Urgent help please.....
A man is standing asymmetrically between two parallel cliffs. He claps his hands and starts hearing a series of echoes at intervals of 1 second. If the speed of sound in air is 340 m/s, then the distance between the two parallel cliffs will be: (a) 170 m (b) 510 m (c) 340 m (d) 680 m
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Answer:
The time taken for the sound to travel to one cliff and back is the time interval between the claps. Let's denote this time interval as \(T\).
The speed of sound (\(v\)) is given as 340 m/s. The formula for distance (\(d\)) is:
\[d = v \times T\]
Since the sound travels to one cliff and back during each echo, the actual distance between the cliffs is half of the total distance traveled by the sound.
\[Actual \ distance = \frac{1}{2} \times v \times T\]
Now, in the given problem, it's mentioned that the time interval between echoes is 1 second (\(T = 1 \ s\)).
\[Actual \ distance = \frac{1}{2} \times 340 \times 1 = 170 \ m\]
So, the correct option is (a) 170 m.
Since the man hears the echoes at intervals of 1 second, it means that the sound is traveling to one cliff and then bouncing back to the other cliff, and then back to the man. So, the total distance the sound travels is twice the distance between the cliffs.
We know that the speed of sound in air is 340 m/s. Since the sound travels the distance between the cliffs twice, we can calculate the distance using the formula:
Distance = (Speed of Sound) × (Time)
In this case, the time between each echo is 1 second. Plugging in the values, we get:
Distance = 340 m/s × 1 s = 340 meters
However, since the sound travels to one cliff and then back to the other cliff, we need to multiply this distance by 2:
Total Distance = 340 meters × 2 = 680 meters
So, the correct answer is option (d) 680 m.
[tex]samay[/tex]