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anonymous
 one year ago
How do you prove that an equation is dimensionally correct?
anonymous
 one year ago
How do you prove that an equation is dimensionally correct?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0By performing the mathematics on the units. For example, consider\[d=v_0t+\frac{ 1 }{ 2 }at^2\]Forgetting about any values and just looking at the units gives\[m=\left( \frac{ m }{ s } \right)\left( s \right) + \left( \frac{ m }{ s^2 } \right)\left( s \right)^2\]\[m = m\]Hence, the equation is dimensionally correct.

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0Donot confuse units with dimensions. \[\sf L=\left( \frac{ \sf L }{ \sf T } \right)\left( \sf T \right) + \left( \frac{ \sf L }{ \sf T^2 } \right)\left( \sf T \right)^2\\ \sf L = \sf L+\sf L\\ \sf L = \sf L\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Right you are @UnkleRhaukas. I'm getting lazy in my old age. However, as long as you are consistent in the use of base units in a single measurement system (SI, Imperial, cgs, etc.) the two approaches are equivalent. Thanks!
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