Answer:
The volume of the rock is 13.368 cubic feet.
Step-by-step explanation:
The volume of the rock ([tex]V_{r}[/tex]), measured in gallons, is equal to the volume of the fish aquarium without the decorative rock ([tex]V'[/tex]) minus the volume of the fish aquarium with the decorative rock ([tex]V''[/tex]), both measured in gallons.
Since the water flow is at constant rate, the volume of the rock is expressed by the following equation:
[tex]V_{r} = V' - V''[/tex]
[tex]V_{r} = \dot V \cdot (\Delta t_{1}-\Delta t_{2})[/tex] (1)
Where:
[tex]\dot V[/tex] - Flow rate, measured in gallons per hour.
[tex]\Delta t_{1}[/tex] - Filling time of the fish aquarium without the decorative rock, measured in hours.
[tex]\Delta t_{2}[/tex] - Filling time of the fish aquarium with the decorative rock, measured in hours.
And the flow rate is:
[tex]\dot V = \frac{2000\,gal}{10\,h}[/tex]
[tex]\dot V = 200\,\frac{gal}{h}[/tex]
The flow rate is 200 gallons per hour.
If we know that [tex]\dot V = 200\,\frac{gal}{h}[/tex], [tex]\Delta t_{1} = 10\,h[/tex] and [tex]\Delta t_{2} = 9.5\,h[/tex], then the volume of the rock is:
[tex]V_{r} = \left(200\,\frac{gal}{h}\right)\cdot (10\,h-9.5\,h)[/tex]
[tex]V_{r} = 100\,gal[/tex]
[tex]V_{r} = 13.368\,ft^{3}[/tex]
The volume of the rock is 13.368 cubic feet.
