91Ó°ÊÓ

According to the principle of continuity, the product of the cross-sectional area and the speed of the fluid remains constant throughout the pipe. So: \(A_1v_1 = A_2v_2\) Here, \(A_1 = 50.0 \,\mathrm{cm}^2 = 5.0 \times 10^{-2} \,\mathrm{m}^2\), \(v_1 = 0.040\, \mathrm{m/s}\), and \(A_2 = 0.500\, \mathrm{cm}^2 = 5.0 \times 10^{-4} \,\mathrm{m}^2\). We will solve for \(v_2\): \(v_2 =\dfrac{A_1v_1}{A_2} = \dfrac{(5.0 \times 10^{-2} \,\mathrm{m}^2)(0.040\, \mathrm{m/s})}{5.0 \times 10^{-4} \,\mathrm{m}^2} = 4\, \mathrm{m/s}\)

Bernoulli's equation states that the total mechanical energy per unit volume remains constant along a streamline, so: \(P_1 + \dfrac{1}{2} \rho v_1^2 = P_2 + \dfrac{1}{2} \rho v_2^2\) Here, \(P_1 = 1.20 \times 10^5\, \mathrm{Pa}\), \(v_1 = 0.040\, \mathrm{m/s}\), and \(v_2 = 4\, \mathrm{m/s}\). The fluid is not specified, but we can assume it's water, with \(\rho = 1000\, \mathrm{kg/m^3}\). We will solve for \(P_2\): \(P_2 = P_1 + \dfrac{1}{2} \rho v_1^2 - \dfrac{1}{2} \rho v_2^2\) \(P_2 = 1.20 \times 10^5\, \mathrm{Pa} + \dfrac{1}{2} (1000\, \mathrm{kg/m^3})(0.040\, \mathrm{m/s})^2 - \dfrac{1}{2} (1000\, \mathrm{kg/m^3})( 4\, \mathrm{m/s})^2\) \(P_2 = 1.20 \times 10^5\, \mathrm{Pa} + 0.8\, \mathrm{Pa} - 8000\, \mathrm{Pa}\) \(P_2 = 1.20 \times 10^5\, \mathrm{Pa} - 7.1992 \times 10^3\, \mathrm{Pa}\) \(P_2 \approx 1.12 \times 10^5\, \mathrm{Pa}\) So the pressure at the narrow end of the pipe segment is approximately \(1.12 \times 10^5\, \mathrm{Pa}\).