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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?

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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A solid sphere has a radius of 0.200 m and a mass of 150.0 kg. how much work is required to get the sphere rolling with an angul

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Here in this case we can use work energy theorem

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Work done by all forces = Change in kinetic Energy of the object

Total kinetic energy of the solid sphere is ZERO initially as it is given at rest.

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also we know that

$I = \frac{2}{5}mR^2$

$w= \frac{v}{R}$

Now kinetic energy is given by

$KE = \frac{1}{2}(\frac{2}{5}mR^2)w^2 +\frac{1}{2} m(Rw)^2$

$KE = \frac{1}{5}mR^2w^2 +\frac{1}{2} mR^2w^2$

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Work done = 10500 - 0 = 10500 J

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3 years ago

If a weight hanging on a string of length 5 feet swings through 5^\circ on either side of the vertical, how long is the arc thro

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3 years ago

A 0.750 kg block is attached to a spring with spring constant 13.0 N/m . While the block is sitting at rest, a student hits it w

trapecia [35]

To solve this problem we will apply the concepts related to energy conservation. From this conservation we will find the magnitude of the amplitude. Later for the second part, we will need to find the period, from which it will be possible to obtain the speed of the body.

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Replacing,

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