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?

Answer 1

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

Answer 2

Thank you so I much!! I understand it now!!