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irinina [24]
3 years ago
9

Sonar is the detection of sound waves reflected off boundaries in water. a region of warm water in a cold lake can produce a ref

lection, as can the bottom of the lake. which would you expect to produce the stronger echo? explain.
Physics
2 answers:
kirill [66]3 years ago
5 0
The stronger echo will originate from the bottom and between the lake since the sound will bounce back harder as the bottom of the lake is solid. Unlike the boundary between the cold and warm water.
svetlana [45]3 years ago
4 0

The echo will be stronger when originated from the bottom of the lake.

Explanation

Echo is created when sound waves get reflected due to difference in the densities of the medium.

As we know that SONAR technique works on the detection of the echo strength to navigate, to determine the thickness of any object in water and also to measure the depth of the water bodies.

A region of warm water and cold water will create very negligible echo as there is no object which can resist the penetration of sound waves in this region.

But the sound waves will get reflected when hits the solid object or the bottom of the lake.

As the bottom of the lake is made up of solid, the sound waves will not be able to penetrate the bottom layer.

This will lead to reflection of sound waves creating echo of stronger energy.

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When several radio telescopes are wired together, the resulting network is called a radio
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(a) 3.1\cdot 10^7 J

The total mechanical energy of the space probe must be constant, so we can write:

E_i = E_f\\K_i + U_i = K_f + U_f (1)

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K_i is the kinetic energy at the surface, when the probe is launched

U_i is the gravitational potential energy at the surface

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U_i is the final gravitational potential energy

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K_i = 5.0 \cdot 10^7 J

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U_i=-G\frac{mM}{R}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{3.3\cdot 10^6 m}=-1.07\cdot 10^8 J

At the final point, the distance of the probe from the centre of Zero is

r=4.0\cdot 10^6 m

so the final potential energy is

U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{4.0\cdot 10^6 m}=-8.8\cdot 10^7 J

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(b) 6.3\cdot 10^7 J

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r=8.0\cdot 10^6 m

which means that at that point, the kinetic energy is zero: (the probe speed has become zero):

K_f = 0

At that point, the gravitational potential energy is

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So now we can use eq.(1) to find the initial kinetic energy:

K_i = K_f + U_f - U_i = 0+(-4.4\cdot 10^7 J)-(-1.07\cdot 10^8 J)=6.3\cdot 10^7 J

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To develop this problem it will be necessary to apply the concepts related to the frequency of a spring mass system, for which it is necessary that its mathematical function is described as

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