• 12 Pages 2011 Summer PS 1 Solutions
    2011 Summer PS 1 Solutions

    School: Stevens

    Course: Materials Processing

    1) Core 2.04 2) Callister 2.6 3) Callister 2.12 4) Callister 2.14 5) Callister 2.15 6) Core 2.21 7) From test 1, spring 2011 8) From test 1, fall 2010 9) Callister 2.23 10) Callister 3.3 11) Callister 3.5 12) Callister 3.7 13) From midterm exam, summer 20

  • 8 Pages 2011 Spring Test 1 Solutions
    2011 Spring Test 1 Solutions

    School: Stevens

    Course: Materials Processing

    This test consists o f 8 pages. There are 5 problems. Questions 6-14 are multiple choice questions. Some constants are given on page 7, and a periodic table is given on page 8. L\~ E -344 Spring 2011 Test 1 3 March, 2011 Name: Honor Pledge: Room: - -~-

  • 11 Pages 2012 Spring Final Solutions
    2012 Spring Final Solutions

    School: Stevens

    Course: Materials Processing

  • 13 Pages 2011 Spring Final Solutions
    2011 Spring Final Solutions

    School: Stevens

    Course: Materials Processing

    C) i . Al, because of need for low density, strength and toughness. i i. Al2 O3, because it is a good thermal insulator. Also, i t is in oxide form, so it is stable at high temperatures. i ii . Al, because fracture cannot be tolerated and Al is much tough

  • 14 Pages 2009 Spring Final Solutions
    2009 Spring Final Solutions

    School: Stevens

    Course: Materials Processing

  • 13 Pages 2010 Fall Final Solutions
    2010 Fall Final Solutions

    School: Stevens

    Course: Materials Processing

  • 10 Pages 2010 Spring Final Solutions
    2010 Spring Final Solutions

    School: Stevens

    Course: Materials Processing

    (0.001m) 2 RA 8 0.2m 10m 1.6 x10 m m L 4 (B) The resistance would increase because: (1) A would get smaller; and (2) would increase slightly because of the defects introduced by cold drawing. (C) The resistivity of pure Nickel wire would be lower than

  • 11 Pages 2011 Fall Final Solutions
    2011 Fall Final Solutions

    School: Stevens

    Course: Materials Processing

  • 9 Pages 2007 Fall Test 2 Solutions
    2007 Fall Test 2 Solutions

    School: Stevens

    Course: Materials Processing

  • 16 Pages 2011 Fall Test 2 Solutions
    2011 Fall Test 2 Solutions

    School: Stevens

    Course: Materials Processing

  • 11 Pages 2011 Spring Test 2 Solutions
    2011 Spring Test 2 Solutions

    School: Stevens

    Course: Materials Processing

  • 8 Pages 2012 Spring Test 2 Solutions
    2012 Spring Test 2 Solutions

    School: Stevens

    Course: Materials Processing

  • 5 Pages 2010 Spring Test 2 Solutions
    2010 Spring Test 2 Solutions

    School: Stevens

    Course: Materials Processing

  • 8 Pages 2010 Fall Test 2 Solutions
    2010 Fall Test 2 Solutions

    School: Stevens

    Course: Materials Processing

  • 10 Pages 2008 Fall Test 2 Solutions
    2008 Fall Test 2 Solutions

    School: Stevens

    Course: Materials Processing

  • 7 Pages 2008 Spring test 2 Solutions
    2008 Spring Test 2 Solutions

    School: Stevens

    Course: Materials Processing

  • 8 Pages 2009 Fall Test 2 Solutions
    2009 Fall Test 2 Solutions

    School: Stevens

    Course: Materials Processing

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  • 11 Pages 2009 Fall Final Solutions
    2009 Fall Final Solutions

    School: Stevens

    Course: Materials Processing

  • 2 Pages Final Exam Sheet
    Final Exam Sheet

    School: Stevens

    Course: Materials Processing

  • 3 Pages Final Exam Sheet
    Final Exam Sheet

    School: Stevens

    Course: Materials Processing

    Rubber has a low elastic modulus, cant be easily recycled due to its crosslinks, typically cant go much plastic deformation prior to failure, is a type of polymer Sharp notches & cracks on the surface of a material loaded in tension are potentionally ver

  • 3 Pages Test 1
    Test 1

    School: Stevens

    Course: Materials Processing

    Metals & alloys shaped by casting, rolling, forging, extrusion Cement hardens by a chemical reaction with water, is a component in both mortar & concrete, can be considered a ceramic material, forms by an exothermic reaction ; doesnt form a drying reacti

  • 1 Page Final sheet page 1
    Final Sheet Page 1

    School: Stevens

    Course: Materials Processing

  • 14 Pages final review sheet
    Final Review Sheet

    School: Stevens

    Course: Materials Processing

  • 9 Pages 2013 Spring PS 11 Solutions
    2013 Spring PS 11 Solutions

    School: Stevens

    Course: Materials Processing

    2013 Spring E344 Solution Set 11 Solution (A) Cementite is much harder but more brittle than ferrite. Increasing the carbon concentration increases the fraction of Fe3C in a steel alloy and will result in a harder and stronger material. (B) Martensite is

  • 9 Pages 2013 Spring PS 12 Solutions
    2013 Spring PS 12 Solutions

    School: Stevens

    Course: Materials Processing

    2013 Spring E344 Solution Set 12 1. (From spring 2006 final) The band gap for pure silicon is 1.12 eV. The donor level for Sb doped into silicon is 0.039 eV below the bottom of the conduction band. If you have samples of pure Si and Sb-doped Si at room te

  • 12 Pages 2013 Spring PS 10 Solutions
    2013 Spring PS 10 Solutions

    School: Stevens

    Course: Materials Processing

    2013 Spring E344 Solution Set 10 1. 4.4 Below, atomic radius, crystal structure, electronegativity, and the most common valence are tabulated, for several elements; for those that are nonmetals, only atomic radii are indicated. Cu Atomic Radius (nm) 0.127

  • 13 Pages 2013 Spring PS 3 Solutions
    2013 Spring PS 3 Solutions

    School: Stevens

    Course: Materials Processing

    2013 Spring E344 Solution Set 3 1. Sodium has the BCC crystal structure. Its density is 0.971 g/cm 3, and its atomic weight is 22.99 g/mole. Use this information to show that the atomic radius of a sodium atom is 0.186 nm. Solution ASSUMPTION: Assuming th

  • 13 Pages 2011 Summer PS 4 Solutions
    2011 Summer PS 4 Solutions

    School: Stevens

    Course: Materials Processing

    6.5 A steel bar 100 mm (4.0 in.) long and having a square cross section 20 mm (0.8 in.) on an edge is pulled in tension with a load of 89,000 N (20,000 lbf), and experiences an elongation of 0.10 mm (4.0 10-3 in.). Assuming that the deformation is entire

  • 15 Pages 2011 Summer PS 3 Solutions
    2011 Summer PS 3 Solutions

    School: Stevens

    Course: Materials Processing

    1. 2.Callister 14.3 3.Callister 14.9 4.Callister 14.10 5.Callister 14.16 6.Callister 14.24 7.Callister 15.15 8.Callister 15.31 9.Callister 15.32 10.From test 1 spring 2011 11.Core 5.21 12.Core 5.18 13.Core 5.19 14.From test 2 spring 2007 15.b 16.b 17.b 18

  • 8 Pages 2013 Spring PS 2 Solutions
    2013 Spring PS 2 Solutions

    School: Stevens

    Course: Materials Processing

    2013 Spring E344 Solution Set 2 1. 2.14 The net potential energy between two adjacent ions, E N, may be represented by the sum of Equations 2.8 and 2.9; that is, EN = AB + r rn Calculate the bonding energy E0 in terms of the parameters A, B, and n using t

  • 8 Pages 2013 Spring PS 1 Solutions
    2013 Spring PS 1 Solutions

    School: Stevens

    Course: Materials Processing

    2013 Spring E344 Solution Set 1 1. Wafers of silicon (Si) are used to make semiconductor devices. The density of Si is 2.33 g/cm 3 and its atomic weight is 28.086 g/mole. Estimate how many Si atoms are contained in a wafer 500 m thick and 50.8 cm in diame

  • 17 Pages 2013 Spring PS 4 Solutions
    2013 Spring PS 4 Solutions

    School: Stevens

    Course: Materials Processing

    2013 Spring E344 Solution Set 4 1 . From Summer 2011 midterm: The -SiC ubit cell is given in the diagram below. A) Show that the SiC unit cell contains 4 Si atoms and 4 C atoms. B) Use the Miller index notation to describe the positions of the 4 C atoms i

  • 7 Pages 2013 Spring PS 8 Solutions
    2013 Spring PS 8 Solutions

    School: Stevens

    Course: Materials Processing

    2013 Spring E344 Solution Set 8 1. 8.4 A polystyrene component must not fail when a tensile stress of 1.25 MPa (180 psi) is applied. Determine the maximum allowable surface crack length if the surface energy of polystyrene is 0.50 J/m 2 (2.86 10-3 in.-lbf

  • 13 Pages 2013 Spring PS 9 Solutions
    2013 Spring PS 9 Solutions

    School: Stevens

    Course: Materials Processing

    2013 Spring E344 Solution Set 9 1. 10.1 Name the two stages involved in the formation of particles of a new phase. Briefly describe each. Solution The two stages involved in the formation of particles of a new phase are nucleation and growth. The nucleati

  • 10 Pages 2013 Spring PS 7 Solutions
    2013 Spring PS 7 Solutions

    School: Stevens

    Course: Materials Processing

    2013 Spring E344 Solution Set 7 1. (From test 2 Fall, 2011) A plain-carbon steel has a yield strength of 35 ksi, a tensile strength of 52 ksi and a modulus of 29,000 ksi. Suppose a 3 inch diameter rod of this steel with a length of 10 feet is loaded in te

  • 10 Pages 2013 Spring PS 5 Solutions
    2013 Spring PS 5 Solutions

    School: Stevens

    Course: Materials Processing

    2013 Spring E344 Solution Set 5 1. 14.2 Compute repeat unit molecular weights for the following: (a) poly(vinyl chloride), (b) poly(ethylene terephthalate), (c) polycarbonate, and (d) polydimethylsiloxane. Solution (c) For polycarbonate, from Table 14.3,

  • 7 Pages 2013 Spring PS 6 Solutions
    2013 Spring PS 6 Solutions

    School: Stevens

    Course: Materials Processing

    2013 Spring E344 Solution Set 6 1. Callister 14.23 For each of the following pairs of polymers, do the following: (1) state whether or not it is possible to determine whether one polymer is more likely to crystallize than the other; (2) if it is possible,

  • 4 Pages Test 1 Sheet
    Test 1 Sheet

    School: Stevens

    Course: Materials Processing

    Metals & alloys can be shaped by casting, rolling, forging, extrusion Cement- hardens by a chemical reaction with water, is a component in both mortar & concrete, can be considered a ceramic material, forms by an exothermic reaction; doesnt form a drying

  • 10 Pages 2008 Fall Final Solutions
    2008 Fall Final Solutions

    School: Stevens

    Course: Materials Processing

  • 1 Page Test 2 open ended sheet 2
    Test 2 Open Ended Sheet 2

    School: Stevens

    Course: Materials Processing

  • 1 Page callister7e_sm_ch02_12
    Callister7e_sm_ch02_12

    School: Stevens

    Course: Materials Processing

    2-12 2.12 (a) The 4f subshell is being filled for the rare earth series of elements. (b) The 5f subshell is being filled for the actinide series of elements. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit bas

  • 1 Page callister7e_sm_ch02_11
    Callister7e_sm_ch02_11

    School: Stevens

    Course: Materials Processing

    2-11 2.11 (a) The 1s22s22p63s23p5 electron configuration is that of a halogen because it is one electron deficient from having a filled p subshell. (b) The 1s22s22p63s23p63d74s2 electron configuration is that of a transition metal because of an incomplete

  • 1 Page callister7e_sm_ch02_10
    Callister7e_sm_ch02_10

    School: Stevens

    Course: Materials Processing

    2-10 2.10 From the periodic table (Figure 2.6) the element having atomic number 112 would belong to group IIB. According to Figure 2.6, Ds, having an atomic number of 110 lies below Pt in the periodic table and in the right-most column of group VIII. Movi

  • 1 Page callister7e_sm_ch02_09
    Callister7e_sm_ch02_09

    School: Stevens

    Course: Materials Processing

    2-9 The Periodic Table 2.9 Each of the elements in Group IIA has two s electrons. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in course

  • 1 Page callister7e_sm_ch02_08
    Callister7e_sm_ch02_08

    School: Stevens

    Course: Materials Processing

    2-8 2.8 The K+ ion is just a potassium atom that has lost one electron; therefore, it has an electron configuration the same as argon (Figure 2.6). The I- ion is a iodine atom that has acquired one extra electron; therefore, it has an electron configurati

  • 1 Page callister7e_sm_ch02_07
    Callister7e_sm_ch02_07

    School: Stevens

    Course: Materials Processing

    2-7 2.7 The electron configurations for the ions are determined using Table 2.2 (and Figure 2.6). P5+: 1s22s22p6 P3-: 1s22s22p63s23p6 Sn4+: 1s22s22p63s23p63d104s24p64d10 Se2-: 1s22s22p63s23p63d104s24p6 I-: 1s22s22p63s23p63d104s24p64d105s25p6 Ni2+: 1s22s22

  • 1 Page callister7e_sm_ch02_06
    Callister7e_sm_ch02_06

    School: Stevens

    Course: Materials Processing

    2-6 2.6 For the L state, n = 2, and eight electron states are possible. Possible l values are 0 and 1, while possible ml values are 0 and 1; and possible ms values are . Therefore, for the s states, the quantum numbers are 1 1 1 1 1 1 200 ( ) and 200 (- )

  • 1 Page callister7e_sm_ch02_05
    Callister7e_sm_ch02_05

    School: Stevens

    Course: Materials Processing

    2-5 2.5 The n quantum number designates the electron shell. The l quantum number designates the electron subshell. The ml quantum number designates the number of electron states in each electron subshell. The ms quantum number designates the spin moment o

  • 1 Page callister7e_sm_ch02_04
    Callister7e_sm_ch02_04

    School: Stevens

    Course: Materials Processing

    2-4 2.4 (a) Two important quantum-mechanical concepts associated with the Bohr model of the atom are (1) that electrons are particles moving in discrete orbitals, and (2) electron energy is quantized into shells. (b) Two important refinements resulting fr

  • 1 Page callister7e_sm_ch02_03
    Callister7e_sm_ch02_03

    School: Stevens

    Course: Materials Processing

    2-3 2.3 (a) In order to determine the number of grams in one amu of material, appropriate manipulation of the amu/atom, g/mol, and atom/mol relationships is all that is necessary, as 1 g / mol 1 mol # g/amu = 23 atoms 1 amu /atom 6.023 x 10 = 1.66 x 10-

  • 1 Page callister7e_sm_ch02_02
    Callister7e_sm_ch02_02

    School: Stevens

    Course: Materials Processing

    2-2 2.2 The average atomic weight of silicon (ASi ) is computed by adding fraction-of-occurrence/atomic weight products for the three isotopes. Thus ASi = f 28 A28 + f 29 A29 + f 30 A30 Si Si Si Si Si Si = (0.9223)(27.9769) + (0.0468)(28.9765) + (0.0309)(

  • 1 Page callister7e_sm_ch02_01
    Callister7e_sm_ch02_01

    School: Stevens

    Course: Materials Processing

    2-1 CHAPTER 2 ATOMIC STRUCTURE AND INTERATOMIC BONDING PROBLEM SOLUTIONS Fundamental Concepts Electrons in Atoms 2.1 Atomic mass is the mass of an individual atom, whereas atomic weight is the average (weighted) of the atomic masses of an atom's naturally

  • 9 Pages ss_11
    Ss_11

    School: Stevens

    1. Callister 9.8 a-e Cite the phases that are present and the phase composition for the following alloys: (a) 15wt% Sn-85 wt% Pb at 100C (212F) (b) 25wt% Pb-75 wt% Mg at 425C (800F) (c) 85wt% Ag-15 wt% Cu at 800C (1470F) (d) 55wt% Zn-45 wt% Cu at 600

  • 3 Pages ps_9
    Ps_9

    School: Stevens

    Problem Set 9 Due: Tuesday, 20 November, 2007 at beginning of lecture E-344 Fall 2007 1. Callister 10.4 Nucleation during nickel solidification 2. Core 9.11 critical size for cube-shaped nucleus 3. Core 9.02 Boron diffusion profiles in silicon 4. C

  • 12 Pages ss_3
    Ss_3

    School: Stevens

    1. Core 8.01 0.75 0.05082 A0 = (0.75in)(0.05082in) = 0.03816in 2 FMAX = 10,000 lbs 10,000lbs ! MAX = = 262ksi 0.03816in 2 ! MAX = 262 > ! T = 65 Thus the tensile testing machine has the capacity to break tensile bars of this steel. 2. Core 8.05

  • 11 Pages ss_2
    Ss_2

    School: Stevens

    1The atomic weight of vanadium (V) is 50.94 g/mole. Its density is =5.96 g/cm3. It has the BCC crystal structure. Calculate the atomic radius of vanadium. Confirm that this matches the value given in Appendix A. 3a = 4 R a 2a V has the BCC crysta

  • 3 Pages ps_2
    Ps_2

    School: Stevens

    Problem Set 2 Due: Wednesday, 30 May 2007 at beginning of lecture E-344 Summer 2007 1. Callister 3.56 Explain why the properties of polycrystalline materials are most often isotropic. 2. Core 3.27 PZT 3. Callister 13.13 viscosity of soda-lime glass

  • 1 Page callister7e_sm_ch02_13
    Callister7e_sm_ch02_13

    School: Stevens

    Course: Materials Processing

    2-13 Bonding Forces and Energies 2.13 The attractive force between two ions FA is just the derivative with respect to the interatomic separation of the attractive energy expression, Equation 2.8, which is just A d- A r = = dr r2 FA = dE A dr The constant

  • 1 Page callister7e_sm_ch02_14
    Callister7e_sm_ch02_14

    School: Stevens

    Course: Materials Processing

    2-14 2.14 (a) Differentiation of Equation 2.11 yields B A d d- r rn = + dr dr dE N dr = nB A - = 0 r (1 + 1) r ( n + 1) (b) Now, solving for r (= r0) A 2 r0 = nB r0( n + 1) or A 1/(1 - n) r0 = nB (c) Substitution for r0 into Equation 2.11 and solving fo

  • 2 Pages callister7e_sm_ch02_15
    Callister7e_sm_ch02_15

    School: Stevens

    Course: Materials Processing

    2-15 2.15 (a) Curves of EA, ER, and EN are shown on the plot below. (b) From this plot r0 = 0.24 nm E0 = 5.3 eV (c) From Equation 2.11 for EN A = 1.436 B = 7.32 x 10-6 n=8 Thus, A 1/(1 - n) r0 = nB 1/(1 - 8) 1.436 = 0.236 nm (8)(7.32 10-6 ) and Excerp

  • 3 Pages Test 2 Sheet
    Test 2 Sheet

    School: Stevens

    Course: Materials Processing

    Rubber has a low elastic modulus, cant be easily recycled due to its crosslinks, typically cant go much plastic deformation prior to failure, is a type of polymer Sharp notches & cracks on the surface of a material loaded in tension are potentionally ver

  • 2 Pages Test 2
    Test 2

    School: Stevens

    Course: Materials Processing

    Rubber has a low elastic modulus, cant be easily recycled due to its crosslinks, typically cant go much plastic deformation prior to failure, is a type of polymer Sharp notches & cracks on the surface of a material loaded in tension are potentionally ver

  • 2 Pages Test 3 Sheet
    Test 3 Sheet

    School: Stevens

    Course: Materials Processing

    Lithography is used in semiconductor device processing for patterning; To create n-type Ge, one could dope using P Consider a parallel plate capacitor where the electrodes are separated by barium titanate. If the barium titanate is removed so that the ga

  • 10 Pages 2007 Fall Final Solutions
    2007 Fall Final Solutions

    School: Stevens

    Course: Materials Processing

  • 1 Page Test 2 open ended sheet 8
    Test 2 Open Ended Sheet 8

    School: Stevens

    Course: Materials Processing

  • 1 Page callister7e_sm_ch04_03
    Callister7e_sm_ch04_03

    School: Stevens

    Course: Materials Processing

    4-3 4.3 This problem calls for the computation of the activation energy for vacancy formation in silver. Upon examination of Equation 4.1, all parameters besides Qv are given except N, the total number of atomic sites. However, N is related to the density

  • 1 Page callister7e_sm_ch04_02
    Callister7e_sm_ch04_02

    School: Stevens

    Course: Materials Processing

    4-2 4.2 Determination of the number of vacancies per cubic meter in gold at 900C (1173 K) requires the utilization of Equations 4.1 and 4.2 as follows: Q Q N A Au N v = N exp v = exp v AAu kT kT = (6.023 10 23atoms / mol)(18.63 g / cm 3) exp 196.9

  • 1 Page callister7e_sm_ch04_01
    Callister7e_sm_ch04_01

    School: Stevens

    Course: Materials Processing

    4-1 CHAPTER 4 IMPERFECTIONS IN SOLIDS PROBLEM SOLUTIONS Vacancies and Self-Interstitials 4.1 In order to compute the fraction of atom sites that are vacant in copper at 1357 K, we must employ Equation 4.1. As stated in the problem, Qv = 0.90 eV/atom. Thus

  • 1 Page callister7e_sm_ch04_04
    Callister7e_sm_ch04_04

    School: Stevens

    Course: Materials Processing

    4-4 Impurities in Solids 4.4 In this problem we are asked to cite which of the elements listed form with Ni the three possible solid solution types. For complete substitutional solubility the following criteria must be met: 1) the difference in atomic rad

  • 1 Page callister7e_sm_ch02_23
    Callister7e_sm_ch02_23

    School: Stevens

    Course: Materials Processing

    2-25 Secondary Bonding or van der Waals Bonding 2.23 The intermolecular bonding for HF is hydrogen, whereas for HCl, the intermolecular bonding is van der Waals. Since the hydrogen bond is stronger than van der Waals, HF will have a higher melting tempera

  • 25 Pages callister7e_sm_ch02_22
    Callister7e_sm_ch02_22

    School: Stevens

    Course: Materials Processing

    2-1 CHAPTER 2 ATOMIC STRUCTURE AND INTERATOMIC BONDING PROBLEM SOLUTIONS Fundamental Concepts Electrons in Atoms 2.1 Atomic mass is the mass of an individual atom, whereas atomic weight is the average (weighted) of the atomic masses of an atom's naturally

  • 1 Page callister7e_sm_ch02_21
    Callister7e_sm_ch02_21

    School: Stevens

    Course: Materials Processing

    2-23 2.21 For silicon, having the valence electron structure 3s23p2, N' = 4; thus, there are 8 N' = 4 covalent bonds per atom. For bromine, having the valence electron structure 4s24p5, N' = 7; thus, there is 8 N' = 1 covalent bond per atom. For nitrogen,

  • 1 Page callister7e_sm_ch02_20
    Callister7e_sm_ch02_20

    School: Stevens

    Course: Materials Processing

    2-22 2.20 Below is plotted the bonding energy versus melting temperature for these four metals. From this plot, the bonding energy for molybdenum (melting temperature of 2617C) should be approximately 7.0 eV. The experimental value is 6.8 eV. Excerpts fro

  • 1 Page callister7e_sm_ch02_19
    Callister7e_sm_ch02_19

    School: Stevens

    Course: Materials Processing

    2-21 2.19 The percent ionic character is a function of the electron negativities of the ions XA and XB according to Equation 2.10. The electronegativities of the elements are found in Figure 2.7. For MgO, XMg = 1.2 and XO = 3.5, and therefore, 2 % IC = 1

  • 1 Page callister7e_sm_ch02_18
    Callister7e_sm_ch02_18

    School: Stevens

    Course: Materials Processing

    2-20 Primary Interatomic Bonds 2.18 (a) The main differences between the various forms of primary bonding are: Ionic-there is electrostatic attraction between oppositely charged ions. Covalent-there is electron sharing between two adjacent atoms such that

  • 1 Page callister7e_sm_ch02_17
    Callister7e_sm_ch02_17

    School: Stevens

    Course: Materials Processing

    2-19 2.17 (a) Differentiating Equation 2.12 with respect to r yields r C dD exp d dE r = dr dr dr C r2 Der / = At r = r0, dE/dr = 0, and (r0 /) C 2 r0 = De (2.12b) Solving for C and substitution into Equation 2.12 yields an expression for E0 as (r0 /)

  • 2 Pages callister7e_sm_ch02_16
    Callister7e_sm_ch02_16

    School: Stevens

    Course: Materials Processing

    2-17 2.16 This problem gives us, for a hypothetical X+-Y- ion pair, values for r0 (0.38 nm), E0 ( 5.37 eV), and n (8), and asks that we determine explicit expressions for attractive and repulsive energies of Equations 2.8 and 2.9. In essence, it is necess

  • 6 Pages ss_8
    Ss_8

    School: Stevens

    1. Callister 6.3 A specimen of copper having a rectangular cross section 15.2 mm x 19.1 mm (0.60 in. x 0.75 in.) is pulled in tension with 44,500 N (10,000 lbf) force, producing only elastic deformation. Calculate the resulting strain. 2. Callister

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