Question

A cube-shaped aquarium has edges that are 3 feet long and is filled with water that has a density of 62 lbs/ft3 Should the aquarium be placed on a table that can support 200 pounds? Why or why not? Would the density of the water change if the aquarium was half full?

Answer 1

Given:

There are given the edge of the cube-shaped aquarium has 3 feet.

Explanation:

To find the value, we need to use the volume of the cube formula:

So,

From the formula of volume:

$V=a^3$

Where

a represents the value of edge.

So,

Put the value of edge into the above formula:

Then,

$\begin{gathered} V=a^3 \\ V=3^3 \\ V=27cube\text{ feet} \end{gathered}$

Now,

The water has a density of 62 pounds per cube foot.

According to the question:

If the weight of 1 cubic foot of water is 62 pounds, then the weight of 27 cube feet water is:

$62\times27=1674pound$

Final answer:

Hence, the water weight of the full aquarium is 1674 pounds, and the table only susupports00 pounds. So the table cannot hold the aquarium.

And,

No, the density of water would not change.