How Do You Convert Seconds to Nanoseconds in Scientific Notation?

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Homework Statement
Please see below
Relevant Equations
Please see below
How do you convert ##1.66 \times 10^{-7} s ## to ##ns##? I know the answer is ##166 ns = 160 \times 10^{-9} s## however what is the thought process?

Many thanks!
 
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Callumnc1 said:
Homework Statement:: Please see below
Relevant Equations:: Please see below

How do you convert ##1.66 \times 10^-7 s ## to ##ns##? I know the answer is ##166 ns## however what is the thought process?

Many thanks!
Do you know what "ns" MEANS ?
 
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phinds said:
And how do you express that in seconds?
THanks for you reply @phinds! ## 1s = 1 \times 10^{-9} ns ##. Oh I think I know where you going with this.
 
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Callumnc1 said:
THanks for you reply @phinds! ## 1s = 1 \times 10^-9 ns ##. Oh I think I know where you going with this.
Sorry yeah it is just basic unit conversion.

##1.66 \times 10^{-7}s \times \frac {1ns} {1 \times10^{-9}s} = 166 ns ##
 
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phinds said:
Right, except you have 1.6 where the original says 1.66
Oh whoops sorry @phinds! I'll fix that up.
 
Callumnc1 said:
Sorry yeah it is just basic unit conversion.

##1.66 \times 10^{-7} \times \frac {1ns} {1 \times10^{-9}s} = 166 ns ##
Yes, the thought process is to multiply the quantity you want to convert by "1", where the "1" is a fraction that is set up to cancel the old units with the denominator and result in the new units of the numerator. (the only thing you are missing in your equation that I quoted is to include the units of "s" in the first quantity, so that can be canceled by the denominator's "s" units in the multiplication)

I still remember my first semester "Introduction to Engineering" class, where a TA explained this trick to the class, and we all looked at each other like "Wow, I never thought of it like that before!". It's a simple trick, but very useful. :smile:
 
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berkeman said:
Yes, the thought process is to multiply the quantity you want to convert by "1", where the "1" is a fraction that is set up to cancel the old units with the denominator and result in the new units of the numerator. (the only thing you are missing in your equation that I quoted is to include the units of "s" in the first quantity, so that can be canceled by the denominator's "s" units in the multiplication)

I still remember my first semester "Introduction to Engineering" class, where a TA explained this trick to the class, and we all looked at each other like "Wow, I never thought of it like that before!". It's a simple trick, but very useful. :smile:
Thank you @berkeman - nice to see the thought process explained!
 
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Callumnc1 said:
THanks for you reply @phinds! ## 1s = 1 \times 10^{-9} ns ##. Oh I think I know where you going with this.

Looks like a sign error! (or a VERY strange second)
 
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Tom.G said:
Looks like a sign error! (or a VERY strange second)
Haha, yes @Tom.G , I'm still working on my LaTex :)