Question

边长为10cm的立方体木块浸在水中,0.25露出水面,1.求下表面受的压力2.浮力是多少3.木块重力

Answer 1

(1.)h^1=10cm=0.1m

h^2=(1-0.25)h^1=0.75*0.1m=0.075m

P=ρgh=1.0*10^3kg/m^3*10N/kg*0.075m=750Pa(g取的10N/kg)

s=(h^1)^2=(0.1m)^2=0.01m^2

F=P*s=750Pa*0.01m^2=7.5N

(2.) V^1=(h^1)^3=(0.1m)^3=0.001m^3

V^排=(1-0.25)V^1=0.00075m^3

F^浮=G^排=ρ^液gV^=1.0*10^3kg/m^3*10N/kg*0.00075m^3=7.5N

(3.)G=F^浮=7.5N