Question

UltraUltrasonic wave from Sheffield rock at the bottom of the sea bed and comes back after 0.5 seconds calculate the distance to the rock from the ship take speed of sound in seawater is 1522 m/s

Answer 1

Answer:

Distance = Speed × Time

As the time taken is for the distance covered during the forward and backard path, it is divided by 2,

d = v × t /2

d = 1522 × 0.5 /2

d = 761 × 0.5

d = 380.5 m

❤️

Answer 2

Answer:

Heya mate ☺️

given,v=1522m/s

t=0.5s

In this condition,

d=vt/2

=1522×0.5/2

=751/2

=375.5m

•°• the distance will be 375.5 m (approx)

Explanation:

Mark as brainliest