A cylindrical blood vessel is partially blocked by the buildup of plaque. At one point, the plaque decreases the diameter of the vessel by 50.0%. The blood approaching the blocked portion has speed v0. Just as the blood enters the blocked portion of the vessel, what is its speed v, as expressed as a multiple of v0?
To determine the speed of blood as it enters the blocked portion of the vessel, we can apply the principle of conservation of mass flow rate.
The equation for mass flow rate is given by:
ρ1 * A1 * v1 = ρ2 * A2 * v2
where ρ is the density of the blood, A is the cross-sectional area of the vessel, and v is the velocity of the blood at each point.
Since the blood is incompressible, the density remains constant.
A1 * v1 = A2 * v2
The cross-sectional area of a cylindrical vessel is given by A = πr^2, where r is the radius of the vessel.
Let the initial radius of the vessel be R and the final radius after the blockage be R/2.
Initial area, A1 = πR^2
Final area, A2 = π(R/2)^2 = π(R^2)/4
Applying the conservation of mass flow rate equation:
A1 * v1 = A2 * v2
πR^2 * v0 = (πR^2)/4 * v
Solving for v:
v = 4v0
Therefore, the speed of the blood as it enters the blocked portion of the vessel is 4 times the initial speed v0.