Question

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`?

Answer 1

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

Explanation:

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

`18.9*"mile"^2-=18.9*"mile"^2xx2.59xx10^6*m^2*"mile"^-2"`

`=49.0xx10^6*m^2` `(i)`

And average depth is `47*ftxx0.3048*m*ft^-1=14.3*m` `(ii)`

And `"volume"=(i)xx(ii)=49.0xx10^6*m^2xx14.3*m=`

`=702xx10^6*m^3`.

And we know that there are `10^3*L*m^-3`

And so we got `702xx10^6*m^3xx1000*L*m^-3-=702xx10^9*L`.

An acquaintance of mine, an engineer, once started using `"sydharbs"-=500000xx10^6*L` as a measurement of LARGE volumes of water. A `"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 `"sydharbs"` have you got here?