Question

物理密度应用题!1.一个容积为3*10^-4m^3的瓶子内盛有0.2kg的水,一只口渴的乌鸦每次将一块质量为0.01kg

Answer 1

1.（1）V水=m/p=0.2kg/1.0*10^3=0.0002m^3

V石=V总-V水=0.0003m^3-0.0002m^3=0.0001m^3;

(2)p石=m石/V=25*0.01kg/0.0001m^3=2500kg/m石/Vm^3.

2.方法一:p=1.06kg/120cm^3=1.06kg/0.00012m^3=833.与P铜=0.8*10^3kg/m^3不符,p