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17 Aug 2010: Since K H+ {0},we have θHa θ. Ka and p. Ha p. ... Ka , and therefore pg p. Ha by Theorem 5. We pose two questions.

30 Dec 2010: By. are shown in Fig. 3. They are related by kAy kBy = 2π α.1Hence, cos(kA(B)y ) = cos(kB(A)y 2π α), so the Hamiltonian (8)is the ... 1Note that the positions kA(B) of these minima are gauge dependent.

4 Apr 2010: The first indication of unusual behaviour in the two-dimensional XY -model (n = 2)appeared in an analysis of the high temperature series expansion by Stanley and Ka-plan (1971).

7 Jun 2010: force of 1.69 kN and a peak current of 8.1 kA. ... a) (b). (c) (d). Figure 3: Resistance spot weld produced using interstitial–free steel with 7 kA current, 30 cyclesweld time [3] (reproduced with permission of B.

7 Jun 2010: ξ = 1 exp{kAtn} (2). where t is the time, kA and n rate constants and ξ is given by.

6 Oct 2010: 1 -. RECOUP Working Paper No. 31 The school governance environment in Uttar Pradesh, India: Implications for teacher accountability and effort Geeta Kingdon Institute of Education, University of London Mohd. Muzammil Department of Economics,

7 Jun 2010: eff. nf Ka. vFfT. = (15). For f the appropriate coefficient of friction from the wear regime applicable for the specific summitshould be used (Eqs.

7 Jun 2010: ξ = 1 exp{kaG3It4}. where ka is a constant.[40%]. END OF PAPER.

7 Jun 2010: Steel 1 / steel 2 (thickness, mm) (coating). Welding data1). I2) (kA).

7 Jun 2010: 3. 1ij ik kj. ka b c. =. = would use do loops both to perform the summation over k, and to run over all the values of i and j, setting each