A SONAR (sound navigation and ranging) uses u | vedclass

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(C) Let the distance between the submarine and the enemy submarine be $S$.The speed of sound in water is $v = 1450\; m/s$.The total time delay between

Vedclass · Accepted Answer

$55.8$

Vedclass · Answer

(C) Let the distance between the submarine and the enemy submarine be $S$.The speed of sound in water is $v = 1450\; m/s$.The total time delay between the transmission and reception of the echo is $t = 77.0\; s$.Since the sound travels to the enemy submarine and back to the source,the total distance covered by the sound wave is $2S$.Using the formula $	ext{Distance} = 	ext{Speed} 	imes 	ext{Time}$,we have $2S = v 	imes t$.Substituting the values: $2S = 1450\; m/s 	imes 77.0\; s = 111650\; m$.Therefore,the distance $S = \frac{111650}{2} = 55825\; m$.Converting to kilometers: $S = 55.825\; km \approx 55.8\; km$.

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