Assume that the height of the block is 9 cm.
(a)
mg=ρgV→ρ0V0g=ρgV→ρ0=ρVV0=mg=\rho gV\to \rho_0V_0g=\rho gV\to\rho_0=\frac{\rho V}{V_0}=mg=ρgV→ρ0V0g=ρgV→ρ0=V0ρV=
=1000⋅0.03⋅0.04⋅0.060.03⋅0.04⋅0.09≈667(kg/m3)=\frac{1000\cdot0.03\cdot0.04\cdot0.06}{0.03\cdot0.04\cdot0.09}\approx667(kg/m^3)=0.03⋅0.04⋅0.091000⋅0.03⋅0.04⋅0.06≈667(kg/m3)
(b)
(W+mg)=ρgV→W=ρgV−mg=(W+mg)=\rho g V\to W=\rho g V-mg=(W+mg)=ρgV→W=ρgV−mg=
=1000⋅9.81⋅0.03⋅0.04⋅0.09−667⋅0.03⋅0.04⋅0.09⋅9.81=0.35(N)=1000\cdot9.81\cdot0.03\cdot0.04\cdot0.09-667\cdot0.03\cdot0.04\cdot0.09\cdot9.81=0.35(N)=1000⋅9.81⋅0.03⋅0.04⋅0.09−667⋅0.03⋅0.04⋅0.09⋅9.81=0.35(N)
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