A submarine is fixed underwater and emits two sound wave pulses into the water i

Question

A submarine is fixed underwater and emits two sound wave pulses into the water in the forward direction and then detects the echoes reflected from a moving object ahead. If the time interval between two emitted pulses is 10 s and the round-trip travel time intervals are 2.0 s and 2.1 s respectively, what is the average speed of the object moving away from the submarine? (Given that the speed of sound wave in water is 1520 m middot s^-1.) (A) 3.8 m middot s^-1 (B) 7.6 m middot s^-1 (C) 15 m middot s^-1 (D) 23 m middot s^-1

Explanation / Answer

for the first pulse

distance travelled by first pulse s1 = vt1/2 = (1520*2)/2 = 1520 m

distance travelled by second pulse s2 = vt2/2 = (1520*2.1)/2 = 1596 m

distance travelled by moving submarine s = s2-s1 = 76 m

time taken T = 10 s

speed of moving submarine = s/T = 76/10 = 7.6 m/s <<<--------ANSWER

OPTION (B)

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