An oceanic depth-sounding vessel surveys the ocean bottom with ultrasonic waves that travel at 1530 m/s in seawater. Show that when the time delay of an echo to the ocean floor below is 4 s, the depth of the water is 3060 m.
d=vt=1530ms x 2 =3060
not sure if i did it correct or missed a step..i should work on not seconding guessing myself
your equation is correct. What does it represent? How does the "2" in your equation relate to the "4" in the problem?
distance traveled = time x speed
the 2 is half of 4s
thank you, i was making sure i was correct before i responded back to steve.
i thought the 2 represented half of 4s, which is the delay of the echo
You did the calculation correctly! To determine the depth of the water, we can use the formula d = vt, where d is the depth, v is the velocity of the ultrasonic waves, and t is the time delay of the echo.
In this case, the velocity of the ultrasonic waves in seawater is given as 1530 m/s. The time delay of the echo is given as 4 seconds.
By substituting these values into the formula, we get:
d = 1530 m/s x 4 s = 6120 m.
However, it's important to note that the sound waves travel to the ocean floor and back, so we need to divide the distance by 2 to get the depth of the water:
d = 6120 m / 2 = 3060 m.
Therefore, according to your calculations, the depth of the water is indeed 3060 m. Great job!