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What happens to the units of a squared variable?

For example, if I squared velocity, would the units, metres per second (${\rm m}/{\rm s}$), change as well?

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  • $\begingroup$ Same for cubed and any other power. If you have fractional power then it get kind of weird but I guess it still holds. $\endgroup$ – paparazzo Aug 26 '15 at 17:53
  • $\begingroup$ @Frisbee I wouldn't call that weird; if squaring squares the units, you'd better be able to reverse the process by taking the square root. Otherwise your units lose all meaning! $\endgroup$ – Kyle Oman Aug 26 '15 at 18:18
  • $\begingroup$ @KyleOman by kind of I meant like 1.38 but I guess it would still hold $\endgroup$ – paparazzo Aug 26 '15 at 18:25
  • $\begingroup$ @Frisbee any rational power is conceptually no different from any other. I would agree that things get kind of weird for irrational powers. I can't imagine what ${\rm m}^\pi$ could possibly mean physically. $\endgroup$ – Kyle Oman Aug 26 '15 at 18:27
  • $\begingroup$ @KyleOman Strange and even irrational units often appear in systems exhibiting fractal behaviors. $\endgroup$ – Cort Ammon Aug 26 '15 at 20:21
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Yes. If you square a variable, its unit of measurement is also squared, in the case of speed $v$ in $m/s$ ($ms^{-1}$), then $v^2$ is expressed in $m^2s^{-2}$. This is true for all physical variables (or constants).

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Yes. Consider the equation for kinetic energy (KE):

$${\rm KE} = \frac{1}{2} mv^{2}$$

the dimensions of KE are:

$${\rm mass} \times {\rm velocity}^{2}=\frac{{\rm mass} \times {\rm length}^{2}}{{\rm time}^{2}}$$

or with SI units:

$$1\,{\rm J} = 1\,{\rm kg}\,{\rm m}^{2}\,{\rm s}^{-2}$$

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    $\begingroup$ Cleaned up your mathjax quite a bit (have a look at 'edit' if you want to see the source). Hope you agree that it looks a bit better. $\endgroup$ – Kyle Oman Aug 26 '15 at 18:22
  • $\begingroup$ @KyleOman Nice I am new at Latex $\endgroup$ – paparazzo Aug 26 '15 at 18:24
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    $\begingroup$ The '\rm' means 'roman', as in upright font (as opposed to 'italic'). Typically format units as roman and variables as italic in formulae, and text looks way better in roman as well. Cheers :) $\endgroup$ – Kyle Oman Aug 26 '15 at 18:25
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Yes.The unit of $(\text{velocity})^2$ is $[\frac{\text{m}}{\text{s}}]^2$ .This is true for all calculations for any physical quantity.On squaring a physical quantity, its dimension gets squared. As a result, the unit is also squared.

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A slight expansion of the question:

When a physical quantity in an equation is raised to a power (like being squared) then all the physical quantities that go into that quantity are also raised to the same power.

So, velocity, $v$, has dimensions of length over time, $l/t$, and velocity squared, $v^2$, has dimensions of length squared over time squared, $l^2/t^2$.

As a consequence of this, the units used for these quantities follow the same rules.

Additionally, in any physics formula the physical quantities on the right hand side must equal the physical quantities on the left hand side.

Google "dimensional analysis"

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