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30 May 2006: 3.20) /Om. where the kernel. and &t) is the zeroth-order Hankel transform of h ( r ) , given by. ... 20, 429-448. Ocean. I n s t. G.F.D. Notes, no. 2, pp.

26 Apr 2006: cooled and solidified from above. The apparatus consisted of a Perspex tank, withwalls 11 mm thick, of dimensions 20 20 37.6 cm with a brass lid as depicted infigure 1. ... E. Huppert. (a). 12. 8. 4. h (cm). (b). T B = – 20 C.

27 Apr 2006: 1998] of 0.20. Thecalculated mean area covered by melt ponds on first-yearice was 0.41. ... Remote Sens., 20(10), 2069–2090. Derksen, C. P., J. M. Piwowar, and E.

27 Apr 2006: Petrol. 20 (1979) 487-554. 2 H,E. H u p p e r t and J.S. ... 100 (1980) 367-384. 20 E. Stolper and D. Walker, Melt density and the average.

26 Apr 2006: 0.4; (d) latex withG104 K m1 [20], slope! 1.0; (e) latex with G4]104 K m1 [21], slope! ... dt. dd0. dd0, (20). where d0(t"0) is the initial "lm thickness at h"0.

30 May 2006: Physics, University of Cambridge, 20 Silver Street, Cambridge CB3 9EW, UK 2Department of Geology, University of Bristol, Bristol BS8 lRJ, UK. ... y = 0 ( t = t,), (3.20). where t, is the time for the initial current to attain the slumping distance x,.

1 Jun 2006: Huppert. To cooler. - I 20 cm l o - 10 cm I 1 Scale. ... A. - 0. -10. T ("C). -20. -30. 100. Source. E D B, n C D.

17 May 2006: 20. The curves for solutions S3 and S4 are identical to within the. ... Liquid rope coiling 291. 10. 20. 30. 40. 50. 7 8 9 10.

26 Apr 2006: and 4190, respectively. The corresponding values for ice were taken as 2.20 and 4190. ... Fig. 9. The heat loss factor against radius for the case k = 0.20 with1600 particles (40 40).

30 May 2006: Substituting (2.19) into (2.20), we obtain. to which must be added the global continuity equation. ... 5 = (fg1Q3/V)-Brt-(3i1)/8 and. h(r> t , = @“(3&V/g’)a $([/EN),. (2.20).