EBK FUNDAMENTALS OF MATERIALS SCIENCE A
5th Edition
ISBN: 9781119175506
Author: RETHWISCH
Publisher: JOHN WILEY+SONS INC.
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Question
Chapter 2.10, Problem 1QP
To determine
The difference between atomic mass and atomic weight
Expert Solution & Answer

Explanation of Solution
Write the difference between atomic mass and atomic weight in tabular form.
S. no | Atomic mass | Atomic weight |
1 | It is defined as adding the number of protons and neutrons in atom. | It is defined as average weight of an element, with respect to all isotopes and their abundance. |
2 | One atomic mass unit (amu) is the one twelfth of the mass of the C-12 isotope. | Ratio of the average mass per atom of the element to 1/12 of the mass of atom of C-12. |
3 |
Expression of atomic mass. Here, A is for atomic mass, Z is for atomic number and N is for neutrons. |
Expression for atomic weight of element. Here, M is for an element,
|
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10.2 For each of the following groups of sources, determineif the three sources constitute a balanced source, and if it is,determine if it has a positive or negative phase sequence.(a) va(t) = 169.7cos(377t +15◦) Vvb(t) = 169.7cos(377t −105◦) Vvc(t) = 169.7sin(377t −135◦) V(b) va(t) = 311cos(wt −12◦) Vvb(t) = 311cos(wt +108◦) Vvc(t) = 311cos(wt +228◦) V(c) V1 = 140 −140◦ VV2 = 114 −20◦ VV3 = 124 100◦ V
Design a typical girder for the floor system shown in the figure below. In addition to the weight of the beam, the dead
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30'
A
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Use the table below.
-
Mn (ft-kips) Mn/ (ft-kips) | Vn (kips) Vn/v (kips)
Shape
W21 × 48
398
265
216
144
W12 × 58
324
216
132
87.8
W16 × 45
309
205
167
111
W18 × 40
294
196
169
113
a. Use LRFD.
Calculate the required moment strength and the maximum shear.
(Express your answers to three significant figures.)
Mu
-
Vu
Select a shape:
-Select-
b. Use ASD.
ft-kips
kips
Calculate the required moment strength and the maximum shear.
(Express your answers to three significant figures.)
Ma =…
Apply single-phase equivalency to determine the linecurrents in the Y-D network shown in Fig. P10.13. The loadimpedances are Zab = Zbc = Zca = (25+ j5) W
Chapter 2 Solutions
EBK FUNDAMENTALS OF MATERIALS SCIENCE A
Ch. 2.10 - Prob. 1QPCh. 2.10 - Prob. 2QPCh. 2.10 - Prob. 3QPCh. 2.10 - Prob. 4QPCh. 2.10 - Prob. 5QPCh. 2.10 - Prob. 6QPCh. 2.10 - Prob. 7QPCh. 2.10 - Prob. 8QPCh. 2.10 - Prob. 9QPCh. 2.10 - Prob. 10QP
Ch. 2.10 - Prob. 11QPCh. 2.10 - Prob. 12QPCh. 2.10 - Prob. 13QPCh. 2.10 - Prob. 14QPCh. 2.10 - Prob. 15QPCh. 2.10 - Prob. 16QPCh. 2.10 - Prob. 17QPCh. 2.10 - Prob. 18QPCh. 2.10 - Prob. 19QPCh. 2.10 - Prob. 20QPCh. 2.10 - Prob. 21QPCh. 2.10 - Prob. 22QPCh. 2.10 - Prob. 23QPCh. 2.10 - Prob. 24QPCh. 2.10 - Prob. 25QPCh. 2.10 - Prob. 26QPCh. 2.10 - Prob. 27QPCh. 2.10 - Prob. 1SSPCh. 2.10 - Prob. 2SSPCh. 2.10 - Prob. 1FEQPCh. 2.10 - Prob. 2FEQPCh. 2.10 - Prob. 3FEQP
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