/*! 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 39 An observer stands \(25 \mathrm{... [FREE SOLUTION] | 91Ó°ÊÓ

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An observer stands \(25 \mathrm{m}\) behind a marksman practicing at a rifle range. The marksman fires the rifle horizontally, the speed of the bullets is \(840 \mathrm{m} / \mathrm{s},\) and the air temperature is \(20^{\circ} \mathrm{C} .\) How far does each bullet travel before the observer hears the report of the rifle? Assume that the bullets encounter no obstacles during this interval, and ignore both air resistance and the vertical component of the bullets' motion.

Short Answer

Expert verified
Each bullet travels approximately 86.32 meters before the observer hears the report of the rifle.

Step by step solution

01

Determine the Speed of Sound

At an air temperature of \(20^{\circ} \mathrm{C}\), the speed of sound is approximately \(343 \mathrm{m/s}\). This value will be used to calculate the time it takes for sound to reach the observer.
02

Calculate the Time for Sound to Travel 25m

Using the speed of sound, we can find the time it takes for the sound to travel \(25 \mathrm{m}\) to the observer. The formula is: \[ t = \frac{\text{distance}}{\text{speed}} = \frac{25}{343}. \] Thus, \( t \approx 0.073 \mathrm{s} \).
03

Find the Distance the Bullet Travels in that Time

Since the bullet travels at \(840 \mathrm{m/s}\), we can calculate the distance it travels in \(0.073\, \mathrm{s}\) using the formula: \[ d = \text{speed} \times \text{time} = 840 \times 0.073 \]. Therefore, \( d \approx 61.32 \mathrm{m} \).
04

Calculate the Total Distance Traveled by Bullet

The total distance the bullet travels is the sum of the distance it travels in the time before the observer hears the gunshot and the initial \(25\, \mathrm{m}\) the bullet covers while the sound also travels: \[ \text{Total distance} = 61.32 + 25 = 86.32 \mathrm{m}. \]

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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 a fundamental concept in physics that refers to how fast sound waves travel through a medium, typically air. It depends on factors such as air temperature and pressure. At an air temperature of \(20^{\circ} \mathrm{C}\), sound travels at approximately \(343 \mathrm{m/s}\). This speed is critical for various calculations, especially when dealing with situations where sound needs to be accurately timed, such as determining when an observer hears a distant event. In the given problem, using the speed of sound helps us compute how quickly the sound from the rifle shot reaches the observer. This value of \(343 \mathrm{m/s}\) is an average used in many physics problems to ensure consistency in calculations.
Projectile Motion
Projectile motion refers to an object moving in a curved path under the influence of gravity alone, after an initial force is applied. In this scenario, though the bullet is a projectile, we focus solely on its horizontal motion because gravity's effect on the bullet's vertical motion and air resistance are ignored for simplicity. In reality, projectile motion is quite complex and involves both horizontal and vertical components. The initial force from the rifle propels the bullet horizontally at a constant speed of \(840 \mathrm{m/s}\), which is crucial for determining how far the bullet travels before sound reaches the observer. Understanding projectile motion allows us to separate motions into independent components that can be analyzed and computed separately.
Horizontal Motion
Horizontal motion is an essential part of the larger projectile motion, described previously. It describes the bullet's consistent movement in a straight line, unaffected by vertical forces, as assumed in this problem. Given that the bullet's horizontal speed is \(840 \mathrm{m/s}\), it highlights the bullet's swift journey across the firing range.
  • Horizontal speed remains constant assuming no air resistance.
  • Distance covered can be calculated simply using the formula \(d = \text{speed} \times \text{time}\).
In this problem, horizontal motion helps determine the precise distance a bullet travels during a specified period before the observer hears the sound. It simplifies the scenario, allowing a straightforward calculation of the bullet's path as a horizontal line.
Time Calculation
Time calculation is at the heart of solving this physics problem. It involves determining how long it takes for sound to travel a certain distance to reach the observer. This calculation is performed using the formula \( t = \frac{\text{distance}}{\text{speed}} \), which helps us find the required time to travel a specific path.
  • Calculate the time it takes for sound to travel \(25 \mathrm{m}\): \( t = \frac{25}{343} \approx 0.073 \mathrm{s}\).
  • Determine the bullet's travel distance in the same period: \( d = 840 \times 0.073 \approx 61.32 \mathrm{m}\).
These time-based calculations are foundational in physics as they enable us to relate speed, distance, and overall motion dynamically. By understanding time calculation, it becomes clearer how dynamic processes unfold and interact over time.

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Most popular questions from this chapter

A sound wave travels twice as far in neon (Ne) as it does in krypton (Kr) in the same time interval. Both neon and krypton can be treated as monatomic ideal gases. The atomic mass of neon is \(20.2 \mathrm{u}\), and the atomic mass of krypton is \(83.8 \mathrm{u}\). The temperature of the krypton is \(293 \mathrm{K} .\) What is the temperature of the neon?

The siren on an ambulance is emitting a sound whose frequency is \(2450 \mathrm{Hz}\). The speed of sound is \(343 \mathrm{m} / \mathrm{s}\). (a) If the ambulance is stationary and you (the "observer") are sitting in a parked car, what are the wavelength and the frequency of the sound you hear? (b) Suppose that the ambulance is moving toward you at a speed of \(26.8 \mathrm{m} / \mathrm{s}\). Determine the wavelength and the frequency of the sound you hear. (c) If the ambulance is moving toward you at a speed of \(26.8 \mathrm{m} / \mathrm{s}\) and you are moving toward it at a speed of \(14.0 \mathrm{m} / \mathrm{s}\), find the wavelength and frequency of the sound you hear.

As a prank, someone drops a water-filled balloon out of a window. The balloon is released from rest at a height of \(10.0 \mathrm{m}\) above the ears of a man who is the target. Then, because of a guilty conscience, the prankster shouts a warning after the balloon is released. The warning will do no good, however, if shouted after the balloon reaches a certain point, even if the man could react infinitely quickly. Assuming that the air temperature is \(20^{\circ} \mathrm{C}\) and ignoring the effect of air resistance on the balloon, determine how far above the man's ears this point is.

A copper wire, whose cross-sectional area is \(1.1 \times 10^{-6} \mathrm{m}^{2}\), has a linear density of \(9.8 \times 10^{-3} \mathrm{kg} / \mathrm{m}\) and is strung between two walls. At the ambient temperature, a transverse wave travels with a speed of \(46 \mathrm{m} / \mathrm{s}\) on this wire. The coefficient of linear expansion for copper is \(17 \times 10^{-6}\left(\mathrm{C}^{\circ}\right)^{-1},\) and Young's modulus for copper is \(1.1 \times 10^{11} \mathrm{N} / \mathrm{m}^{2} .\) What will be the speed of the wave when the temperature is lowered by \(14 \mathrm{C}^{\circ}\) ? Ignore any change in the linear density caused by the change in temperature.

A loudspeaker has a circular opening with a radius of \(0.0950 \mathrm{m}\). The electrical power needed to operate the speaker is \(25.0 \mathrm{W}\). The average sound intensity at the opening is \(17.5 \mathrm{W} / \mathrm{m}^{2} .\) What percentage of the electrical power is converted by the speaker into sound power?

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