Question

Both questions plz............. and explain

Answer 1

30. The sound wave travels at 340 m/s and takes 20 seconds to travel to the obstacle and back. This means it travelled 340 * 20 = 6800 m = 6.8 km.

The distance between the submarine and obstacle is half of that, 3.4 km.

31. Regardless of how fast the submarine is moving, the sound is emitted when the island is 2040 metres away. This means that the first half of the journey takes 2040/340 = 6 seconds.

In this 6 seconds, the submarine travelled 6*6 = 36 metres. So, when the sound reaches the island, the distance between the submarine and the island is now 2040 - 36 = 2004 metres.

On the return journey, the sound is travelling towards the submarine at 340 m/s and the submarine is travelling towards the sound at 6 m/s. This means that they have a combined approach speed of 340 + 6 = 346 m/s.

To cover 2004 metres at 346 m/s, you need 2004/346 = 5.79 seconds (2 d.p.)

In total, it takes 6 + 5.79 = 11.8 s (1 d.p.) for the signal to be detected.