Question

A reservoir has a surface area of 30 km^2 and an average depth of 49 m. What mass of water is held behind the dam?

Answer 1

Mass of water,
`m=1.47* 10^(12)\ kg`

Step-by-step explanation:

Given that,

Surface area of the reservoir, A = 30 km² = 3 × 10⁷ m²

Average depth, d = 49 m

The volume of the reservoir is V such that,

`V=A* d`

`V=3* 10^7\ m^2* 49\ m`

`V=1.47* 10^9\ m^3`

We have to find the mass of water is held behind the dam. It can be calculated using the expression for density. We know that density of water, d = 1000 kg/m³

Density,
`d=(m)/(V)`

`m=d* V`

`m=1000\ kg/m^3* 1.47* 10^9\ m^3`

`m=1.47* 10^(12)\ kg`

Hence, this is the required solution.