Asonar echo returns to a submarine 3.20 s after being emitted. what is the distance to the object creating the echo? (assume that the submarine is in the ocean, not in fresh water.)

2,464m

Explanation:

Echo made by a source is dependent on the velocity of the sound in air/water, the distance between the source and the reflector and the time taken

Using the echo formula

2x = vt

Where x is the distance between the source and the reflector

v is the velocity of the sound in sea water is approximately= 1540m/s (This varies from place to place and depending on the season and nature of water)

t is the time taken

t = 3.20s

Substituting this datas into the equation to get 'x'

2x = 1540 × 3.20

x = 1540×3.20/2

x = 2,464m

The distance between the object creating the echo is 2,464m

Well that's a problem.

The answer depends on the speed of the sonar signal through the water, but the speed of sound in seawater is not a constant value. It varies by a few percent from place to place, from season to season, from morning to evening, and also with the depth of the water.

I'll use the round figure of 1,500 m/s just to show how to handle the problem.

Traveling at an average speed of 1,500 m/s for 3.2 seconds, the ping of the sonar covers

(1,500 m/s) x (3.2 sec) = 4,800 meters

But that's not the distance to the object that reflects the ping back.

Don't forget that the sound had to cover the distance twice . . . once from the sub to the target, and then RETURN to the sub.  So the actual distance from the sub to the target is half of that.

Distance = (4,800 meters / 2) = 2,400 meters .

2b2t

explanation:

2b2t