a) Equilíbrio de forças: Peso = Empuxo Peixe dentro do barco: $\left(m_{\mathrm{B}}+m_{\mathrm{P}}\right) g=\rho_{\mathrm{H}_{2} \mathrm{O}} V_{\mathrm{B}} g \Rightarrow\left(m_{\mathrm{B}}+m_{\mathrm{P}}\right)=\rho_{\mathrm{H}_{2} \mathrm{O}} V$ Peixe fora do barco: $\left(m_{\mathrm{B}}+m_{\mathrm{P}}\right) g=\rho_{\mathrm{H}_{2} \mathrm{O}} \alpha \cdot\left(V_{\mathrm{B}}+V_{\mathrm{P}}\right) g \Rightarrow\left(m_{\mathrm{B}}+m_{\mathrm{P}}\right)=\rho_{\mathrm{H}_{2} \mathrm{O}} \alpha \cdot\left(V_{\mathrm{B}}+V_{\mathrm{P}}\right)$ De (1) e de (2) temos $\rho_{\mathrm{H}_{2} \mathrm{O}} V_{\mathrm{B}}=\rho_{\mathrm{H}_{2} \mathrm{O}} \alpha \cdot\left(V_{\mathrm{B}}+V_{\mathrm{P}}\right)$; assim: $V_{\mathrm{B}}=\alpha \cdot\left(V_{\mathrm{B}}+V_{\mathrm{P}}\right) \Rightarrow \alpha=\frac{V_{\mathrm{B}}}{\left(V_{\mathrm{B}}+V_{\mathrm{P}}\right)}=\frac{1}{\left(1+\frac{V_{\mathrm{P}}}{V_{\mathrm{B}}}\right)}=\frac{1}{(1+0,25)}=\frac{1}{1,25}=\frac{4}{5}=0,8$ b) Da Lei de Hooke, temos: $T=F_{\text {Hooke }}=k \Delta L=\left(E \frac{A}{L}\right) \Delta L=E A\left(\frac{\Delta L}{L}\right)$ $\Rightarrow A=\frac{T}{E\left(\frac{\Delta L}{L}\right)}=\frac{9,0 \times 10^{2} \mathrm{~N}}{\left(3,0 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}\right) \times(0,10)}=\frac{9,0}{3,0} \times 10^{-6} \mathrm{~m}^{2}=3,0 \times 10^{-6} \mathrm{~m}^{2}$ $A=\pi\left(\frac{d}{2}\right)^{2} \Rightarrow d^{2}=\frac{4 \times A}{\pi}=\frac{4 \times 3,0 \times 10^{-6} \mathrm{~m}^{2}}{3}=4,0 \times 10^{-6} \mathrm{~m}^{2} \Rightarrow d=2,0 \times 10^{-3} \mathrm{~m}=2,0 \mathrm{~mm}$