The discussion centers on calculating the time it takes for a sonar signal to return from the ocean floor, specifically at a depth of 4000 meters with a sound speed of 1500 m/s in seawater. The correct formula to determine the time elapsed is Δt = Δd / ^V{}_{s}, where Δd is the distance traveled (8000 m for a round trip) and ^V{}_{s} is the speed of sound. The total time for the signal to travel to the ocean floor and back is calculated as 8000 m / 1500 m/s, resulting in approximately 5.33 seconds. The discussion highlights the importance of correctly applying the formula for time based on distance and speed.
PREREQUISITESStudents studying physics, marine scientists, sonar technicians, and anyone interested in understanding the principles of sound propagation in water and its applications in underwater navigation and exploration.
Given: [tex]^V{}_{s}[/tex] = 1500 m/s
Δd = 4000 m (The speed of sound in the water in the given conditions
Required: Δt
Analysis: [tex]^V{}_{s}[/tex]= [tex]\frac{\Delta d}{\Delta t}[/tex]
I drop the [tex]\Delta d[/tex] down to the denominator and also bring it across to be the denominator on the other side.
[tex]\frac{^V{}_{s}}{\Delta d}[/tex] =[tex]\frac{\Delta d}{\Delta t \Delta d}[/tex]
I cancel out the [tex]\Delta d[/tex] and end up with
[tex]\Delta d[/tex]= [tex]\frac{^V{}_{s}}{\Delta t}[/tex]
I know my math isn't correct here because I have the correct answer in front of me
I just don't know how they did it. This is what the textbook has
[tex]\Delta t[/tex]= [tex]\frac{\Delta d}{^V{}_{s}}[/tex]
Can you please explain to me how they did this.
Thank you
Scott