Final answer:
Using the principle of continuity, the velocity of water in the narrow section of a pipe with half the area of the wide section is found to be 6.4 m/s, double the velocity in the wide section.
Step-by-step explanation:
The question pertains to the physics concept of fluid dynamics, specifically concerning the flow rate of water through a pipe with varying cross-sectional areas. According to the principle of continuity for incompressible fluids, the flow rate (Q) must remain constant throughout the pipe. This principle can be expressed mathematically as Q = A₁v₁ = A₂v₂, where A is the cross-sectional area and v is the fluid velocity at any two points (1 and 2) along the pipe.
Given that the area of the narrow section is one-half the area of the wide section (A₂ = ½A₁), and the velocity in the wide section is v₁ = 3.2 m/s, we can rearrange the equation to solve for v₂ (the velocity in the narrow section) as follows: v₂ = (A₁/A₂) × v₁. Substituting the known values, we get: v₂ = (2/1) × 3.2 m/s = 6.4 m/s. Therefore, the velocity of the fluid in the narrow section is 6.4 m/s.