Static fluids; Density; Buoyancy

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SUMMARY

The discussion revolves around the buoyancy and density calculations of a toy dinosaur floating in a fluid system consisting of water and oil. The density of the dinosaur is established at 900 kg/m³. When oil with a density of 869 kg/m³ is added, the challenge is to determine the percentage of the dinosaur submerged in water after the oil is introduced. The key principle involves using the relationship between the densities of the fluids and the submerged volume of the dinosaur to solve for the submerged percentage.

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Homework Statement



A toy dinosaur floats on water in a container and is partially submerged with 90 % of its volume underwater (submerged).

(a) What is the density of the dinosaur?
I got this answer right. It is 900 kg/m^3

A layer of oil, with a density of 869 kg/m3, is now poured into the container on top of the water to a depth that more than covers the dinosaur.

Find the percentage of the dinosaur submerged in the water after the oil was added. (Ignore the buoyancy of the air before the oil is added.)

This answer I did not get right, and I really don't know how to do it.(Hint: the dinosaur is suspended at rest in the fluids with x% of its volume in the water and (1-x)% of its volume in the oil

Homework Equations


The Attempt at a Solution

i tried 900 ÷ 869 and that gives 1.03567 then i thought to subtract that from 100 percent. I know that doesn't make much sense, but I was grasping for any kind of path that might get me the answer. (sound desperate enough to you?) lol help is very much appreciated!
 
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hi awertag! :smile:
awertag said:
Hint: the dinosaur is suspended at rest in the fluids with x% of its volume in the water and (1-x)% of its volume in the oil

what happened to x ?? :confused:

follow the hint! …

what principle is involved? use x to calculate it! :smile:

(and remember that the dinosaur is still floating … it's lighter than the water but heavier than the oil)
 

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