Static fluids; Density; Buoyancy

AI Thread Summary
The discussion focuses on a physics problem involving buoyancy and density, where a toy dinosaur floats in water and is partially submerged. The density of the dinosaur is established at 900 kg/m^3, while a layer of oil with a density of 869 kg/m^3 is added. Participants emphasize the need to apply the principle of buoyancy to determine the percentage of the dinosaur submerged in both fluids after the oil is introduced. The hint suggests using variables to represent the submerged volumes in water and oil, leading to a calculation based on their respective densities. The conversation highlights the importance of understanding how different fluid densities affect buoyancy and the equilibrium of floating objects.
awertag
Messages
63
Reaction score
0

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!
 
Last edited:
Physics news on Phys.org
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)
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top