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9 Feb 2004: From the dynaboundary condition2.20!, we find thatm(12 n) 5 2; whilefrom 2.18!, we find thatmn5 2. ... 4H P1h 1 hU P1H 0!85 0, 4.20! 14U P11 H P18 1.

9 Mar 2004: Solidification and compositional convection of a ternary alloy 175. 0 20 40 60 80. ... Aitta. (a). NaN. O3. (wt %. ). H2O (wt %). 0 20 40 60 80 100.

9 Mar 2004: The present-day symmetricalstratocone (summit altitude 6,350 m and 20–25 km3 involume; Fig. ... 20 km from Parinacota Volcano) show-ing deformed and faulted fluvioglacial deposits by the upper ava-.

9 Mar 2004: 10. 15. 20. 25. l (cm). BFR. BFR. Figure 5. (a) The drainage of the fluid layer and (b) the depth of the fluid layer within the porousmedium as functions of ... 1 23.3 31 7.2 0.67 46.4 158.32 21.6 64 9.4 0.85 75.2 263.23 20.7 48 9.5 0.77

9 Mar 2004: 30. 20. 10. 03 0 3 0 3 0 3 0 3 0 3 0. ... 0.4. – 0.6. 80. 60. 40. 20. 0 50 100 150 200. –

9 Feb 2004: xN =Kt2/3[1τX1 τ2X2 ], (20)u= 2. 3Kt1/3. [U0(y)τU1(y)τ2U2(y)], (21). h= 49K2t2/3. ... h1/20 ). Substitutingtsl in (19) yieldsτsl which isexpected to be the lower bound of the intervalτ for which the present approximation is valid.

9 Feb 2004: 15 and 20 kmalong the ground at velocities of around 100 m s1 (Sparks et al. ... 20 m)pipe held horizontally a few centimetres above a bed of loose sand and silt.

9 Mar 2004: ft 1 20. The volume of the current is conserved during the propagation, which on account of Eqs. ... 1718) and (20) can be expressed as. 2Vt r2NhN 1. 6C2r2N rN02 21.

9 Mar 2004: solute flux condition for B atthe cotectic boundary (equation (4.20)), the modified marginal equilibrium condition(equation (4.22)), and the flux balance at η = λE (equation (4.27)). ... M. Anderson. has approximately 20 points in the solutal boundary

10 Jun 2004: 20. 15. 10. 50 10 20 30 40 50. a. r (cm).