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25 Nov 2011: T̄. (νB2H 3/2W 3/2. gV 20. )1= 4. 9 sin θ(a1/2 a1/2 tan1/3 θ )3,. ... V̄ = V0(t/T̄ )1/2, (5)using Eqs. (2)–(4). Note that T̄ V 20 from Eq.

26 Sep 2011: 5) Huppert, H. E.; Manins, P. C. Deep-sea Res. 1973, 20, 315-23. ... 7 = 0.287. The cross-terms are 20 times less than the smallest main term.

25 Nov 2011: A cutoff of either 20% or 40% of the total concentration was used to determine the position of the front. ... A cutoff of either 20% or 40% was used to determine the front position.

13 Apr 2011: H 20 m,motivated by the Sleipner site in the North sea [Bickle et al.,2007]. ... FCO2 DDCO2. HSh DCO2. kgD. 4=5 DH. 1=5. ’ 20 kg m2 yr1.

24 Aug 2011: 3.3mm per year, remarkably uniformly for at least the last 20 years. ... so to influence the result in 10, 20, … years’ time we need to act now.

21 Oct 2011: r10 20 30 40. r. t. 0. 0.4. 0.8. 1.2. Vel. ... 5 10 15 20 25 30. rf. (b)(a). Figure 5. Experimental comparison.

13 Apr 2011: y (mm). u (m. m/s. ). t=3 mint=5 mint=12 mint=20 min. ... FIG. 20. The minimum flow rate required for steady flow isestimated by Eq.

26 Sep 2011: 20 = 0"874 K25 , (1)t. where C2o (c25) signifies the diffusivity at 20C (25C). ... 22). The system (20) to (22) is now linear and has solutions o f the form.

21 Oct 2011: m). 8. 4. 0. 0. 20 40 60 80. 40 6020 80 100. ... 4. 4.2. 10. 15. 20. 25. 8 12 16. 5 10 15.

18 Jan 2011: φ+. x. x=0. φx. x=0. 4. 1 y2=. (23π. )1/2λ̄φ. 1/20. ... 3.20). Substituting the similarity solution for h from (3.16) and (3.15), we find that.