Suggested languages for you:

Americas

Europe

Q9PE

Expert-verified
Found in: Page 428

### College Physics (Urone)

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

# (a) Estimate the time it would take to fill a private swimming pool with a capacity of 80,000 L using a garden hose delivering 60 L/min. (b) How long would it take to fill if you could divert a moderate size river, flowing at $${\bf{5000}}\;{{\bf{m}}^{\bf{3}}}{\bf{/s}}$$, into it?

(a) The time to fill the swimming pool with a garden hose is $${\bf{1333}}{\bf{.33}}\;{\bf{min}}$$

(b) The time to fill the swimming pool with a moderate size river is $${\bf{1}}{\bf{.6 \times 1}}{{\bf{0}}^{{\bf{ - 2}}}}\;{\bf{sec}}$$

See the step by step solution

## Step 1: Determination of time for filling swimming pool with garden hose

(a)

Given Data:

The capacity of the swimming pool is $$V = 80000\;{\rm{L}}$$

The flow rate of water from garden hose is $$q = 60\;{\rm{L}}/\min$$

The flow rate of moderate size river is $$Q = 5000\;{{\rm{m}}^3}/{\rm{s}}$$

The flow rate of water is the volume of the water passing through the garden hose in per second.

The time to fill the swimming pool by garden hose is given as:

$$t = \frac{V}{q}$$

Here, $$t$$ is the duration to fill swimming pool by garden hose

\begin{aligned}t &= \frac{{80000\;{\rm{L}}}}{{60\;{\rm{L}}/\min }}\\t &= 1333.33\;\min \end{aligned}

Therefore, the time to fill the swimming pool by garden hose is $$1333.33\;\min$$.

## Step 2: Determination of time for filling swimming pool by river

(b)

The time to fill the swimming pool by moderate size river is given as:

$$T = \frac{V}{Q}$$

Here, $$t$$ is the duration to fill the swimming pool by a garden hose

\begin{aligned}T &= \frac{{\left( {80000\;{\rm{L}}} \right)\left( {\frac{{1\;{{\rm{m}}^3}}}{{1000\;{\rm{L}}}}} \right)}}{{5000\;{{\rm{m}}^3}/{\rm{s}}}}\\T &= 1.6 \times {10^{ - 2}}\;{\rm{sec}}\end{aligned}

Therefore, the time to fill the swimming pool by moderate size river is $$1.6 \times {10^{ - 2}}\;{\rm{sec}}$$.