A bat flying in a cave emits a sound and receives its echo 0.1 s later. Show that it's distance from the cave wall is 19 m.

To solve this problem, we can utilize the equation for distance, speed, and time. The speed of sound in air is approximately 343 meters per second.

Let's denote the distance of the bat from the wall as d. When the bat emits a sound, it travels towards the wall and gets reflected back as an echo. The total distance the sound wave travels is then 2d (going towards the wall and coming back).

Using the equation for speed (v), distance (d), and time (t), we have:

Speed = Distance / Time

Therefore,

343 m/s = 2d / 0.1 s

To find the value of d, let's rearrange the equation:

2d = 343 m/s * 0.1 s

2d = 34.3 m/s * s

Now divide both sides of the equation by 2:

d = 34.3 m/s * s / 2

d = 17.15 m/s * s

So, the distance of the bat from the cave wall (d) is approximately 17.15 meters.

However, this is only half the distance traveled by the sound wave (since it goes to the wall and back). Therefore, the total distance from the bat to the cave wall is twice the distance we found:

Total distance (2d) = 2 * 17.15 meters
Total distance (2d) = 34.3 meters

Therefore, the distance of the bat from the cave wall is approximately 19 meters (rounded to the nearest meter).