As the density of copper is greater than either fluid, the piece of copper is totally immersed. The net force on the copper in water is mg - V_{Cu} x d_{H2O} x g = 2.65, where m is the mass of the copper. The net force in the second fluid is mg - V_{Cu} x d_{fluid} x g = 2.50. Solving each equation for mg and equating the results gives 2.65 + V_{Cu} x d_{H2O} x g = 2.50 + V_{Cu} x d_{fluid} x g. This can be rearranged and solved for V_{Cu} to give 0.15/(g(d_{fluid} - d_{H2O})) = 3.06 x 10^{-5} mm^{3}. This value of V_{Cu} is substituted into the first equation given and solved for mass which gives 3.01 x10^{-1} kg. The density can then be calculated to give 9.84 x 10^{3} kg/m^{3}. This result indicates that the copper is not pure.

Gabriel L.

asked • 10/29/20# A piece of copper immersed in water has an apparent weight of 2.65 N and when it is immersed in a liquid of density 1.50 x 103 kg/m3 its apparent weight is 2.50 N.

Is it pure copper or does it contain any other metal?

Copper density 8.9 X 103 kg/m3.

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