An object with height h, mass M, and a uniform cross-sectional area A floats upright in a liquid with density ρ.

** Missing information: The vertical distance from surface of liquid to bottom of the object is sought in this question, with the condition that the object is at equilibrium **

Ans: The vertical distance = y = M/(ρA)

Explanation:

Support the vertical distance = y

Object's density = M/(A*h) (since A*h = volume)

By applying the condition,

(M/(Ah))/ρ = y/h

M/(ρAh) = y/h

y = M/(ρA)

Answer Link

okpalawalter8 okpalawalter8 · Answer 2 · 2022-04-12T15:08:48+02:00

The vertical distance from the surface of the liquid to the bottom of the object is mathematically given as

y = M/(ρA)

What is the vertical distance from the surface of the liquid to the bottom of the object?

Question Parameter(s):

An object with height h,

mass M,

and a uniform cross-sectional area A

density ρ

Generally, the equation for the Object's density is mathematically given as

d = M/(A*h)

Therefore

(M/(Ah))/ρ = y/h

M/(ρAh) = y/h

y = M/(ρA)

In conclusion, the Object's density is

y = M/(ρA)