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The amount of electricity required to run a 1 W device for 1 s. The energy required to accelerate a 1 kg mass at 1 m/s 2 through a distance of 1 m. The kinetic energy of a 2 kg mass travelling at 1 m/s, or a 1 kg mass travelling at 1.41 m/s. The energy required to lift an apple up 1 m, assuming the apple has a mass of 101.97 g.
Continuous charge distribution. The volume charge density ρ is the amount of charge per unit volume (cube), surface charge density σ is amount per unit surface area (circle) with outward unit normal nĚ‚, d is the dipole moment between two point charges, the volume density of these is the polarization density P.
A unit of electrical energy, particularly for utility bills, is the kilowatt-hour (kWh); [3] one kilowatt-hour is equivalent to 3.6 megajoules. Electricity usage is often given in units of kilowatt-hours per year or other periods. [4] This is a measurement of average power consumption, meaning the average rate at which energy is transferred ...
The electron was discovered in 1897 by J. J. Thomson, and it was quickly realized that it was the particle (charge carrier) that carried electric currents in electric circuits. In 1900, the first ( classical ) model of electrical conduction, the Drude model , was proposed by Paul Drude , which finally gave a scientific explanation for Ohm's law.
The SI unit of electric potential energy is joule (named after the English physicist James Prescott Joule). In the CGS system the erg is the unit of energy, being equal to 10 −7 Joules. Also electronvolts may be used, 1 eV = 1.602×10 −19 Joules.
Radiant energy density: w e: joule per cubic metre J/m 3: M⋅L −1 ⋅T −2: Radiant energy per unit volume. Radiant flux: Φ e [nb 2] watt: W = J/s M⋅L 2 ⋅T −3: Radiant energy emitted, reflected, transmitted or received, per unit time. This is sometimes also called "radiant power", and called luminosity in Astronomy. Spectral flux: Φ ...
Jefimenko equations; Larmor formula; ... such as kinetic energy, into heat. This Joule heating reduces efficiency of ... k is a constant equal to 1 for a thin sheet ...
Power output = energy / time 1 terawatt hour per year = 1 × 10 12 W·h / (365 days × 24 hours per day) ≈ 114 million watts, equivalent to approximately 114 megawatts of constant power output. The watt-second is a unit of energy, equal to the joule. One kilowatt hour is 3,600,000 watt seconds.