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Kinetic energy is the movement energy of an object. Kinetic energy can be transferred between objects and transformed into other kinds of energy. [10] Kinetic energy may be best understood by examples that demonstrate how it is transformed to and from other forms of energy.
In a set of curvilinear coordinates ξ = (ξ 1, ξ 2, ξ 3), the law in tensor index notation is the "Lagrangian form" [18] [19] = (+) = (˙), ˙, where F a is the a-th contravariant component of the resultant force acting on the particle, Γ a bc are the Christoffel symbols of the second kind, = is the kinetic energy of the particle, and g bc ...
Since the speed v is likewise unchanging, the areal velocity 1 ⁄ 2 vr ⊥ is a constant of motion; the particle sweeps out equal areas in equal times. The area A of a circular sector equals 1 ⁄ 2 r 2 φ = 1 ⁄ 2 r 2 ωt = 1 ⁄ 2 r v φ t. Hence, the areal velocity dA/dt equals 1 ⁄ 2 r v φ = 1 ⁄ 2 h.
The Euler equations can be generalized to any simple Lie algebra. [1] The original Euler equations come from fixing the Lie algebra to be s o ( 3 ) {\displaystyle {\mathfrak {so}}(3)} , with generators t 1 , t 2 , t 3 {\displaystyle {t_{1},t_{2},t_{3}}} satisfying the relation [ t a , t b ] = ϵ a b c t c {\displaystyle [t_{a},t_{b}]=\epsilon ...
In the center of mass frame the kinetic energy is the lowest and the total energy becomes = ˙ + The coordinates x 1 and x 2 can be expressed as = = and in a similar way the energy E is related to the energies E 1 and E 2 that separately contain the kinetic energy of each body: = = ˙ + = = ˙ + = +
The sum over r covers other degrees of freedom specific for the field, such as polarization or spin; it usually comes out as a sum from 1 to 2 or from 1 to 3. E p is the relativistic energy for a momentum p quantum of the field, = m 2 c 4 + c 2 p 2 {\textstyle ={\sqrt {m^{2}c^{4}+c^{2}\mathbf {p} ^{2}}}} when the rest mass is m .
Turbulence kinetic energy is then transferred down the turbulence energy cascade, and is dissipated by viscous forces at the Kolmogorov scale. This process of production, transport and dissipation can be expressed as: D k D t + ∇ ⋅ T ′ = P − ε , {\displaystyle {\frac {Dk}{Dt}}+\nabla \cdot T'=P-\varepsilon ,} where: [ 1 ]
Thus, the ratio of the kinetic energy to the absolute temperature of an ideal monatomic gas can be calculated easily: per mole: 12.47 J/K; per molecule: 20.7 yJ/K = 129 μeV/K; At standard temperature (273.15 K), the kinetic energy can also be obtained: per mole: 3406 J; per molecule: 5.65 zJ = 35.2 meV.