Question

Submarine produce the ultrasonic waves of velocity 1500 m/s the officer receives signal after 50 sec of emission of ultrasonic waves.find the distance of object which is present at bottom of sea

Answer 1

Explanation:

s=2d/t

s or v=1500m/s

time=50s

2d=1500*50

d=(1500*50)/2

d=37500m

d=37.5km

Answer 2

The distance of object which is present at bottom of sea is 37.5 km.

Explanation:

Given that,

Velocity of the ultrasonic waves, v = 1500 m/s

The officer receives signal after 50 sec of emission of ultrasonic waves, t = 50 s

We need to find the distance of object which is present at bottom of sea . The sound traveled a distance of 2d. The velocity of sound is given by :

$v=\dfrac{d}{t}\\\\2d=vt\\\\d=\dfrac{vt}{2}\\\\d=\dfrac{1500\ m/s\times 50\ s}{2}\\\\d=37500\ m\\\\d=37.5\ km$

So, the distance of object which is present at bottom of sea is 37.5 km.

Learn more,

Ultrasonic waves

https://brainly.in/question/15753374