Water (density = 1.0 x10³ kg/m³)flows through a horizontal tapered pipe.At the wide end its speed is 4.0 m/s.The difference in pressure between the two ends is 4.5 x10³ Pa.The speed of the water at the narrow end is: A)2.6 m/s B)3.2 m/s C)4.0 m/s D)4.5 m/s E)5.0 m/s

Question

Quizplus · Accepted Answer

The Bernoulli's equation, which relates the speed of a fluid in a pipe to the pressure and height at different points along the pipe, can be used to solve this problem. Since the pipe is horizontal, the height term can be ignored. The equation simplifies to $P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$, where $P$ is pressure, $\rho$ is the density of the fluid, and $v$ is the speed of the fluid. Given that the difference in pressure $\Delta P = P_1 - P_2 = 4.5 \times 10^4$ Pa, the density of water $\rho = 1000$ kg/m$^3$, and the initial speed $v_1 = 4.0$ m/s, we can solve for $v_2$, the speed of the water at the narrow end. Rearranging the equation and solving for $v_2$ gives us $v_2 = \sqrt{v_1^2 + 2\Delta P/\rho}$. Plugging in the given values, we find $v_2 = \sqrt{4.0^2 + 2 \times 4.5 \times 10^4 / 1000} = 5.0$ m/s.

Water (Density = 1