Density of Moon Rock: Solve Mass with Pressure

Click For Summary

Homework Help Overview

The discussion revolves around calculating the density of a moon rock based on its mass and apparent mass when submerged in water. The problem involves concepts from fluid mechanics and density calculations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between mass, volume, and density, questioning the need for additional information such as the volume of water. They discuss the concept of buoyancy and the role of displaced water in determining effective weight.

Discussion Status

Participants are actively engaging with the problem, offering various approaches to calculate density. Some have provided calculations based on the principles of buoyancy and density, while others are checking the validity of their reasoning and calculations. There is no explicit consensus yet, but several productive lines of inquiry are being explored.

Contextual Notes

Participants note the negligible effect of air density in this context and emphasize the importance of using the density of water to understand the buoyant force acting on the moon rock.

needhelp83
Messages
193
Reaction score
0
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?
 
Physics news on Phys.org
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?
 
Wouldn't I need the volume of water it was placed into?
 
Not if you're dealing with densities...
 
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?
 
Do I now need to use the density of air?
 
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...
 
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
 
Am I on the right track now?
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 6 ·
Replies
6
Views
7K
  • · Replies 15 ·
Replies
15
Views
2K
Replies
5
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
2
Views
1K