Final answer:
The diameter of the constricted portion of the pipe is approximately 1.34 cm.
Step-by-step explanation:
To find the diameter of the constricted portion of the pipe, we can use the principle of continuity equation. According to the continuity equation, the product of the cross-sectional area and the fluid velocity is constant along the pipe.
Let's denote the initial diameter of the pipe as D1 and the final diameter as D2. We can write the continuity equation as:
A1 * V1 = A2 * V2
Where A1 and V1 are the initial cross-sectional area and velocity, and A2 and V2 are the final cross-sectional area and velocity.
We are given that the initial diameter, D1, is 3.0 cm and the initial velocity, V1, is 3.0 m/s. We are also given that the final velocity, V2, is 15 m/s. Let's plug in these values to solve for the final cross-sectional area and then calculate the final diameter, D2.
Solving the equation:
A1 * V1 = A2 * V2
=> (π * (D1/2)^2) * V1 = (π * (D2/2)^2) * V2
=> (π * (3.0/2)^2) * 3.0 = (π * (D2/2)^2) * 15
=> (π * (9/4)) * 3.0 = (π * (D2/2)^2) * 15
=> (π * (9/4)) * 3.0 = (π * (D2/2)^2) * 15
Simplifying the equation:
(9/4) * 3.0 = (D2/2)^2 * 15
=> (27/4) = (D2/2)^2 * 15
=> (D2/2)^2 = (27/4) / 15
=> (D2/2)^2 = 27/60
=> (D2/2)^2 = 9/20
=> D2/2 = sqrt(9/20)
=> D2/2 ≈ 0.6708
=> D2 ≈ 2 * 0.6708
=> D2 ≈ 1.3416 cm
Therefore, the diameter of the constricted portion of the pipe is approximately 1.34 cm.