Question

answer that question as soon as possible

Answer 1

Hey chika ❤

$\huge \mathfrak{solution}$

Given,

✴distance from the closest wall = 660
✴time taken to hear first echo = 4s
✴ time taken to hear second echo = 2s

CASE I
We know that,
$echo \: = 2 \times distance \: \div \: time$
So,
$echo \: = \: 2 \times 660 \div 4$
$echo \: = 1320 \div 4 \\ echo = 330ms { - }^{1}$
Hence, the velocity of sound is 330m/s.

CASE II
Let the distance be x from the second wall.

So, echo = 2x ÷ (time taken)

Time taken = 4s + 2s = 6s

Now, finding x
=> x = echo × time ÷ 2
=> x = 330 × 6 ÷ 2
=> x = 1980 ÷ 2
=> x = 990m

Hence, the distance from second wall is 990m .

DISTANCE B/W TWO WALLS
= 990 + 660
= 1650 m

Hope it helps ✔