用弹簧测力计测得某金属块的重力为8.9N,把一半的体积浸在水中时,弹簧测力计拉力为8.4N,求密度

解
F浮=ρgV排
V排=F浮/ρg
V排=0.4/(1.0x10^3x10)
V排=4x10^-5 m^3
V物=2V排
V物=8x10^-5 m^3
m物=G/g
m物=0.89Kg
ρ物=m物/V物
ρ物=0.89/8x10^-5
ρ物=11125Kg/m^3
答：金属块的密度是11125Kg/m^3.