What is the density of an iceberg and how much of it will show in fresh water?

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Homework Help Overview

The discussion revolves around the density of icebergs and how much of an iceberg would be visible when submerged in fresh water. The original poster presents two parts to the problem, focusing on the relationship between the portion of the iceberg above water and its density, as well as the implications of moving the iceberg into fresh water.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply Newton's first law to equate upthrust and gravitational force to find the density of ice. They express confusion about whether atmospheric pressure should be considered in this context. Other participants engage by questioning the relevance of atmospheric pressure in the scenario, drawing analogies to everyday experiences.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the forces acting on the iceberg. Some guidance is provided regarding the application of Newton's laws, but there is no explicit consensus on the role of atmospheric pressure in the calculations.

Contextual Notes

The problem involves specific densities of sea water and fresh water, which are provided, but the discussion also highlights potential assumptions regarding atmospheric pressure and its effects on buoyancy.

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Homework Statement



a) If the part of an iceberg above sea level is one ninth of the whole, what is the density of ice?

b)How much of the iceberg would show if it moved into a fresh water region?

Homework Equations


Density of sea water=1025kgm^-3
Density of fresh water 1000kgm^-3

Weight displaced=upthrust
Force due to gravity = mg


The Attempt at a Solution



For part a) I equated the upthrust and the force due to gravity, by applying Newtons first law. However I am confused as to whether I should also take into account the atmospheric pressure pressing the part of the iceberg above sea level down. In all the stuff I've read no one seems to take it into account when finding the density of an iceberg. Is there a reason for this?

In part b) I think you again apply Newtons first law and use the ice density calculated in part a),

[tex]\rho_{ice}*g*v_{ice}=\rho_{water}*g*v_{water}[/tex]

[tex]\frac{\rho_{ice}}{\rho_{water}}=\frac{v_{water}}{v_{ice}}[/tex]

Which gives the proportion of ice under the water.

Please could someone help me?
 
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Vuldoraq said:
However I am confused as to whether I should also take into account the atmospheric pressure pressing the part of the iceberg above sea level down.


If you pick up a piece of paper, do you feel the 1400 pounds of force that air pressure exerts on it?
 
Vanadium 50 said:
If you pick up a piece of paper, do you feel the 1400 pounds of force that air pressure exerts on it?

I do :(
 
Vanadium 50 said:
If you pick up a piece of paper, do you feel the 1400 pounds of force that air pressure exerts on it?

I geuss not, but isn't that because as soon as you pick it up air rushes underneath the paper, very quickly, and the air underneath exerts an equal and oppisite force to the air above, thus we don't feel the pressure force. In the sea this situation is clearly impossible.
 
Last edited:

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