I found an answer from www.britannica.com

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

E = mc^**2**, **equation** in Einstein's theory of special relativity that expresses the ... E
= mc^{2}, **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 (c^{2}) is equal to the **kinetic energy** (E) of that body. E =
mc ...

For more information, see E = mc² | **Equation**, Explanation, & Proof | Britannica

I found an answer from en.wikipedia.org

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 (**kg**
⋅**m**/**s**). ... **1 Mathematical** derivation in the case of an object of constant mass; **2** ...

For more information, see Impulse (physics) - Wikipedia

I found an answer from www.quora.com

Why does [**math**] **kg** * (**m**^**2** / **s**^**2**) **[/math**] 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 ...

For more information, see Why does [**math**] **kg** * (**m**^**2** / **s**^**2**) **[/math**] equal **Joules**? - Quora

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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
...

For more information, see **Units** and **Dimensions** - **Dimensional Analysis**, **Formula**, Applications

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 [math] K = [ML^2T^{-2}][/math]

Step 1: Identify the dimensionally correct and incorrect equations

a) K = m^2 v^3

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

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

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

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

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

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

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

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

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

c) K = ma

[math] K = [M] [LT^{-2}] [/math]

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

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

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

d) [math] K = \frac{3}{16} mv^2

[math] K = \frac{3}{16}[M] [LT^{-1}]^2 [/math]

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

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

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

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

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

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

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)