What is the Density of a 6.8-kg Piece of Metal That Displaces 1 Liter of Water?

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Homework Help Overview

The discussion revolves around calculating the density of a piece of metal based on its mass and the volume of water it displaces. The subject area pertains to concepts of density, mass, and volume in physics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster expresses confusion about how displacement relates to density and seeks suggestions for starting points. Some participants clarify the relationship between mass, volume, and density, while others question the distinction between mass and weight.

Discussion Status

Participants have provided guidance on the formula for density and the relationship between mass and volume. There is an ongoing exploration of the concepts involved, with some participants clarifying definitions and calculations without reaching a consensus.

Contextual Notes

There is a mention of the original poster's self-identified lack of experience with the topic, which may influence the discussion dynamics.

sharonita
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Heres the question:
A 6.8-kg piece of metal displaces 1 liter of water when submerged. Calculate its density.

So I have no idea where to begin or how the displacement relates to the density. What can you suggest? Thanks!
 
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i reallllyy need help =\
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Density equals mass divided by volume. If the object displaces 1 liter of water, then it has a volume of 1 liter. You also have to know that 1 liter = 0.001 cubic meters. The objects density is 6.8kg/0.001m^3 or 6800kg/m^3.
 
But wasnt the weight 6.8kg?
 
Yes the mass is 6.8kg and the volume is 1liter which is equal to 0.001m^3. The density is mass divided by volume which is 6.8kg/0.001m^3. Do 6.8 divided by 0.001 on a calculator; what answer do you get?
 
sharonita said:
But wasnt the weight 6.8kg?


Nope, the mass was 6.8 kg. The kilogram is a unit of mass, not weight.
 
6800...thanks!
 
ohhhh. okay thanks a bunch!
im such a newbie, that's all. :)
 

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