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Today, 05:37

A ship on the surface of the ocean receives a sonar echo back from an underwater object. The echo path from the object to the ship is at an angle of 35 degrees above the horizontal and takes 3s to return. (Velocity of sound in sea water = 1500m/s). a) What is the distance of the object from the ship? b) What is the depth of the object in the sea?

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  1. Today, 08:05
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    distance of the object from the ship is 2250 m

    depth of the object in the sea is 1289 m

    Explanation:

    Given data

    angle = 35 degrees

    Velocity of sound in sea water = 1500 m/s

    time = 3 s

    to find out

    distance of the object and depth of the object

    solution

    we know that here object make angle 35 degree with ship under the sea

    so height from top sea level to object is D and object to sea distance is d

    so we can say

    2d = velocity * time

    2 d = 1500 * 3 = 4500 m

    so d = 2250 m

    distance of the object from the ship is 2250 m

    and

    we know

    depth = distance * sinθ

    depth = 2250 * sin35

    depth = 2250 * 0.573

    depth = 1289 m

    depth of the object in the sea is 1289 m
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