Answer:
The viscosity [tex]\mathbf{\eta = 7.416 \times 10^{-4} \ N.s/m^2}[/tex]
Explanation:
From the given information:
Time t = 1.55 s
The radius of capillary = 5.00 μm /2
The pressure drop ΔP = 2.45 kPa
The length of the capillary = 2.00 mm
∴
The viscosity of the blood flow can be calculated by using the formula:
[tex]\eta = \dfrac{r^2 \Delta P }{8Lv}[/tex]
where;
v = L/t
Then;
[tex]\eta = \dfrac{r^2 \Delta P }{8L(\dfrac{L}{t})}[/tex]
[tex]\eta = \dfrac{(\dfrac{5 \times 10^{-6} \ m}{2})^2(2.45 \times 10^3 \ Pa) }{8(2.0 \times 10^{-3} \ m ) (\dfrac{2.0 \times 10^{-3} \ m }{1.55 \ s })}[/tex]
[tex]\eta = 7.416 \times 10^{-4} \ Pa.s[/tex]
To (N.s/m²)
[tex]\mathbf{\eta = 7.416 \times 10^{-4} \ N.s/m^2}[/tex]
