推荐回答（2个）
解:
1．由图可知，金属块被提高2cm时，弹簧测力计的示数开始变大，此时金属块的上表面刚露出液面，金属块被提升6cm时，弹簧测力计的示数不再变化，此时金属块的下表面开始露出液面，故金属块的边长为6cm- 2cm= =4cm
2.金属块完全露出液面时拉力等于重力，由图象可知，该金属块的重力: G=F拉=3.5N
V= (4cm) 3=64cm3=6.4x10 -5m3
G=ρgV
ρ=G/gV=3.5/10x6.4x10-5=5.47x103(kg/m3)
3.在高度h由0变化到2cm的过程中，金属块完全浸没在液体中，此时排开液体的体积: V 排=V= (4cm) 3=6.4x10 -5m3此过程中弹簧测力计的示数为F拉'=2.5N
所以，物体完全浸没在液体中受到的浮力为
F浮=G-F拉'=3.5N - 2.5N=1N
由F浮=ρ液gV排得，液体的密度:
ρ液=F浮/gV排=1N/10N/kgX6.4x10 -5m3≈1 56x103kg/m3.
4.由图象可知，甲图中刚提起金属块时上表面距离液面的高度为h'=2cm=0.02m,则上表面受到的液体压强:
p=ρ液gh'=1. 56x103kg/m3x10N/kgx0.02m=312Pa
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