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  1. Class 9
  2. Chapter 12 Class 9 - Sound

Transcript

Question 1 Page 172 A submarine emits a sonar pulse, which returns from an underwater cliff in 1.02 s. If the speed of sound in salt water is 1531 m/s, how far away is the cliff? Time taken by the sound to return = t = 1.02s Speed of sound in water = v = 1531 m/s Let the distance between the submarine and the cliff = d Hence, Total distance covered by the sound = 2d Now, Speed of sound = (๐ท๐‘–๐‘ ๐‘ก๐‘Ž๐‘›๐‘๐‘’ ๐‘ก๐‘Ÿ๐‘Ž๐‘ฃ๐‘’๐‘™๐‘™๐‘’๐‘‘ ๐‘๐‘ฆ ๐‘ ๐‘œ๐‘ข๐‘›๐‘‘)/(๐‘‡๐‘–๐‘š๐‘’ ๐‘‡๐‘Ž๐‘˜๐‘’๐‘›) v = 2๐‘‘/๐‘ก v ร— t = 2d 2d = v ร— t d = (๐‘ฃ ร— ๐‘ก)/2 d = (1531 ร— 1.02)/2 d = (1531 ร— 102)/(2 ร— 100) d = (1531 ร— 51)/100 d = 78081/100 d = 780.81 m Distance between submarine and cliff is 780.81 m.

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