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Pendulum (**mathematics**) - Wikipedia

Contents · **1** Simple **gravity** pendulum · **2** Small-angle approximation. 2.1 Rule of
thumb **for** pendulum length · 3 Arbitrary-amplitude period · 4 Approximate
formulae ...

For more information, see Pendulum (**mathematics**) - Wikipedia

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Why does the formula **for** kinetic energy have a half in it? - Quora

Newton's second law says **that** force equals **mass** times **acceleration**, [**math**]F =
ma[/**math**]. ... What will the value of kinetic energy of moving **body** of **mass m** and
**its speed** is ... **Consider** the work done as an object moves from momentum zero
to **MV due** ... Why is the **equation for** kinetic energy **1/2mv**^**2** when the **equation**
**for** ...

For more information, see Why does the formula **for** kinetic energy have a half in it? - Quora

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**Units** and **Dimensions** - **Dimensional Analysis**, **Formula**, Applications

**Units** and **dimensions** - Understand **Dimensional analysis** with Limitations and
Applications. Know **Dimensional Formulas** of Quantities and Quantities with
Same ... **Dimensional analysis** is the practice of **checking** relations between
physical ... For example, I can compare **kinetic energy** with **potential energy** and
say they ...

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.

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Given that

\frac{1}{2} mv^2 = mgh

Step 1: Set up a dimensional formula for kinetic energy

K.E = \frac{1}{2} mv^2 where, m - mass and v - velocity

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

[math]=\ [ML^2T^{-2}][/math] ....................(1)

Step 1: Set up a dimensional formula for potential energy

P.E = mgh , where m - mass, g - acceleration due to gravity ( 9.8 m/s^2 ).

Dimensional formula [math] = [M] [LT^{-2}] [L] [/math]

[math] = [M L^2 T^{-2}] [/math] ...................(2)

From equation (1) and (2)

\frac{1}{2} mv^2 = mgh

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

L.H.S = R.H.S

Since the dimensions of the LHS and RHS are the same, the equation is dimensionally correct.