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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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In which of the following is radiation the only type of heat transfer?
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Answer:

C.Vacuum

Explanation:

There are three methods of transfer of heat:

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2) Convection: convection is the transfer of heat by mass movement of molecules. This occurs in fluids (liquids or gases), when an external source of heat is applied to the fluid. As a result, the part of the fluid closer to the source gets warmer, so it becomes less dense and rises, while the colder part sinks and replaces the hotter part, forming a convective current. The process continues until the heat source is removed.

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8 0
3 years ago
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At the very end of Wagner's series of operas The Ring of Nibelung, Brunnhilde takes the golden ring form the finger of the dead
Blababa [14]

Answer:

a) 404 m² b) apparent height = 7.5 m

Explanation:

This question is about refraction and total internal refraction.

Here I will take refractive index of air and water

n_{air}=1\\ n_{water}=1.33=4/3

Now let's look at the diagram I have attached here

At some angle A, the light from the ring (yellow point) under water will be totally internally refracted (B = 90°), which means that rays of light (yellow arrow) that make large enough angle A will not be able to escape from the water. Since we assumed that the ring is a point, there will be a critical cone of angle A with the ring at its apex which traces a circle of radius R on the surface of water, which, beyond this radius, no light could escape.

According to snell's law

\frac{sin(B)}{sin(A)} = \frac{n_{water}}{n_{air}} = 4/3

At critical angle B = 90°

\frac{3)}{4}sin(B) = [tex]\frac{3}{4} sin(90^\circ ) = 0.75 = sin(A)

Therefore

A = 48.6^\circ

With this, we can find the radius of the circle (refer to my diagram)

h* tan (A) = R\\R =11.3 m

And with that we can find the area

A = \pi R^2=404\ m^2

Additional Problem

For apparent depth from above, we can think that, since we are accustomed to seeing light at the speed of c in air, our brain interpret light from <em>any</em> source to be traveling at c. This causes light that originated under water, which has the speed of

v_{water} = \frac{c}{n_{water}} = 0.75c

to appear as if it has traveled with the same duration as light with speed c

In order for this to happen our brain perceive shortened length  which is the apparent depth.

To put it in mathematical term

t_{travel}=\frac{h_{apparent}}{v_{water}} =\frac{h}{c}

So we get apparent depth

h_{apparent}=0.75h = 7.5\ m

4 0
3 years ago
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Solve to find velocity. Remember, the spring is displaced .15 m, not 15!

To find the acceleration, use F = ma. The force being applied to the car is kx, and you know the mass. You do the math.

For problem C I don't know, haven't done that yet in my class. Sorry!
3 0
3 years ago
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