Ebook Lange pathology flash cards (2nd edition) Part 2

... located near the ostial openings on the intima, thus partly occluding them 95 Systemic Pathology Exercise 23 : Blood Vessels and Lymphatics FIGURE 23 .2: A, Diagrammatic view of the histologic appearance ... FIGURE 27 .5: Chronic peptic ulcer Histologic zones of the ulcer are illustrated The mucosal surface shows necrosis, ulceration, and inflammation 1 12 Exercise 27 : GIT I Systemic Pathology FIGURE 27 .6: ... (Fig 24 .5) iii Areas of brown atrophy are seen 100 FIGURE 24 .4: Myocardial infarction (old) The infarcted area on right shows ingrowth of granulation tissue Exercise 24 : Heart Systemic Pathology...

... Ansari Road, Daryaganj New Delhi 11 0 002, India Phones: + 91- 11- 2327 214 3, + 91- 11- 23272703, + 91- 11- 232820 21, + 91- 11- 23245672, Rel: 32558559 Fax: + 91- 11- 23276490, + 91- 11- 23245683 e-mail: jaypee@jaypeebrothers.com ... Meningitis, Meningioma, Schwannoma, Astrocytoma 95 99 10 2 10 6 11 0 11 4 11 8 12 1 12 6 13 0 13 4 13 7 14 0 14 3 14 6 14 9 15 3 15 6 16 0 Contents xi SECTION V: CYTOPATHOLOGY Exercise 42 Exercise 43 Exercise 44 Basic ... Bone Marrow Examination Bone Marrow Aspiration, Trephine Biopsy 17 7 18 1 18 5 18 7 19 1 19 4 19 9 202 206 213 218 2 21 SECTION VII: AUTOPSY PATHOLOGY Exercise 57 Exercise 58 Introduction to Autopsy Protocol...

... solutions 7 12 24 28 32 35 44 48 50 71 71 72 73 75 80 91 102 10 7 11 4 14 1 14 1 14 4 14 5 14 7 15 1 15 3 16 0 viii The Finite Element Method Further topics in the finite element method 5 .1 The variational ... Consider the equation (2 .13 ) −u = f, < x < 1, 21 Weighted residual and variational methods Table 2.3 The integrals for the Galerkin method i, j vi Lvj dx and 10 15 21 15 10 10 5 84 10 10 35 14 84 15 10 5 ... 24 c0 c1 = , giving c0 = 12 , c1 = 16 The approximate solution is then u 1 (x) = x (1 − x) (1 + 2x) 12 Example 2.6 The final method to be considered is the Galerkin method In this method the integral...

... + 32 (1 + 2U2 + U3 )U2 − 32 (U2 + U3 )U3 = − , − 32 (U2 + U3 ) + 32 (U2 + 2U3 )U3 = − Solving these yields U 22 = 49 , U 32 = 19 21 2 The Finite Element Method Just as in Example 5.5, the positive ... after the conditions U11 = 0, U21 = 29 , U31 = 89 , U41 = 2, U 12 = 1, U 42 = are enforced, becomes ⎤ ⎥ ⎢ ⎢ U 22 ⎥ + ⎣ U 32 ⎦ ⎡ which yields U 22 = 11 , U 32 = 17 29 46 29 29 46 29 ⎡ ⎤ ⎡ ⎤ 0 2 ... Performing the integrations, ⎤ ⎡ 156 22 h 54 −13h kh ⎢ 4h2 13h −3h2 ⎥ ⎥ + EI ⎢ ke = ⎣ 156 22 h ⎦ h3 420 sym 4h2 fe = fh [−6 12 −h h dξ, 4EI N N h2 i j ⎡ ⎢ ⎢ ⎣ 12 − 12 −6h 12 6h 4h2 sym ⎤ 6h 2h2 ⎥ ⎥,...

... metal-metal bonding is present MX2 compounds either have layer structures (e.g TiS2, TiSe2, TiTe2, all CdI2 types) or structures containing diatomic ions (e.g FeS2 has S 22 units and so is formally ... E-E bonding are known (see Topic D5) CaC2 has C 22 ions (isoelectronic with N2) and reacts with water to give ethyne C2H2 On the other hand, KSi and CaSi2 are Zintl compounds with single-bonded ... polyanionic species (e.g P24− isoelectronic with S 22 in Sr2P2, and P73− in Na3P7), although the bonding is certainly not fully ionic Section F Chemistry of nonmetals F7 OXYGEN Key Notes The element...