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Q11PE

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Found in: Page 428

### College Physics (Urone)

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

# (a) What is the fluid speed in a fire hose with a 9.00-cm diameter carrying 80.0 L of water per second? (b) What is the flow rate in cubic meters per second? (c) Would your answers be different if salt water replaced the fresh water in the fire hose?

(a) The speed of fluid is $${\bf{12}}{\bf{.6}}\;{\bf{m/s}}$$.

(b) The flow rate in cubic meter per second is $${\bf{8 \times 1}}{{\bf{0}}^{{\bf{ - 2}}}}\;{{\bf{m}}^{\bf{3}}}{\bf{/s}}$$.

(c) The answer for flow rate of salt water would not change.

See the step by step solution

## Step 1: Determination of fluid speed

(a)

Given Data:

The diameter of fire hose is $$d = 9\;{\rm{cm}} = 0.09\;{\rm{m}}$$

The flow rate of water is $$Q = 80\;{\rm{L/s}}$$

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 fluid is given as:

$$v = \frac{{4Q}}{{\pi {d^2}}}$$

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

\begin{aligned}v &= \frac{{4\left( {80\;{\rm{L}}/{\rm{s}}} \right)\left( {\frac{{1\;{{\rm{m}}^3}}}{{1000\;{\rm{L}}}}} \right)}}{{\pi {{\left( {0.09\;{\rm{m}}} \right)}^2}}}\\v &= 12.6\;{\rm{m}}/{\rm{s}}\end{aligned}

Therefore, the speed of fluidis $$12.6\;{\rm{m}}/{\rm{s}}$$.

## Step 2: Determination of flow rate in cubic meter per second

The flow rate in cubic meter per second is given as:

\begin{aligned}Q &= \left( {80\;{\rm{L}}/{\rm{s}}} \right)\left( {\frac{{1\;{{\rm{m}}^3}}}{{{{10}^3}\;{\rm{L}}}}} \right)\\Q &= 8 \times {10^{ - 2}}\;{{\rm{m}}^3}/{\rm{s}}\end{aligned}

Therefore, the flow rate in cubic meter per second is $$8 \times {10^{ - 2}}\;{{\rm{m}}^3}/{\rm{s}}$$.

## Step 3: Determination of change in answer for salt water

(c)

The flow rate of fluid is independent from the density of fluid so the flow rate for salt water in fire hose remains same.