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Q10PE

Expert-verifiedFound in: Page 629

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**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.**

- The distance at which the explosion occurs is \(138.2\;{\rm{m}}\).T
- he distance to explosion is \(138.2\;{\rm{m}}\).

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

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

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

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 \(138.2\;{\rm{m}}\).

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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