Question

A piece of metal weighs 50.0 N in air, 36.0 N in water, an 41.0 N in an unknown liquid. Find the density of the unknown liquid.

Answer 1

find volume of metal first
Buoyant force in water = 14 N = weight of water displaced in 1 g gravitational field
mass of water displaced = 14 N/9.8 m/s^2 = 1.43 kg of water displaced
the density of water = 1,000 kg/m^3
so
volume of metal = 1.43*10^-3 m^3

Buoyant force in unknown = 50 - 41 = 9 N
so mass of unknown displace = 9/9.8 = .918 kg

so the density of unknown = .918kg/1.43*10^-3 = 642 kg/m^3
or 64.2% of the water density

Answer 2

I'm confused about the "1 g gravitational field". What's the 1 g referring to and why is it 1 g?

Answer 3

g is the acceleration of gravity which is about 9.8 m/s^2 on the surface of earth.
Since the weights were given in Newtons, a unit of force, not mass, I used weight in Newtons = mass in kilograms * acceleration in meters / second squared to find the mass in kilograms.

Answer 4

where did you got the buoyant force in water?

Answer 5

It's well known it doesn't need to be given (like the gravitational force)

Answer 6

Take the actual weight (in air) of the object and subtract the weight(apparent) the objects seems to have in the unknown fluid ie take 50N and subtract 36N to get 14N which is the buoyant force/upthrust.