Question

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

Answer 1

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.