Question 70909
An equation used very often in physics is that Distance = Rate * Time
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For this problem we are given that the rate or speed of light is {{{3*10^5}}} km/s. 
We are also told that light coming to the Earth from the moon takes {{{1.28* 10^0}}} 
seconds to make the trip.
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We can substitute these two values into the distance equation and use a units check to
make sure the units check out.
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We write the equation as {{{D = (3*10^5)*(1.28*10^0)(km/s)*(s/1)}}}
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Let's clear the units first. Note that the s in the numerator (over 1) cancels with the
s in the denominator of {{{km/s}}} and we are left with km as the units of the answer.
We hold in mind that the units of the answer will be km and now drop the units from
the equation.
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Next we group the numerical multiplications together into {{{3*1.28}}}.  The result of that
multiplication is {{{3.84}}}. We in turn multiply that by the product of the exponential
tens terms ... that product being {{{10^5*10^0}}}. When we multiply common base terms that
have exponents, we do so by just adding the exponents.  So the multiplication becomes:
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{{{10^5*10^0=10^(5+0) = 10^5}}}
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Putting everything together we get an answer of {{{3.84*10^5}}} km
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Hope this helps you to understand how to handle numbers that are associated with powers
of 10 and also how to use units to see if the dimensions check out OK.