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A cylindrical blood vessel is partially blocked by the buildup of plaque. At one point, the plaque decreases the diameter of th...Asked by Sara
A cylindrical blood vessel is partially blocked by the buildup of plaque. At one point, the plaque decreases the diameter of the vessel by 62.0%. 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, expressed as a multiple of v0?
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Answered by
Damon
If you decrease the diameter by 62% you have 38% of the diameter left open.
Assume constant density of blood
Q = volume flow through per second, the same in the narrow and the wide.
Q = v A where v is the speed and A is the cross sectional area pi D^2/4
so
v * pi (.38* D)^2 /4 = v0 * pi D^2/4
v = v0 /.38^2 = about 6.9 times as fast
Assume constant density of blood
Q = volume flow through per second, the same in the narrow and the wide.
Q = v A where v is the speed and A is the cross sectional area pi D^2/4
so
v * pi (.38* D)^2 /4 = v0 * pi D^2/4
v = v0 /.38^2 = about 6.9 times as fast
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