Đáp án:
$\begin{array}{l} h_1=0,044m \\ h_3=0,09(3)m \end{array}$
Giải:
a) Phần chìm của khối gỗ trong nước:
`P=F_{A_1}+F_{A_2}`
→ `10D_0V=10D_1V_1+10D_2V_2`
→ `D_0V=D_1V_1+D_2V_2`
→ `D_0Sa=D_1Sh_1+D_2Sh_2`
→ `D_0a=D_1h_1+D_2h_2`
→ `h_1=\frac{D_0a-D_2h_2}{D_1}=\frac{600.0,1-800.0,02}{1000}=0,044 \ (m)`
b) Chiều cao lớp chất lỏng đổ vào:
`P=F_{A_1}+F_{A_2}+F_{A_3}`
→ `10D_0V=10D_1V_1+10D_2V_2+10D_3V_3`
→ `D_0V=D_1V_1+D_2V_2+D_3V_3`
→ `D_0Sa=D_1Sh_1+D_2Sh_2+D_3Sh_3`
→ `D_0a=D_1(a-h_3)+D_2h_2+D_3h_3`
→ `D_0a=D_1a-D_1h_3+D_2h_2+D_3h_3`
→ `(D_1-D_3)h_3=D_1a-D_0a+D_2h_2`
→ `h_3=\frac{D_1a-D_0a+D_2h_2}{D_1-D_3}=\frac{1000.0,1-600.0,1+800.0,02}{1000-400}=0,09(3) \ (m)`
