/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 It was a dark and stormy night, ... [FREE SOLUTION] | 91Ó°ÊÓ

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It was a dark and stormy night, when suddenly you saw a flash of lightning. Three-and-a-half seconds later you heard the thunder. Given that the speed of sound in air is about \(340 \mathrm{m} / \mathrm{s}\) how far away was the lightning bolt?

Short Answer

Expert verified
The lightning bolt was 1190 meters away.

Step by step solution

01

Understand the data available

You know that there is a time delay of 3.5 seconds between seeing the lightning and hearing the thunder. You also know that the speed of sound in air is approximately 340 m/s.
02

Determine the formula to use

The distance that sound travels can be found using the formula \[ \text{Distance} = \text{Speed} \times \text{Time}. \]
03

Plug the known values into the formula

In this case, the speed is 340 m/s and the time is 3.5 seconds. By substituting these values into the formula, we get:\[ \text{Distance} = 340 \mathrm{m/s} \times 3.5 \mathrm{s}. \]
04

Calculate the result

Multiply the values to find the distance:\[ \text{Distance} = 1190 \mathrm{meters}. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Speed of Sound
The speed of sound is the rate at which sound waves travel through a medium, like air. This speed can vary depending on factors such as temperature, humidity, and altitude. In general, sound travels faster in warmer conditions because molecules move more quickly and can transmit sound waves more efficiently. The typical speed of sound in air at room temperature is about 340 meters per second (m/s).

Understanding the speed of sound is crucial in many practical scenarios, such as determining how far away a storm is by counting the seconds between seeing lightning and hearing thunder. The speed allows us to calculate distances using sound, a handy tool in everyday life and in scientific applications.
Distance Calculation
Calculating distance with speed and time is a straightforward process that involves multiplying the two values. When you know the speed at which an object is moving, and the duration it has been moving, you can easily find the distance it has covered using the formula:
  • \[ \text{Distance} = \text{Speed} \times \text{Time} \]
This method is widely used in physics for solving various motion-related problems. In this exercise, we apply this formula to sound traveling through air. By knowing the speed of sound and the time delay, we determine how far away the lightning bolt struck by multiplying the two.

For example, if you heard thunder 3.5 seconds after seeing lightning, and the speed of sound in air is 340 m/s, the calculation would be:
  • \[ \text{Distance} = 340 \text{ m/s} \times 3.5 \text{ s} = 1190 \text{ meters} \]
Time Delay
Time delay refers to the time it takes for sound to travel from one point to another. In the context of thunder and lightning, the delay between seeing the lightning flash and hearing the thunder clap allows us to estimate the distance to the storm. This delay occurs because light travels much faster than sound.

Light speed is about 299,792 kilometers per second, which means you see the flash almost instantly, but the sound takes longer to reach you. By measuring this delay, you can calculate how far away the storm is. In our example, calculating the 3.5-second delay between the lightning and thunder helped determine the storm was 1190 meters away. Being able to interpret time delays like this is a valuable skill for understanding natural phenomena and applying mathematical principles.

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Most popular questions from this chapter

A glaucous-winged gull, ascending straight upward at \(5.20 \mathrm{m} / \mathrm{s},\) drops a shell when it is \(12.5 \mathrm{m}\) above the ground. \((a)\) What are the magnitude and direction of the shell's acceleration just after it is released? (b) Find the maximum height above the ground reached by the shell. (c) How long does it take for the shell to reach the ground? (d) What is the speed of the shell at this time?

A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of \(3.5 \mathrm{m} / \mathrm{s}\). Two seconds later the bicyclist hops on his bike and accelerates at \(2.4 \mathrm{m} / \mathrm{s}^{2}\) until he catches his friend. (a) How much time does it take until he catches his friend? (b) How far has he traveled in this time? \((c)\) What is his speed when he catches up?

A model rocket rises with constant acceleration to a height of \(3.2 \mathrm{m},\) at which point its speed is \(26.0 \mathrm{m} / \mathrm{s} .\) (a) How much time does it take for the rocket to reach this height? (b) What was the magnitude of the rocket's acceleration? (c) Find the height and speed of the rocket \(0.10 \mathrm{s}\) after launch.

A hot-air balloon is descending at a rate of \(2.0 \mathrm{m} / \mathrm{s}\) when a passenger drops a camera. If the camera is \(45 \mathrm{m}\) above the ground when it is dropped, (a) how long does it take for the camera to reach the ground, and (b) what is its velocity just before it lands? Let upward be the positive direction for this problem.

A child rides a pony on a circular track whose radius is \(4.5 \mathrm{m}\). (a) Find the distance traveled and the displacement after the child has gone halfway around the track. (b) Does the distance traveled increase, decrease, or stay the same when the child completes one circuit of the track? Explain. (c) Does the displacement increase, decrease, or stay the same when the child completes one circuit of the track? Explain. (d) Find the distance and displacement after a complete circuit of the track.

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