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Sagot :
ANSWER:
a) 1.91 cm/s
b) 0.039 cm/s
STEP-BY-STEP EXPLANATION:
a)
Here, since we need the velocity in cm/s, we assume that a small volume of blood passes through the section of the arteriole in a given time.
So, we have to divide the blood flow by the area of the arteriole section, therefore:
[tex]A=\pi\cdot\mleft(\frac{d}{2}\mright)^2[/tex]d = 0.08 mm = 0.008 cm
Replacing:
[tex]A=3.14\cdot\mleft(\frac{0.008}{2}\mright)^2=0.00005024=5.024\cdot10^{-5}cm^2[/tex]We calculate the speed by dividing the rate by the previous calculated area, like this:
[tex]\begin{gathered} v=\frac{q}{A} \\ \text{Replacing} \\ v=\frac{9.6\cdot10^{-5}}{5.024\cdot10^{-5}} \\ v=1.91\text{ cm/s} \end{gathered}[/tex]b)
First we find the area of the section of the capillaries:
[tex]\begin{gathered} d=6\cdot10^{-6}m=6\cdot10^{-4}cm \\ R\text{eplacing} \\ A=3.14\cdot\mleft(\frac{6\cdot10^{-4}}{2}\mright)^2 \\ A=0.0000002826=2.826\cdot10^{-7}cm^2 \end{gathered}[/tex]Now we have to remember that the flow is dividid in equal parts, so the volume by seconds is:
[tex]\begin{gathered} q=\frac{9.6\cdot10^{-5}}{8800} \\ q=1.09\cdot10^{-8}\frac{cm^3}{s} \end{gathered}[/tex]So the speed in this case is:
[tex]\begin{gathered} v=\frac{q}{A} \\ v=\frac{1.09\cdot10^{-8}}{2.826\cdot10^{-7}} \\ v=0.039\text{ cm/s} \end{gathered}[/tex]This latter speed is less than in the main arteriole.
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