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2 years ago

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A small artery has a length of 1.10 × 10-3 m and a radius of 2.50 × 10-5 m. If the pressure drop across the artery is 1.15 kPa,

what is the flow rate through the artery? Assume that the temperature is 37°C and the viscosity of whole blood is 2.084 × 10-3 Pa·s.

1 answer:

levacccp [35]2 years ago

8 0

Answer:

$7.69533\times 10^{-11}\ m^3/s$

Explanation:

P = Pressure difference = 1.15 kPa

r = Radius = $2.5\times 10^{-5}\ m$

$\eta$ = Viscosity of liquid = $2.084\times 10^{-3}\ Pas$

l = Length of artery = $1.1\times 10^{-3}\ m$

From Poiseuille's equation we have

$Q=\frac{\pi Pr^4}{8\eta l}\\\Rightarrow Q=\frac{\pi 1.15\times 10^3\times (2.5\times 10^{-5})^4}{8\times 2.084\times 10^{-3}\times 1.1\times 10^{-3}}\\\Rightarrow Q=7.69533\times 10^{-11}\ m^3/s$

The flow rate of blood is $7.69533\times 10^{-11}\ m^3/s$

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<u>Electrostatic Force</u>

Two point-charges $q_1$ and $q_2$ separated a distance d will exert a force on each other of a magnitude given by the Coulomb's formula

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The diagram provided in the question shows four identical charges (let's assume their value is Q) separated by identical distance (of value d). The force between the charges next to others is

$\displaystyle F_1=\frac{k\ Q\ Q}{d^2}$

$\displaystyle F_1=\frac{k\ Q^2}{d^2}$

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$\displaystyle F_2=\frac{k\ Q^2}{4d^2}$

And the force between the charges A and D is

$\displaystyle F_3=\frac{k\ Q^2}{(3d)^2}$

$\displaystyle F_3=\frac{k\ Q^2}{9d^2}$

Now, let's analyze each charge and the force applied to them by the others

Let's recall equally signed charges repel each other and differently signed charges attrach each other

Charge A. It receives force to the left from B and C and to the right from D

$\displaystyle F_A=-F_1-F_2+F_3=-\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{9d^2}$

$\displaystyle F_A=\frac{k\ Q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})$

$\displaystyle F_A=-\frac{41}{36}F_1$

Charge B. It receives force to the right from A and D and to the left from C

$\displaystyle F_B=F_1-F_1+F_2=\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{d^2}+\frac{k\ Q^2}{4d^2}$

$\displaystyle F_B=\frac{1}{4}F_1$

Charge C. It receives forces to the right from all charges.

$\displaystyle F_C=F_2+F_1+F_1=\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{d^2}+\frac{k\ Q^2}{d^2}$

$\displaystyle F_C=\frac{9}{4}F_1$

Charge D. It receives forces to the left from all charges

$\displaystyle F_D=-F_3-F_2-F_1=-\frac{k\ Q^2}{9d^2}-\frac{k\ Q^2}{4d^2}-\frac{k\ Q^2}{d^2}$

$\displaystyle F_D=-\frac{49}{36}F_1$

Comparing the magnitudes of each force is just a matter of computing the fractions

$\displaystyle \frac{41}{36}=1.13,\ \frac{1}{4}=0.25,\ \frac{9}{4}=2.25,\ \frac{49}{36}=1.36$

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