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### College Physics (Urone)

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

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# A physicist at a fireworks display times the lag between seeing an explosion and hearing its sound, and finds it to be $$0.400\;{\rm{s}}$$. (a) How far away is the explosion if air temperature is $$24\;{\rm{^\circ C}}$$ and if you neglect the time taken for light to reach the physicist? (b) Calculate the distance to the explosion taking the speed of light into account. Note that this distance is negligibly greater.

1. The distance at which the explosion occurs is $$138.2\;{\rm{m}}$$.T
2. he distance to explosion is $$138.2\;{\rm{m}}$$.
See the step by step solution

## Step 1: Definition of speed of sound

A sound wave's speed is defined as the distance it travels per unit of time as it propagates through an elastic medium.

## Step 2: Given Data

The temperature is $$T = 24.0\;{\rm{^\circ C}}$$.

The time of travel is $$t = 0.400\;{\rm{s}}$$.

## Step 3: Calculation of the distance

The distance of the object using the speed of sound is,

$$d = vT$$

And

$$v = 331 + 0.6T$$

Plugging the values,

\begin{align}d &= \left( {1331 + 0.6 \times 24} \right)0.400\\ &= 138.2\;{\rm{m}}\end{align}

Therefore, the distance where the explosion occurs is $$138.2\;{\rm{m}}$$.

## Step 4: Determination of distance to explosion

The speed of sound in air at that temperature is,

\begin{align}{v_{air}} &= 331 + 0.6 \times 24\\ &= 345.4\;{\rm{m/s}}\end{align}

The time taken by light is,

$${t_l} = \frac{d}{{3 \times {{10}^8}}}$$

The time taken by sound is,

$${t_s} = \frac{d}{{345.4}}$$

Total time is,

$$t = {t_s} - {t_l}$$

According to the question,

\begin{align}0.400 &= \frac{d}{{345.4}} - \frac{d}{{3 \times {{10}^8}}}\\d &= 138.2\;{\rm{m}}\end{align}

Hence, The distance to explosion is $$138.2\;{\rm{m}}$$.

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