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A cube-shaped aquarium has edges that are 1.75 ft long. the aquarium is filled with water that has a density of 72" lb/f" "t" ^3.

(a) should the aquarium be placed on a table that can support a maximum weight of 150 lb? explain why or why not.
(b) would the density of the water change if the aquarium was only half full? explain.

User OArnarsson
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2 Answers

6 votes

Answer:

No and the density changes along with the weight.

User Jogojapan
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5 votes

Answer:

(a) No. The weight is more than 150 lb.

(b) No. Density is not a function of volume.

Explanation:

The weight of the aquarium is the product of the volume and the density of the contents.

(a)

The volume of the aquarium is the product of its dimensions:

V = LWH = (1.75 ft)(1.75 ft)(1.75 ft) = 1.75³ ft³ = 5.359375 ft³

The weight of its contents, at 72 lb/ft³ will be ...

(72 lb/ft³)(5.359375 ft³) = 385.875 lb

The aquarium should not be placed on the table. The weight of the aquarium is too great for the load limit of the table.

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(b)

The density of water is the ratio of its mass to its volume. Density does not change with depth or volume. It is essentially constant. (Water is considered to be virtually incompressible. Its density changes slightly with temperature,)

No, the density of the water would not change.

_____

Additional comments

The half-full aquarium would weigh about 385.875/2 lb ≈ 194 lb. This is still too heavy for the table.

The density of water is actually about 62.4 lb/ft³. We assume the extra 10 lb/ft³ might take into account the weight of the aquarium and any gravel or other non-water content.

User Markus Safar
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