Answer:
The total weight of the pool when it is full of water is 690.2 pounds.
Step-by-step explanation:
We know that total weight of the wading pool ([tex]m_{T}[/tex]) is the sum of the weight of empty pool ([tex]m_{pool}[/tex]) and the weight of water ([tex]m_{w}[/tex]), all measured in pounds, that is:
[tex]m_{T} = m_{pool} + m_{w}[/tex]
The weight of water can be determined by definition of density, as we know the geometry of the wading pool and density of water:
[tex]m_{w}=\rho_{w}\cdot (w\cdot h\cdot l)[/tex]
Where:
[tex]\rho_{w}[/tex] - Density, measured in pounds per cubic foot.
[tex]w[/tex], [tex]h[/tex], [tex]l[/tex] - Width, height and length of the wading pool, measured in feet.
Now, we are noticed that [tex]\rho_{w} = 6.24\,\frac{pd}{ft^{3}}[/tex], [tex]w = 7\,ft[/tex], [tex]h = 6\,ft[/tex], [tex]l = 2.5\,ft[/tex] and [tex]m_{pool} = 35\,pd[/tex] and the total weight of the pool when it is full of water is:
[tex]m_{w} = \left(6.24\,\frac{pd}{ft^{3}} \right)\cdot (7\,ft)\cdot (6\,ft)\cdot (2.5\,ft)[/tex]
[tex]m_{w} = 655.2\,pd[/tex]
[tex]m_{T} = 35\,pd + 655.2\,pd[/tex]
[tex]m_{T} = 690.2\,pd[/tex]
The total weight of the pool when it is full of water is 690.2 pounds.
