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E = mc² | Equation, Explanation, & Proof | Britannica

E = mc^2, equation in Einstein's theory of special relativity that expresses the ... E = mc2, equation in German-born physicist Albert Einstein's theory of special relativity that ... In the equation, the increased relativistic mass (m) of a body times the speed of light squared (c2) is equal to the kinetic energy (E) of that body. E = mc ...

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Impulse (physics) - Wikipedia

In classical mechanics, impulse is the integral of a force, F, over the time interval, t, for which it ... The SI unit of impulse is the newton second (N⋅s), and the dimensionally equivalent unit of momentum is the kilogram meter per second (kgm/s). ... 1 Mathematical derivation in the case of an object of constant mass; 2 ...

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Why does $kg * (m^2 / s^2)$ equal Joules? - Quora

F = m . a Then, the units (using SI): N = kg . m/s² Energy, work = N . m Then, energy = kg , m/s² . m = kg . m² / s². ... Part of the Work physics formulas: mass = kg, distance = m, time = s, velocity ... Also we know, kinetic energy is equal to 1/2( m*v^2). ... Richard J Breen, BS physics U ND '631/2 Ms Ed LI U 70 tested Nukes" in ...

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Units and Dimensions - Dimensional Analysis, Formula, Applications

Know Dimensional Formulas of Quantities and Quantities with Same ... The units that are used to measure these fundamental quantities are called ... practice of checking relations between physical quantities by identifying the dimensions of ... In any correct equation representing the relation between physical quantities, the  ...

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According to the Principle of Homogeneity, the dimensions of each term of a dimensional equation on both sides should be the same. This theory allows us to convert the units from one form to another.

Given that SI units

Energy J = kgm^2s^{-2}

Speed v = ms^{-1}

Acceleration a = ms^{-2}

Dimensional formula for kinetic energy $K = [ML^2T^{-2}]$

Step 1: Identify the dimensionally correct and incorrect equations

a) K = m^2 v^3

$K = [M^2] [LT^{-1}]^3$

$K = [M^2 L^3 T^{-3}]$

$[ML^2T^{-2}] \ne [M^2 L^3 T^{-3}]$

Hence, L.H.S \neq R.H.S , dimensionally incorrect equation

b) K = \frac{1}{2} mv^2

$K = \frac{1}{2} [M] [LT^{-1}]^2$

$K = [M L^2 T^{-2}]$

$[ML^2T^{-2}] = [M L^2 T^{-2}]$

Hence, L.H.S = R.H.S , dimensionally correct equation

c) K = ma

$K = [M] [LT^{-2}]$

$K = [M L T^{-2}]$

$[ML^2T^{-2}] \neq [M L T^{-2}]$

Hence, L.H.S \neq R.H.S , dimensionally incorrect equation

d) $K = \frac{3}{16} mv^2 [math] K = \frac{3}{16}[M] [LT^{-1}]^2$

$K = [M L^2 T^{-2}]$

$[ML^2T^{-2}] = [M L^2 T^{-2}]$

Hence, L.H.S = R.H.S , dimensionally correct equation

e) K = \frac{1}{2} mv^2 + ma

$K = [M] [LT^{-1}]^2 + [M] [LT^{-2}]$

Since two substances of various dimensions have been added there are no proper dimensions.    $[ML^2T^{-2}] \neq [M] [LT^{-1}]^2 + [M] [LT^{-2}]$

Hence, L.H.S \neq R.H.S , dimensionally incorrect equation

The formulas (a), (c), and (e) are ruled out. Dimensional arguments cannot say which of the two formulas, (b) or (d), is right. For that, the actual definition of kinetic energy must be explained. The right kinetic energy formula is (b)