To solve the problem, we first need to find out how far the sound traveled in the water before it returned to the boat. Given that the time taken for the echo in the water to return is 0.40 seconds, we can use the speed of sound in water to find the distance.
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Calculate the speed of sound in water: The speed of sound in saltwater is approximately 1500 m/s.
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Calculate the total distance traveled in water: \[ \text{Distance} = \text{Speed} \times \text{Time} \] Since the echo takes 0.40 seconds to return, the time to travel to the cliff is half of this: \[ \text{Time to cliff} = 0.40 , \text{s} / 2 = 0.20 , \text{s} \] Thus the distance to the cliff is: \[ \text{Distance to cliff} = 1500 , \text{m/s} \cdot 0.20 , \text{s} = 300 , \text{m} \]
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Calculate the time for the echo to return in air: The sound has to travel the same distance (300 m) back in air. Using the speed of sound in air (340 m/s), we calculate the time it takes to travel this distance: \[ \text{Time in air} = \frac{\text{Distance}}{\text{Speed}} = \frac{300 , \text{m}}{340 , \text{m/s}} \approx 0.8824 , \text{s} \]
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Calculate the time for the echo to return in air after traveling to the cliff: Since this time is for a one-way trip to the cliff, we want the total time for the round trip: \[ \text{Total time in air} = 0.8824 , \text{s} \times 2 \approx 1.7648 , \text{s} \]
Therefore, the additional time it will take the echo in the air to return (after the water echo time of 0.40 seconds) is: \[ \text{Additional time in air} = 1.7648 , \text{s} - 0.40 , \text{s} \approx 1.3648 , \text{s} \]
To summarize, the echo in air takes approximately 1.36 seconds longer to return compared to the echo in water.
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