π₯ = π΄ β πππ (ππ‘) Acceleration Maximum speed πΌ = −π΄π2 β πππ (ππ‘) π£πππ₯ = π΄π π£πππ₯ = 2ππ΄π Frecuency (In function of π₯ and π) π= 1 π √ 2π π₯ π₯ π π = 2π√ Mass-spring system Frecuency Period Restoring force (Hooke’s Law) Acceleration (In function of π and π) πΉ = −ππ₯ π= Maximum speed π£ππππ₯ Period (In function of π₯ and π) 1 π √ 2π π π π π = 2π√ Elastic Potential Energy 1 π = ππ₯ 2 2 ππ₯0 2 =√ π π= Law of Conservation of Energy (In relation with the elastic forces) 1 1 1 1 ππ₯0 2 + ππ£0 2 = ππ₯π 2 + ππ£π 2 2 2 2 2 Simple Pendulum Frecuency π= 1 π √ 2π πΏ ππ₯ π Period πΏ π = 2π√ π Topic 4: Waves Propagation speed of a wave π£ = ππ Linear density of a string π= π πΏ π π£= π Speed of a wave on a string under tension πΉ π£=√ π πΉπΏ π£=√ π Wave lenght of the π harmonic ππ = 2πΏ π Harmonics on a string Resonant Frecuencies for the π harmonic ππ = ππ£ 2πΏ π = 1, 2, 3, … Resonant Frecuencies for the π harmonic in function of π ππ = π πΉ √ 2πΏ π π = 1, 2, 3, … π = 1, 2, 3, … Energy in a wave: Intensity πΌ = 2π 2 π£ππ 2 π΄2 Topic 5: Sound Sound speed through the air at a specific temperature π£ = (331 Sound speed through a liquid Sound speed through a solid (Bar) π΅ π£=√ π π π )√ π 273°πΎ π π£=√ π Absolute temperature π = °πΆ + 273.15 Sound speed through an extended solid (Body) Sound speed through a gas Sound speed through an ideal gas πΎπ π£=√ π 4 π΅ + 3π √ π£= π πΎπ π π£=√ π Harmonic frecuencies through open pipes Wave lenght of the π harmonic Frecuencies of the π harmonic ππ = 2πΏ π ππ = ππ£ 2πΏ π = 1, 2, 3, … π = 1, 2, 3, … Harmonic frecuencies through closed pipes Wave lenght of the π harmonic Frecuencies of the π harmonic ππ = 4πΏ 2π + 1 ππ = π = 1, 2, 3, … Sound intensity πΌ= π π΄ πΉππ π = 1, π = 0 πΉππ π = 2, π = 1 πΉππ π = 3, π = 2 Sound level (dB) πΌ π½ = 10πππ ( ) πΌ0 πΌ0 = 1 × 10−12 (2π + 1) π£ 4 πΏ π π2 Doppler effect (Approaching) ππΏ = ππ ( π£ + π£πΏ ) π£ − π£π π£ = 343 π π Doppler effect (Moving away) ππΏ = ππ ( π£ − π£πΏ ) π£ + π£π π£ = 343 π π Topic 6: The charge and its interactions Coulomb’s Law (Electrostatic force) πΉ=πΎ |π1 π2 | π2 Electrostatic constant πΎ = 9 × 109 ππ2 πΆ2 Scientific notation 1 ππΆ = 1 × 10−6 πΆ 1 ππΆ = 1 × 10−9 πΆ 1 πΎπ = 1 × 103 π Topic 7: Electric field Electric field πΉ πΈ= π πΈ=πΎ One electron charge π = 1.6 × 10−19 πΆ |π| π2 Topic 8: Gauss’ Law Electric flow Gauss’ Law ΦπΈ = πΈπ΄πππ π ΦπΈ = ππππ‘ π0 π0 = 8.854 × 10−12 Linear charge density π= π πΏ Surface charge density π= π π΄ πΆ2 ππ Volumetric charge density π= π π Electric field for different charge distributions Charge distribution Electric field point Electric field magnitude 1 π Single point charge Distance π from π πΈ=( ) 2 4ππ0 π 1 π Outside the sphere, π > π πΈ=( ) Charge π in a conductor sphere 4ππ0 π 2 surface of radius π Inside the sphere, π < π πΈ=0 1 π Infinite wire, charge per unit Distance π from the wire πΈ=( ) lenght π 2ππ0 π 1 π Infinite conductor cylinder with Outside the cylinder, π > π πΈ=( ) radius π , charge per unit lenght 2ππ0 π π Inside the cylinder, π < π πΈ=0 1 π Solid insulated sphere with Outside the sphere, π > π πΈ=( ) 2 radius π , with charge π 4ππ0 π 1 ππ distributed uniformly through Inside the sphere, π < π πΈ=( ) all volume 4ππ0 π 3 π Infinite charged plate with πΈ= Any point 2π0 uniform charge per unit area π Two conductor plates with π πΈ= opposite charges uniformly At any point between plates π0 distributed +π and −π Topic 9: Electric potential energy Electric Potential Energy Electric Work π = π0 πΈπ π = ππΈπ Electric Potential Energy due to two-point charges π=πΎ π1 π2 π Electric Potential Energy in a system with three-point charges π = πΎ[ π1 π2 π1 π3 π2 π3 + + ] π12 π13 π23 Electric Potential π= Electric Potential in the point B π π ππ΅ = ππ΄π΅ π Topic 10: Electric Potential Electric Potential π=πΎ π π π Δπ = ππ΅ − ππ΄ ππ π1π π2π ππ = πΎ ∑ = πΎ +πΎ +β― ππ π1π π2π π=1 Potential Difference ππ΄π΅ = ππ΅ − ππ΄ π0 Potential difference between two charged plates with opposite charges and the same magnitude ππ΅ − ππ΄ = πΈπ Δπ = ππ΅ − ππ΄ π0 Conservation of Energy Law in Electrostatics (Applied to the movement of one charge between two parallel charged plates) π1 + πΎ1 + π = π2 + πΎ2 Electron-Volt 1 ππ = 1.6 × 10−19 π½ Topic 11: Electric current 1 Electric Current πΌ= Current Density π π‘ π½= πΌ π΄ Electric Current (Drift velocity) πΌ = πππ£π π΄ π½ = πππ£π Electric Resistance π =π Temperature Coefficient πΏ π΄ πΌ= πΌ= Δπ Ohm’s Law π = πΌπ π ππππ‘πππ β Δπ π πππππ − π ππππ‘πππ π ππππ‘πππ β [ππππππ − πππππ‘πππ ] Electric Power π = ππΌ π= π2 π Ohm´s Law and Electric Power π = πΌ2 π Topic 12: Electric Current 2 Series connection π ππ = π 1 + π 2 + β― + π π πΌπ = πΌ1 = πΌ2 = β― = πΌπ ππ΅ππ‘πππ¦ = π1 + π2 + β― + ππ Parallel connection π ππ = [ 1 1 1 −1 + + β―+ ] π 1 π 2 π π πΌπ = πΌ1 + πΌ2 + β― + πΌπ ππ΅ππ‘πππ¦ = π1 = π2 = β― = ππ Kirchhoff’s Laws Currents Law Voltages Law ∑ πΌπΈππ‘πππππ = ∑ πΌπΏπππ£πππ ∑π =0 Topic 13: Introduction to Magnetism Magnetic Force Magnetic force on a moving charge Magnetic force on a conductor carrying a current ββ πΉβ = ππ£β × π΅ ββ πΉβ = πΌβπΏ × π΅ πΉ = ππ£π΅π πππ πΉ = πΌπΏπ΅π πππ Topic 14: Magnetism Current distribution Long and straight conductor Point in the magnetic field Distance π from the conductor At the center of the coil Circular coil of radius π On the coil axis Inside the conductor, π < π Long cylindric conductor of radius π Outside the conductor, π > π Magnitude of magnetic field π0 πΌ π΅= 2ππ π0 πΌ π΅= 2π π0 πΌπ2 π΅= 3 2(π₯ 2 + π2 )2 π0 πΌ π π΅= 2π π 2 π0 πΌ π΅= 2ππ π0 ππΌ π΅= πΏ π΅≈0 Inside the solenoid, near the center Outside the solenoid Inside the space enclosed by the π0 ππΌ coil, at a distance π of the axis of π΅= Toroidal solenoid (toroid) with 2ππ symmetry compact coil and π turns Outside the space enclosed by π΅=0 the coil Permeability Constant in vacuum Absolut Permeability Long solenoid, with compact coil and π turns per lenght πΏ π0 = 4π × 10−7 π π π΄ π0 = 1.2567 × 10−6 π π = ππ π0 π π΄ Topic 15: Introduction to Modern Physiscs Lorentz Transformations π₯ ′ = πΎ(π₯ − π£π‘) Gamma Factor πΎ= π¦′ = π¦ 1 Speed of light π = 3 × 108 2 √1 − π£ 2 π π π π§′ = π§ π‘ ′ = πΎ (π‘ − π£π₯ ) π2 Time Dilation Lenght Contraction Δπ = πΎΔπ0 Δπ = ΔπΏ = Δπ0 πΈ = ππ 2 ΔπΏ0 πΎ 2 √1 − π£2 π Mass-energy Relation ΔπΏ = ΔπΏ0 √1 − π£2 π2 Electromagnetic Energy Planck’s constant πΈ = βπ β = 6.63 × 10−34 π½π DeBroglie’s equation (Wave lenght) Frecuency π= β ππ£ π= π π Supporting formulas Theorem of Pitagoras Trigonometric functions π πππ = Μ Μ Μ Μ ππππ ππ‘ πΏππ π΅πΆ = π»π¦πππ‘πππ’π π Μ Μ Μ Μ π΄π΅ πππ π = π΄πππππππ‘ πΏππ Μ Μ Μ Μ π΄πΆ = Μ Μ Μ Μ π»π¦πππ‘πππ’π π π΄π΅ π‘πππ = Μ Μ Μ Μ ππππ ππ‘ πΏππ π΅πΆ = Μ Μ Μ Μ π΄πππππππ‘ πΏππ π΄πΆ Areas Circle Volumes Cube π΄ = ππ 2 1 π΄ = ππ· 2 4 Square π΄ = π2 Rectangle π = π3 Prism/Cylinder π = π΄πππ π β Sphere 4 π = ππ 3 3 π΄ = πβ Venturi tube π1 − π2 = ππβ 1 1 π1 − π2 = ππ£2 2 − ππ£1 2 2 2 2(π1 − π2 ) π£1 = π΄2 √ π(π΄1 2 − π΄2 2 )
