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This manual includes procedure for maintenance, adjustment, service operation and removal and installation of components. All information, illustrations and specifications contained in this manual are based on the latest product information available at the time of manual approval. The right is reserved to make changes at any time without notice.

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1. What two conditions must be met before an entity can be classified as a weak entity? Give an example of a weak entity. To be classified as a weak entity, two conditions must be met: 1. The entity must be existence-dependent on its parent entity. 2. The entity must inherit at least part of its primary key from its parent entity. entity: 1. CLASS is clearly existence-dependent on COURSE. (You can database course exists.) 2. CRS_CODE. The CRS_CODE attribute is also the PK of COURSE. The conditions that define a weak entity are the same as those for a strong relationship between an entity and its parent. In short, the existence of a weak entity produces a strong relationship. And if the entity is strong, its relationship to the other entity is weak. (Note the solid relationship line in the tex decisions. For instance, if the database designer had decided to use a single-attribute as shown in 8, the CLASS entity wou CLASS_CODE, which is not derived from the COURSE entity.) In this case, the relationship between COURSE and CLASS is weak. However, regardless of how the designer classifies the relationship weak or strong CLASS is always existence-dependent on COURSE. 2. A strong relationship exists when en entity is existence-dependent on another entity and inherits at least part of its primary key from that entity. The Visio Professional software shows the strong relationship as a solid line. In other words, a strong relationship exists when a weak entity is related to its parent entity. (Note the discussion in question 1.)

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FUll list Test Bank and Solution Manual 2020-2021 ( Student Saver Team ) -Part 9

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#‎Test_Bank #‎TB
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#‎Appendix
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#‎TestBank #‎Solutionmanual #‎Statistics #‎Calculus
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Referencing has become an integral part of all sorts of academic writing, the major purposes of which are to discourage plagiarism and give credit to the scholars, researchers etc. for their efforts in the growth of knowledge.
There are a number of referencing styles used throughout the world, some of which are more common than the others. These styles differ from each other in a lot of ways. I have tried to include the names of all the common and well-known styles with a detailed description of a few more common and widely used styles (Harvard, APA, Vancouver, MLA, Chicago); and brief description of some less commons (ACS, AMA, AGLC, CSE, IEEE).
Although, automatic reference generators are now available, nevertheless there is no parallel to manual creation of references for it makes you far more familiar with the format of the style. In your hunt of knowledge you come across with a number of sources that are referenced in quite distinct ways, so it is very important to know which way your source is referenced and how you have to convert it into your desired style. Unless you know about the differences in various styles you cannot perform the task. The present manual has been designed in a way to elaborate the major differences that exist among various types of referencing styles so the conversion from one style to the other becomes easier.
The manual begins with the introduction of referencing, defining it, its purposes and various types of styles.
Afterwards the manual breaks up into various styles. Each section of the respective style can broadly be broken into four parts: first part carries an introduction of the style describing the origin and usage of style; the second part is about the general principals of the citation style; the third part carries the information on how to deal with varying situations and in the last part
general format of referencing for various types of sources is provided.
The above division is for the five more common referencing styles, while the five less common styles described in this manual are divided into only two parts: the introduction, and the general principals.

Foreword This collection of problems and solutions is intended to aid students taking our course in Solid State Physics. Exercises are an integral part of a course and the reader is urged to attempt most of them. The problems are selected from areas usually covered in a first course and are of a type most often assigned for class work and given on examinations. No arrangement in order of complexity has been attempted and for some problems only answers are given.
Figure A.1: Scattering from (a) a single electron, (b) two electrons. (c) The scattering
vector s. Note that the vectors k0, k, and s form an isosceles triangle.
where δ is the phase lag of the wave from electron 1 behind that of electron 2 1. (The time
factor has been omitted for the sake of brevity, but its presence is implied.) Referring to
the figure, we may write
δ = (P1M − P1N) 2π/λ = (r · S − r · S0) k
where r is the vector radius of electron 2 relative to electron 1, and S0 and S are the unit
vectors in the incident and scattered directions, respectively. The expression for δ can be
set forth in the form
δ = s · r, (A.3)
where the scattering vector s is defined as
s = k (S − S0) = k − k0. (A.4)
As seen from Fig A.1c, the magnitude of the scattering vector is given by
s = 2k sin(θ) (A.5)
where θ is half of the scattering angle. Substituting the expression Eq (A.3) for δ into
Eq. (A.2), one finds
u′ = fe
A
D
eikD
h
1 + eis·r
i
(A.6)
In deriving this we have chosen the origin of our coordinates at electron 1. But it is now
more convenient to choose the origin at an arbitrary point, and in this manner treat the
two electrons on equal footing. The ensuing expression for the scattered field is then
u′ = fe
A
D
eikD
h
eis·r1 + eis·r2
i
(A.7)
where r1 and r2 are the position vectors of the two electrons relative to a new origin.
Equation (A.6) is a special case of (A.7), where r1 = 0, that is, where the origin is chosen
1The distance D to the field point is assumed to be large, otherwise the denominator D in Eq. (A.2)
would not be the same for the two electrons. This conditions simplifies the calculations, and is the reason
why the detector is usually placed far from the crystal
1. SCATTERING FROM AN ATOM 3
at electron 1, as pointed out above. The generalisation of (A.7) to an arbitrary number
of scatterers is now immediate, and the result is
u′ = fe
A
D
eikD
X
l
eis·rl
where rl is the position of the l th electron, and the sum is carried out over all the electrons.
By analogy with the case of the single electron, equation (A.1), the scattering length for
the system as a whole is now given by the sum
f = fe
X
l
eis·rl . (A.8)
That is, the total scattering length is the sum of individual lengths with the phases taken
properly into account. The intensity I of the scattered beam is proportional to the square
of the magnitude of the field, and therefore
I ∼| f |2= f2
e |
X
l
eis·rl |2 . (A.9)
Results (A.8) and (A.9) are the basic equations in the treatment of scattering and diffraction
processes, and we shall use them time and again in the following pages.
We may digress briefly to point out an important aspect of the scattering process: the
coherence property involved in the scattering. This property means that the scatterers
maintain definite phase relationships with each other. Consequently we can speak of
interference between the partial rays. By contrast, if the scatterers were to oscillate
randomly, or incoherently, the partial rays would not interfere, and the intensity at the
detector would simply be the sum of the partial intensities, that is,
I ∼ Nf2
e ,
where N is the number of scatterers. Note the marked difference between this result and
that of coherent scattering in eq. (A.9).
The scattering length of the electron is well known , and can be found in books on
electromagnetism. Its value is
fe =
h
1 + cos2(2θ)

/2
i1/2
re,
where re, the so–called classical radius of the electron, has a value of about 10−15m. 2
We can now apply these results to the case of a single free atom. In attempting to
apply eq. (A.8), where the sum over the electrons appears, we note that the electrons do
not have discrete positions, but are spread as a continuous charge cloud over the volume
of the atom. It is therefore necessary to convert the discrete sum to the corresponding
integral. This readily leads to
f = fe
X
l
eis·rl → fe
Z
ρ(r)eis·rd3r,
2For the sake of visual thinking, consider the electron to be in the form of a sphere whose radius is
roughly equal to the scattering length fe. Thus the electron ”appears” to the radius as a circular obstacle
of cross section f2
e .
4 A. X-RAY DIFFRACTION
where ρ(r) is the density of the cloud (in electrons per unit volume), and the integral
is over the atomic volume. The atomic scattering factor fa is defined as the integral
appearing in the above expression, i.e,
fa =
Z
ρ(r)eis·rd3r,
(Note that fa is a dimensionless quantity.) The integral can be simplified when the
density ρ(r) is spherically symmetric about the nucleus, because then the integration
over the angular part of the element of volume can be readily performed. The resulting
expression is
fa =
Z R
0
4πr2ρ(r)
sin sr
sr
dr, (A.10)
where R is the radius of the atom (the nucleus being located at the origin). As seen
from eq. (A.10), the scattering factor fa depends on the scattering angle (recall that
s = 2k sin θ), and this comes about from the presence of the oscillating factor (sin sr)/sr
in the integrand. The wavelength of oscillation is inversely proportional to s in Fig A.2a,
and the faster the oscillation – i.e., the shorter the wavelength – the smaller is fa, due
to the interference between the partial beams scattered by different regions of the charge
cloud. Recalling that s = 2k sin θ, Eq (A.5), we see that as the scattering angle 2θ
increases, so also does s, and this results in a decreasing scattering factor fa.
0 3 6 9 12 15
-1
-0.5
0

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Moment-Curvature Relationship: s = c r E = a 0.005 10 b C200A10 9 B D = 100 MPa 1 r = 1 c s EI 1 r = M EI however, M = I c s •12-1. An A-36 steel strap having a thickness of 10 mm and a width of 20 mm is bent into a circular arc of radius 10 m. Determine the maximum bending stress in the strap. r = Ans. s = c r E = a 0.02 4.5 b C131A10 9 B D = 582 MPa 1 r = I c s EI 1 r = M EI however, M = I c s 12-2. A picture is taken of a man performing a pole vault, and the minimum radius of curvature of the pole is estimated by measurement to be 4.5 m. If the pole is 40 mm in diameter and it is made of a glass-reinforced plastic for which GPa, determine the maximum bending stress in the pole. E g = 131 r ϭ 4.5 m 12 Solutions 46060 6/11/10 11:52 AM Page 883

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