2. a cube-shaped aquarium has edges that are 3 ft long. the aquarium is filled with water that has a density of 62 lb/ft³. (a) should the aquarium be placed on a table that can support a maximum weight of 200 lb? explain why or why not. (b) would the density of the water change if the aquarium was only half full? explain.

Part (a)

Let's find the volume of the tank. 

Volume of a rectangular prism = length*width*height
V = L*W*H
V = 3*3*3
V = 27 cubic feet

There is 27 cubic feet of water in the tank if the tank is completely full. Use the fact that
1 cubic foot of water = 62 pounds
so we can convert from volume to weight

(27 cubic feet)*(62 pounds/1 cubic foot) = 27*62 = 1674 pounds

If the tank is completely full, then the water weighs 1674 pounds, which is far greater than the 200 lb max weight limit for the table. The table will fall apart if you place a full aquarium on top of it. Note: this is not including anything else that would go inside the tank such as rocks, plant matter, fish, the glass of the tank itself, etc

In short, the aquarium should not be placed on the table as it exceeds the table's weight limit.



Part (b)

The density will not change because the density is how packed in a substance is for any cubic unit of volume. Saying "the density of water is 62 pounds per cubic foot" means that for a cubic foot, water naturally weighs 62 pounds .We can scale this to any size we want and the density will stay the same. 

It's like saying that the fractions 1/2 and 2/4 are the same. The 2/4 is the scaled version of 1/2. So back to the water example, if we had 2 cubic feet of water then it would weigh 2*62 = 124 pounds. So the ratio 
2 cubic feet: 124 pounds
simplifies back to
1 cubic foot: 62 pounds

note: the larger the density, the heavier the object is assuming the volume is held constant

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