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### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# The flow rate of blood through a $${\bf{2}}{\bf{.00 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}\;{\bf{m}}$$-radius capillary is $${\bf{3}}{\bf{.80 \times 1}}{{\bf{0}}^{\bf{9}}}\;{\bf{c}}{{\bf{m}}^{\bf{3}}}{\bf{/s}}$$ . (a) What is the speed of the blood flow? (This small speed allows time for diffusion of materials to and from the blood.) (b) Assuming all the blood in the body passes through capillaries, how many of them must there be to carry a total flow of$${\bf{90}}\;{\bf{c}}{{\bf{m}}^{\bf{3}}}{\bf{/s}}$$? (The large number obtained is an overestimate, but it is still reasonable.)

(a) The speed of blood flow is $${\bf{3}}{\bf{.02 \times 1}}{{\bf{0}}^{{\bf{12}}}}\;{\bf{m/s}}$$.

(b) The number of capillaries for total flow is $${\bf{4}}{\bf{.22 \times 1}}{{\bf{0}}^{\bf{7}}}$$.

See the step by step solution

## Step 1: Determination of speed of blood flow

(a)

Given Data:

The radius of capillary is $$r = 2 \times {10^{ - 6}}\;{\rm{m}}$$

The total flow rate is $$q = 90\;{\rm{c}}{{\rm{m}}^3}/{\rm{s}}$$

The flow rate of blood is $$Q = 3.80 \times {10^9}\;{\rm{c}}{{\rm{m}}^3}/{\rm{s}}$$

The flow rate of blood is found by using the formula for flow rate which is equal to product of cross sectional area and speed of blood.

The speed of blood flow is given as:

$$v = \frac{Q}{{\pi {r^2}}}$$

Here, $$v$$ is the speed of blood flow

\begin{aligned}v &= \frac{{\left( {3.80 \times {{10}^9}\;{\rm{c}}{{\rm{m}}^3}/{\rm{s}}} \right)\left( {\frac{{1\;{{\rm{m}}^3}}}{{{{10}^6}\;{\rm{c}}{{\rm{m}}^3}}}} \right)}}{{\pi {{\left( {2 \times {{10}^{ - 6}}\;{\rm{m}}} \right)}^2}}}\\v &= 3.02 \times {10^{12}}\;{\rm{m}}/{\rm{s}}\end{aligned}

Therefore, the speed of blood flow is $$3.02 \times {10^{12}}\;{\rm{m}}/{\rm{s}}$$.

## Step 2: Determination of number of capillaries to carry total flow

(b)

The number of capillaries for total flow is given as:

$$n = \frac{Q}{q}$$

\begin{aligned}n &= \frac{{3.80 \times {{10}^9}\;{\rm{c}}{{\rm{m}}^3}/{\rm{s}}}}{{90\;{\rm{c}}{{\rm{m}}^3}/{\rm{s}}}}\\n &= 4.22 \times {10^7}\end{aligned}

Therefore, the number of capillaries for total flow is $$4.22 \times {10^7}$$.

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