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第112页

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$ 解：(1)m=ρV=0.6×10^{3}\ \text {kg/m}^3× 200×10^{-6}\ \text {m}^3=0.12\ \text {kg}$

$F_浮=G=mg=0.12\ \text {kg}×10\ \text {N/kg}=1.2\ \text {N}$

$(2)ρ_油=\frac {F_浮}{gV_排}=\frac {1.2\ \text {N}}{10\ \text {N/kg}× \frac 23×200×10^{-6}\ \text {m}^3}=0.9×10^{3}\ \text {kg/m}^3$

$(3)V_总=\frac {m_总}{ρ_油}=\frac {1×10^{3}\ \text {kg}}{0.9×10^{3}\ \text {kg/m}^3}=\frac {10}9\ \text {m}^3$

$1\ \text {t}原有可换成桶数n=\frac {\frac {10}9\ \text {m}^3}{\frac 16\ \text {m}^3}≈6.7$

$解：(1)G=12\ \text {N}$

$(2)F_{浮}=G-F=12\ \text {N}- 4\ \text {N}= 8\ \text {N}$

$(3)V=\frac {F_浮}{ρ_水g}=\frac {8\ \text {N}}{1×10^{3}\ \text {kg/m}^3×10\ \text {N/kg}}=8×10^{-4}\ \text {m}^3$

$(4)m=\frac {G}{g}=\frac {12\ \text {N}}{ 10\ \text {N/kg}}=1.2\ \text {kg}$

$ρ=\frac {m }{V}=\frac {1.2\ \text {kg}}{8×10^{-4}\ \text {m}^3}=1.5×10^{3}\ \text {kg/m}^3$