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Grade: 10
        
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 ms^-1.)
(A) 3.8 ms^-1 (B) 7.6 m s^-1
(C) 15 m s^-1 (D) 23 m s^-1
 
one year ago

Answers : (1)

ANS B
11 Points
							
distance travelled by sound wave = vt/2 = 2*1520/2 = 1520m
distance travelled by sound wave2 = vt/2 = 2.1*1520/2 = 1596m
d2 – d1 = D = 76m'
v   = distance/time = 76/10 = 7.6m/s
therefore (b) is the correct option
11 months ago
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