Answer:
The speed of the blood in the region of the aorta with plaque buildup is approximately 15.73 cm/s.
Step-by-step explanation:
To solve this problem, we can use the principle of the continuity equation, which states that the flow rate of fluid is constant in a closed system. The formula for the continuity equation is:
A1 * V1 = A2 * V2
Where:
A1 = cross-sectional area at the first point (before plaque buildup)
V1 = velocity at the first point
A2 = cross-sectional area at the second point (with plaque buildup)
V2 = velocity at the second point (what we need to find)
We are given:
Diameter at the first point = 1.8 cm
Diameter at the second point = 1.2 cm
Velocity at the first point = 22 cm/s
First, we need to calculate the cross-sectional areas at the two points using the formula for the area of a circle:
Area = π * (radius)^2
At the first point:
Radius = (diameter/2) = 0.9 cm
A1 = π * (0.9 cm)^2
At the second point:
Radius = (diameter/2) = 0.6 cm
A2 = π * (0.6 cm)^2
Now we can substitute the values into the continuity equation:
A1 * V1 = A2 * V2
(π * (0.9 cm)^2) * 22 cm/s = (π * (0.6 cm)^2) * V2
(0.81π) * 22 cm/s = (0.36π) * V2
17.82 cm/s = 0.36π * V2
V2 = 17.82 cm/s / (0.36π)
V2 ≈ 15.73 cm/s
Therefore, the speed of the blood in the region of the aorta with plaque buildup is approximately 15.73 cm/s.