Question

A 515 cm3(cm cubed) block of aluminum (D = 2.7 g/cm3) is lowered carefully into a completely full beaker of water. What is the weight of the water, in newtons, that spills out of the beaker?
When the aluminum block is lowered, the relative volume of water spills out of a breaker.
Weight of water is D•V•g= =1000•515•10^-6•9.8=5.047 N
Thank you so I much!! I understand it now!!

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