Calculating Al Foil Thickness in mm

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Discussion Overview

The discussion revolves around calculating the thickness of a sheet of aluminum foil based on its area and mass, utilizing the density of aluminum. The context is primarily homework-related, focusing on the application of density and volume equations.

Discussion Character

  • Homework-related, Mathematical reasoning

Main Points Raised

  • One participant presents a problem involving the mass and area of aluminum foil to calculate its thickness using the density of aluminum.
  • Another participant questions the initial calculation provided, suggesting that the answer may be correct in a different system of units and seeks clarification on the variables used in the density equation.
  • There is a lack of clarity regarding the definitions of variables in the relevant equations, particularly what "d" represents and how to derive volume from the given data.
  • A third participant expresses gratitude, which may imply they found the discussion helpful or resolved their query.

Areas of Agreement / Disagreement

The discussion remains unresolved, with no consensus on the correctness of the initial calculation or the definitions of the variables involved.

Contextual Notes

There are limitations in the clarity of the variables used in the density equation, and the initial attempt lacks units and detailed calculations, which may affect the understanding of the problem.

mightyhealthy
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Homework Statement



A Sheet of aluminum(Al) foil has a total area of 1.000ft^2 and a mass of 3.636g.

What is the thickness of the foil in millimeters (density of Al=2.699g/cm^3)?

Homework Equations



d=m/v



The Attempt at a Solution



1.347=totally wrong... help please
 
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1.347 is probably the right answer in some system of units :-p

Also, what is the d in your "relevant equations"? Is it the d for density? Or is it the d that is sometimes used for small distances (such as, say, thickness?) And what is v? How do you calculate it from the given data?

Your "attempt" is not so much an attempt from which we can clearly locate some error, as some numerical guess, without units or any other relevant information.
 
thank you
 
You're welcome.
I take it this means you solved it?
 

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