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This equation states that the kinetic energy (E k) is equal to the integral of the dot product of the momentum (p) of a body and the infinitesimal change of the velocity (v) of the body. It is assumed that the body starts with no kinetic energy when it is at rest (motionless).
The solution of this differential equation is useful in calculating the concentration after the administration of a single dose of drug via IV bolus injection: = C t is concentration after time t; C 0 is the initial concentration (t=0) K is the elimination rate constant
As a body accelerates or decelerates, its massic kinetic energy changes accordingly, reflecting alterations in its state of motion. Knowledge of massic kinetic energy is fundamental in fields such as physics, engineering, and transportation. For example, in aerodynamics, it is crucial for analyzing the motion of aircraft and projectiles.
The change in kinetic energy is hence: = = + | | where μ is the reduced mass and u rel is the relative velocity of the bodies before collision. With time reversed we have the situation of two objects pushed away from each other, e.g. shooting a projectile , or a rocket applying thrust (compare the derivation of the Tsiolkovsky rocket equation ).
For example, in the kinetic energy (KE) formula, if g c = 1 is used, then KE is expressed in foot-poundals; but if g c = 32.174 is used, then KE is expressed in foot-pounds. Motivations [ edit ]
This equation holds for a body or system, such as one or more particles, with total energy E, invariant mass m 0, and momentum of magnitude p; the constant c is the speed of light. It assumes the special relativity case of flat spacetime [ 1 ] [ 2 ] [ 3 ] and that the particles are free.
The exact k-ε equations contain many unknown and unmeasurable terms. For a much more practical approach, the standard k-ε turbulence model (Launder and Spalding, 1974 [3]) is used which is based on our best understanding of the relevant processes, thus minimizing unknowns and presenting a set of equations which can be applied to a large number of turbulent applications.
Then the time-rate of change of the specific energy of the rocket is : an amount () for the kinetic energy and an amount for the potential energy. The change of the specific energy of the rocket per unit change of delta-v is v ⋅ a | a | {\displaystyle {\frac {\mathbf {v\cdot a} }{|\mathbf {a} |}}} which is | v | times the cosine of the angle ...