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NNADVOKAT [17]
3 years ago
6

Can someone please illustrate how the refracted ray will look like?

Physics
1 answer:
andrew11 [14]3 years ago
8 0

Answer

As the angle of incidence increases in Figure 2.8, a point is finally reached where the refracted ray does not emerge at the second layer but lie along the interface. This particular angle of incidence at which the angle of refraction is 90° and the refracted ray lies along the interface is known as the critical angle. At and beyond the critical angle, there is no transmitted ray and therefore a very high reflected ray will be recorded .

Therefore,

sinθisin90=Vp1Vp2

But, sin 90 = 1.

At critical angle,

sinθcritical=Vp1Vp2

A critical refracted wave travels along the interface between layers and is refracted back into the upper layer at the critical angle. The waves refracted back into the upper layer are called head waves or first-break refractions because at certain distances from a source, they are the first arriving energy. Recorded first-break refraction is shown in Figure 2.10.

Note that these first-break refractions can give us important information about the shallow velocities on land seismic data.

Note also that seismic data are acquired in such a way that reflections from horizons of interest are in the pre-critical region, even at the farthest offset in the data.

In reality, part of the seismic energy arriving at an interface is transmitted and refracted, and another part of the energy is reflected at that same interface. Given that there are many reflectors in the subsurface, there will be many paths from source to receiver, each of them with a different travel time. The proportion of energy reflected depends on the material properties of the two bounding layers and on the angle of incidence

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klasskru [66]

ans: B

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= \frac{1}{2} (4)(5^{2} )

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ps. the height will only affect its potential energy, not kinetic.

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What is the net force on an object when you combine a force of 10n north with a force of 5n south?
svetoff [14.1K]
Let north as positive
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4 years ago
how fast will and in what direction will a 20kg object accelerate if one force pushes at a 30 degree angle and another pushes at
DiKsa [7]

Answer:

|a|=2.83\ m/s^2

\theta=75^o

Explanation:

<u>Net Force And Acceleration </u>

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\vec F_n=m\vec a

The net force is the vector sum of all forces. In this problem, we are not given the magnitude of each force, only their angles. For the sake of solving the problem and giving a good guide on how to proceed with similar problems, we'll assume both forces have equal magnitudes of F=40 N

The components of the first force are

\vec F_1=

\vec F_1=\ N

The components of the second force are

\vec F_2=

\vec F_2=\ N

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\vec F_n=

\vec F_n=\ N

The magnitude of the net force is

|F_n|=\sqrt{14.64^2+54.64^2}

|F_n|=\sqrt{3200}=56.57\ N

The acceleration has a magnitude of

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\displaystyle |a|=\frac{56.57}{20}

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The direction of the acceleration is the same as the net force:

\displaystyle tan\theta=\frac{54.64}{14.64}

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Distance 350 Km

Time 1 hour

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350Km/h

your answer is a

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3 years ago
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