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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
avatar+698 
0

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
avatar+99331 
0

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
avatar+698 
0

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
avatar+99331 
+2

Yes that is right, the Kg cancel out leaving you just with 1metre.    laugh

 

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
avatar+698 
0

Ok, I wasn't 100% sure if I was supposed to cancel out the Kg since the answer was supposed to include units. indecision

PartialMathematician  Dec 17, 2018
 #6
avatar+99331 
0

You can always cancel units. They work just like numbers, it can make difficult problems much easier.  !

Melody  Dec 17, 2018

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