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Vera_Pavlovna [14]
3 years ago
11

A research vessel is mapping the bottom of the ocean using sonar. It emits a short sound pulse called "ping" downward. The frequ

ency of the sound is 5600 Hz. In water sound propagates at a speed of 1474 m/s. The sound pulse is then reflected back from the bottom of the ocean and it is detected by the vessel 8.88 s after it was emitted. How deep is the ocean just below the vessel?
Physics
1 answer:
Elena L [17]3 years ago
7 0

Answer:

The ocean is 6485.6m deep when measured from the vessel

Explanation:

v=1474m/s

t=8.88s

let d represent distance from the vessel to the ocean bottom.

an echo travels a distance equivalent to 2d, that is to and fro after it reflects from the obstacle.

velocity=\frac{distance}{time}\\ v=\frac{2d}{t} \\vt=2d\\d=\frac{vt}{2}

d=\frac{1474*8.8}{2}

d= 6485.6m

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Answer:

If it is moving 34 m/s it will take 100 seconds, or 1:40 to reach 3400 meters.

Explanation:

I found this answer by dividing 3400 by 34 and converting seconds to minutes

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3 years ago
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What is the resistance of a voltage of 65 V and a current of 2.2 A? Include units.
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Answer:

29.5 ohms

Explanation:

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3 years ago
The next four questions refer to the situation below.
Anna11 [10]

Answer:

 t_{out} = \frac{v_s - v_r}{v_s+v_r} t_{in},      t_{out} = \frac{D}{v_s +v_r}

Explanation:

This in a relative velocity exercise in one dimension,

let's start with the swimmer going downstream

its speed is

         v_{sg 1} = v_{sr} + v_{rg}

The subscripts are s for the swimmer, r for the river and g for the Earth

with the velocity constant we can use the relations of uniform motion

           v_{sg1} = D / t_{out}

           D = v_{sg1}  t_{out}

now let's analyze when the swimmer turns around and returns to the starting point

        v_{sg 2} =  v_{sr}  - v_{rg}

         v_{sg 2} = D / t_{in}

         D = v_{sg 2}  t_{in}

with the distance is the same we can equalize

           v_{sg1} t_{out} = v_{sg2} t_{in}

          t_{out} =  t_{in}

           t_{out} = \frac{v_s - v_r}{v_s+v_r} t_{in}

This must be the answer since the return time is known. If you want to delete this time

            t_{in}= D / v_{sg2}

we substitute

            t_{out} = \frac{v_s - v_r}{v_s+v_r} ()

            t_{out} = \frac{D}{v_s +v_r}

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Basically, melting ice is an endothermic reaction because the ice absorbs (heat) energy, which causes a change to occur.

I hope it's helpful!!

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