Question

. A cube-shaped aquarium has edges that are 3 ft long. The aquarium is filled with water that has a density of 62 Ib/ft^3. Should the aquarium be placed on a table that can support a maximum weight of 200 lb? Explain why or why not.

Answer 1

No, since the filled aquarium weights a lot more than `200lb`.

Explanation:

The first thing we need to know is what is the volume that our aquarium can hold. Since we know it has a side of 3 ft, we multiply that by itself three times:

`3ft xx 3 ft xx 3 ft = 27 ft^3`

So, now, we have that the density of water is `(62lb)/(1ft^3)`, which means that each cubic foot (`ft^3`) of water weights `62lb`. So, to know the total weight of the aquarium, we multiply the `27ft^3`, by the density to get:

`27ft^3 xx (62lb)/(1ft^3)`

`27cancel(ft^3) xx (62lb)/cancel(1ft^3)`

`27 xx 62lb = 1674 lb`

Since `1674lb` is a lot more than the table can sustain, we should not put the aquarium on top of it.