Solve Difficult Problem: Average Density of Rectangular Parallelopiped

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To find the average density of a rectangular parallelopiped floating in water, the dimensions provided are 28.0 cm x 19.1 cm x 2.000 m, with 3.17 cm above the water. The problem relates to buoyancy, which can be analyzed using Archimedes' principle. The volume of the submerged part can be calculated to determine the buoyant force, which is equal to the weight of the water displaced. The average density can then be found by dividing the weight of the object by its total volume. Understanding these principles is key to solving the problem effectively.
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Homework Statement



A solid student in the shape of a rectangular parallelopiped of dimensions 28.0 cm x 19.1 cm x 2.000 m floats horizontally in a calm hot tub filled with warm water; 3.17 cm of the parallelopiped (along the 19.1 cm side) is above the water surface. Find the average density of the student.

Homework Equations



maybe density, volume of parallelopiped

The Attempt at a Solution



I have no idea how to solve this problem. I tried using geometry to find certain things but it did not help. Any help is appreciated! o:)
 
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i have a feeling it has something to do with the buoyancy force. you know anything about that?
 
… eureka … !

jrzygrl said:
I tried using geometry to find certain things but it did not help.

Hi jrzygrl! :smile:

A rectangular parallelepiped (no "o") (also called a cuboid) is just an ordinary rectangular box!

Hint: have you tried thinking about this in the bath, like Archimedes, and then running down the street completely naked, shouting "Eureka!" ? :biggrin:
 

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