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# help with question on units

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a = b/e

b = 1 kg m-1

e = 1 kg m-2

what is a? including units

Dec 16, 2018

#1
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I do not know if the variable m stands for meters or not. If it does, how can something be both kilograms and meters?

$$a = \dfrac{1 (kg) m^{-1}}{1 (kg) m^{-2}} \Rightarrow a = \dfrac{1 (kg) m^{-1}}{\dfrac{1}{1 (kg) m^{2}}}$$. if we multiply the numerator and denominator $$1 (kg) m^{2}$$by , we have $$a = {1 (kg^2) m^{1}}$$. You can adjust my answer according to your variables.

- PM

Dec 17, 2018
edited by PartialMathematician  Dec 17, 2018
#2
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Note to  PM

a = b/e

b = 1 kg m-1

e = 1 kg m-2

what is a? including units

PM, your current answer is       $$a = {1 (kg^2) m^{1}}$$

It is NOT correct.   Can you work out why?

Dec 17, 2018
#3
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I guess you could also do this:

Take away kg from the numerator and denominator of $$a = \dfrac{(kg)m^{-1}}{(kg)m^{-2}}$$ to become $$a = \dfrac{m^{-1}}{m^{-2}}$$.

This equals $$a = {m^1}$$ because $$m^{-1} / m^{-2} = m^{-1} \cdot m^{2} \Rightarrow m^1$$.

- PM

PartialMathematician  Dec 17, 2018
#4
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Yes that is right, the Kg cancel out leaving you just with 1metre. 1m^1=m

if you want to do it just with the indices rules you have

$$a=\frac{1Kg\times m^{-1}}{1Kg\times m^{-2}}\\ a=1m^{-1--2}\\ a=1m$$

Melody  Dec 17, 2018
edited by Melody  Dec 17, 2018
#5
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Ok, I wasn't 100% sure if I was supposed to cancel out the Kg since the answer was supposed to include units. PartialMathematician  Dec 17, 2018
#6
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You can always cancel units. They work just like numbers, it can make difficult problems much easier.  !

Melody  Dec 17, 2018