Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 1 - Introduction, Measurement, Estimating - General Problems: 49


$4.8\times10^5$ metric tons of water fell on the city. $1.3\times10^8$ gallons of water fell on the city.

Work Step by Step

First let's calculate the volume of water that fell on the city: $V = (6000 ~m)(8000 ~m)(0.01 ~m) = 4.8\times10^5 ~m^3$ Water has a mass of $10^3$ kg per $m^3$: $mass = (4.8\times10^5 ~m^3)(10^3 ~kg/m^3) = 4.8\times10^8 ~kg$ $(4.8\times10^8 ~kg)(\frac{1 ~ton}{10^3 ~kg}) = 4.8\times10^5 ~tons$ $(4.8\times10^5 ~m^3)(10^3 ~L/m^3)(\frac{1 ~gal}{3.8 ~L}) = 1.3\times10^8$ gallons
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