# What volume of water in LITRES is contained in a dam whose area was 18.9*"mile"^2, and whose average depth was 47*ft?

Sep 11, 2017

Well, let's convert to kosher units first.......

#### Explanation:

We gots ..................

$18.9 \cdot \text{mile"^2-=18.9*"mile"^2xx2.59xx10^6*m^2*"mile"^-2}$

$= 49.0 \times {10}^{6} \cdot {m}^{2}$ $\left(i\right)$

And average depth is $47 \cdot f t \times 0.3048 \cdot m \cdot f {t}^{-} 1 = 14.3 \cdot m$ $\left(i i\right)$

And $\text{volume} = \left(i\right) \times \left(i i\right) = 49.0 \times {10}^{6} \cdot {m}^{2} \times 14.3 \cdot m =$

$= 702 \times {10}^{6} \cdot {m}^{3}$.

And we know that there are ${10}^{3} \cdot L \cdot {m}^{-} 3$

And so we got $702 \times {10}^{6} \cdot {m}^{3} \times 1000 \cdot L \cdot {m}^{-} 3 \equiv 702 \times {10}^{9} \cdot L$.

An acquaintance of mine, an engineer, once started using $\text{sydharbs} \equiv 500000 \times {10}^{6} \cdot L$ as a measurement of LARGE volumes of water. A $\text{sydharb}$ was the average volume of water in Sydney Harbour. This helped us vizualize the quantity, because we had all taken the ferry from Circular Quay to Manly across Sydney Harbour. How many $\text{sydharbs}$ have you got here?