What is the density of a 7.85 kg moon rock submerged in water?

[needhelp83]

A geologist finds that a moon rock whose mass is 7.85 kg has an apparent mass of 6.18 kg when submerged in water. What is the density of the rock

Would I use pressure to solve for the mass of the moon rock?

 

[berkeman]

When submerged in water, the object displaces an equivalent volume of water. That displaced water weighs something, and it not being there provides some effective bouyancy. Like, if you have a balloon under water, the upward force is equivalent to the weight of the missing water. Does that help you solve this problem?

 

[needhelp83]

Wouldn't I need the volume of water it was placed into?

 

[berkeman]

Not if you're dealing with densities...

 

[needhelp83]

p=m/V

1000 kg/m^3=7.85 kg/V
1000V=7.85
V=0.00785 m^3

p=m/V
p=7.85 kg/0.00785 m^3
p= 1000 density of rock

Did I get off somewhere?

 

[needhelp83]

Do I now need to use the density of air?

 

[berkeman]

The mass density of the air is negligible in this problem, and you need to use the mass density of water to give you the lift component force that lightens the moon rock when it is under water. Remember that the density of water is 1kg/liter. So the difference between the 7.85kg weight and the 6.18kg weight is the weight of the displaced water...

 

[needhelp83]

This should be better...

pwaterVg=FB
(1000 kg/m3)(9.8 m/s)V=(7.85 kg – 6.18 kg)(9.8 m/s)\
V=0.00167 m3

p=m/V
p = 7.85 kg/0.00167 m3= 4700.6 kg/m3

 

[needhelp83]

Am I on the right track now?