Fluid Mechanics Density problems

[Susan]

Is this correct?

A 6-kg piece of metal displaces 1 liter of water when submerged. What is its density?
Density = mass/volume 1 liter of water = 1000 cm3 = 1 kg
Ans: 6 kg / 1000 cm3 = .006 cm3


How would this one work out?
When a 2.0-kg object is suspended in water, it ‘masses’ 1.5 kg. What is the density of the object?
I'm guessing:
Density = mass/volume = 2.0 kg/1.5 kg = 1.333
Or 2.0 kg/ 1500 cm3 = .00133 cm3

Now I'm totally lost on what to do:

A vacationer floats lazily in the ocean with 90% of his body below the surface. The density of the ocean water is 1.025 kg/m3. What is the vacationer’s average density?

 

[Tide]

The buoyant force is equal to the weight of the water displaced (Archimedes principle). That allows you to do a force balance from which you can ultimately determine the volume of the swimmer!

 

[wais]

Your calculations for the first two problems are correct. For the third problem, we need to use the equation Density = mass/volume. We know the density of the ocean water (1.025 kg/m3) and we can assume that the vacationer's body has the same density as water (since they are floating). So, we can set up the equation as follows:

1.025 kg/m3 = mass/volume

We know that 90% of the vacationer's body is below the surface, so we can assume that 90% of their volume is also below the surface. This means that their volume above the surface is 10% of their total volume. We can represent this mathematically as:

Volume = 0.1 x total volume

Now we can substitute this into our equation to solve for the vacationer's average density:

1.025 kg/m3 = mass/(0.1 x total volume)

We can rearrange this to solve for mass:

mass = 1.025 x 0.1 x total volume

And we know that mass is equal to the weight of the vacationer, so we can represent this as:

weight = 1.025 x 0.1 x total volume

Finally, we can divide both sides by the total volume to solve for the vacationer's average density:

average density = weight/total volume = (1.025 x 0.1 x total volume)/total volume = 0.1025 kg/m3

So, the vacationer's average density is 0.1025 kg/m3.