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How much is the mass of aluminum with a volume of 1 m3 less than the mass of lead of the same volume?

User Whoughton
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The aluminum sphere must be larger in volume to compensate for its lower density. We require equal masses:m A​ =m Fe​ or ρ A1​ V A1​ =ρ
User Jeremywoertink
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Answer: the masses:mA​=mFe or ρA1​VA1​=ρFe​VFe​then use the volume of a sphere. By substitution. The answer is (2.70∗103kg/m37.86∗103kg/m3​)(2.00cm)3=23.3cm3

Explanation: you must go from left to right when solving substitutuing the first side then dividing

User WoogieNoogie
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