Final answer:
The average velocity in the narrower section of the artery is four times the initial velocity and hence, twice the original critical velocity. Since the flow velocity was half of the critical velocity in the wider section, after narrowing, it doubles due to the reduced cross-sectional area.
Step-by-step explanation:
The question relates to the flow of blood through an artery, and it involves applying principles of fluid dynamics to calculate velocities. According to the principle of conservation of mass, the product of cross-sectional area (A) and velocity (v) remains constant for a steady flow (i.e., Q = Av, where Q is the flow rate). If the flow velocity of blood is half of the critical velocity in the first region of the artery with a 4 mm inner diameter, and if the diameter is reduced by half in a second region, we can deduce that the cross-sectional area of the second region will be reduced by a factor of four (since area is proportional to the square of the diameter).
Because the flow rate must remain constant, the velocity in the narrower section of the artery must increase by a factor of four to compensate for the reduced area. Therefore, the average velocity in the narrower section of the artery would be twice the original critical velocity, since the original velocity is half the critical velocity. In terms of the critical velocity in both regions, we cannot calculate the specific values without knowing the initial critical velocity or additional information about the flow conditions. However, we can say that the critical velocity will also scale according to the changes in diameter, since critical velocity is dependent on factors such as the Reynolds number, which in turn depends on the characteristics of the fluid and the geometry of the flow.