Echo-based Valley Width Calculation - Knowing the Speed of Sound

Click For Summary
The discussion focuses on calculating the width of a valley using the speed of sound and echo timing. The commonly accepted speed of sound in air is approximately 343 meters per second, though it can vary based on environmental factors. In the scenario presented, the echoes from the valley walls are heard 6.0 seconds apart, indicating a total travel time of 6.0 seconds for sound to reach both walls and return. This results in a calculated valley width of 1029 meters using the formula d = vt, where d is distance, v is speed, and t is time. The conversation emphasizes the importance of using accurate speed of sound data for precise calculations.
Kdawg
Messages
34
Reaction score
0
Need some help with this question. I think I know how to do it but I don't know the speed of sound. Could someone let me know what it is?

A rifle is fired in a valley with parallel vertical walls. The echo from one wall is heard 6.0 s after the rifle was fired. The echo from the other wall is heard 6.0 s after the first echo. How wide is the valley?
 
Physics news on Phys.org
The generally used value for the speed of sound is 1100ft/s. In reality, the actual value is dependant on barometric pressure, altitude and temperature. Interestingly, 1100 ft/s has become common in textbooks because of an idealized calculation of the speed of sound in introductory fluid dynamics. The calculation is actually wrong, it ignores the pressure effect of compressive heating. But for your problem 1100ft/s will do fine.
 


Sure, I can help you with this question. The speed of sound can vary depending on factors such as temperature, humidity, and altitude. However, a commonly accepted average speed of sound in air at sea level is approximately 343 meters per second (m/s). With this information, we can use the formula d = vt, where d is the distance, v is the speed of sound, and t is the time taken for the sound to travel.

In this scenario, we have two echoes, each taking 6.0 seconds to travel back to the source. Since the sound travels to and from the walls, we can divide the total time by 2 to get the time taken for the sound to travel to one wall and back. This gives us a time of 3.0 seconds for the sound to travel to each wall and back.

Now, we can plug in the known values into the formula: d = (343 m/s)(3.0 s) = 1029 meters. This means that the distance between the two walls is 1029 meters, which is also the width of the valley.

I hope this helps you with your calculation. Remember, the speed of sound can vary, so if you have access to more accurate data, you can use that to get a more precise answer. Good luck!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
2
Views
874
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
4
Views
4K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
16
Views
2K