Answer: So, just as the blood enters the blocked portion of the vessel, its speed is approximately 7.31 times its initial speed v0
Explanation: The flow rate of a fluid is given by the equation Q = A * v, where Q is the flow rate, A is the cross-sectional area of the vessel, and v is the speed of the fluid. Since the blood is incompressible, its flow rate must remain constant. Therefore, if the cross-sectional area of the vessel decreases, the speed of the blood must increase to maintain a constant flow rate.
The cross-sectional area of a cylinder is given by the equation A = πr^2, where r is the radius of the cylinder. If the diameter of the vessel decreases by 63.0%, then its radius decreases by 63.0% as well. Therefore, the new radius of the vessel is r_new = r_old * (1 - 0.63) = 0.37 * r_old.
Substituting this into the equation for the cross-sectional area, we get:
A_new = π * (0.37 * r_old)^2
A_new = (0.37^2) * π * r_old^2
A_new = 0.1369 * A_old
Since the flow rate must remain constant, we have:
Q = A_old * v0 = A_new * v_new
Substituting in the values for A_old and A_new, we get:
v_new = (A_old / A_new) * v0
v_new = (1 / 0.1369) * v0
v_new ≈ 7.31 * v0
So, just as the blood enters the blocked portion of the vessel, its speed is approximately 7.31 times its initial speed v0.