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Figure 24.5 is the wave traveling? Section 24.2 4. Suppose you stand in front of a tall rock wall that is 170 meters away. If you yell, how long does it take for the echo to get back to your ears if the speed of sound is 340 m/s? 5. A sound wave has a speed of 340 m/s and a wavelength of 10 meters. What is its frequency? Would you be able to hear this sound? 6. The range of human hearing is between 20 Hz and 20,000 Hz. If the speed of sound is 340 m/s, what is the longest wavelength you can hear? What is the shortest? Section 24.3 7. The note E has a frequency of 330 Hz. What is the frequency of the E note one octave higher? als in 8. A note has a frequency of 988 Hz. What is the frequency of the note one octave lower? What note is this?​

User Mndhr
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Answer: B4

(Not including the beginning Figure 24.5 is the wave traveling? Section 24.2 4.)

Step-by-step explanation:

If you yell in front of a tall rock wall that is 170 meters away, it will take approximately 1 second for the echo to get back to your ears. This is because the sound has to travel the distance to the wall and back, so the total distance traveled by the sound is 340 meters (170 meters to the wall and 170 meters back), and sound travels at a speed of 340 m/s, so the time it takes is:

Time = Distance/Speed = 340 meters / 340 m/s = 1 second

The frequency of the sound wave can be calculated using the formula:

frequency = speed/wavelength

So, the frequency of the sound wave with speed 340 m/s and wavelength 10 meters is:

frequency = 340 m/s / 10 meters = 34 Hz

This is a very low frequency sound, which is below the range of human hearing.

The longest wavelength that can be heard by humans is when the frequency is 20 Hz, using the formula:

wavelength = speed/frequency = 340 m/s / 20 Hz = 17 meters

The shortest wavelength that can be heard by humans is when the frequency is 20,000 Hz, using the same formula:

wavelength = speed/frequency = 340 m/s / 20,000 Hz = 0.017 meters or 1.7 centimeters

To find the frequency of the E note one octave higher, we need to double the frequency of the original note. Doubling the frequency is equivalent to increasing it by one octave. Therefore, the frequency of the E note one octave higher is:

330 Hz x 2 = 660 Hz

To find the frequency of the note one octave lower than a frequency of 988 Hz, we need to halve the frequency. Halving the frequency is equivalent to decreasing it by one octave. Therefore, the frequency of the note one octave lower is:

988 Hz / 2 = 494 Hz

This note is B4.

User Lee Loftiss
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