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stabilising short-range interaction, Π̃2(h̃) = B/h̃6. Here h̃ is the dimensional film thickness,. ... Using cylindrical polar coordinates (r, θ, z), such that y1 = r cos θ, y2 = r sin θ and.

The equations of motion for the elemental free body travelling at constant belt speed v, from (b) are. ... tangentially : ( F δF ) cos. δφ. /. 2. - F cos.

B. C. D. E. F. X. 0.100. 0.200. 0.1 cos (22.5). ... sin. cos. sin cos. sin cos. sin cos. cos. sin. 2.

elements represents a sector of. /8 radians. Pressure being in “Cos. “, ... 2. 0. p. (cos. 1. cos. 2. ). 2). Vertical gravity according to X (Ug field).

F. C. D. X 0.1.0.2.0.1 cos (45) 0.2 cos (45) 0.1 cos (22.5) 0.2 cos (22.5). ... sin. cos. sin cos. sin cos. sin cos. cos. sin. 2.

ii) Let (a,b) B = and assume that for Lebesgue almost all E (a,b) there is a subordinatesolution wE,n to (2.7) at E such that. ... For part (ii) note that for (a,b) B = we can apply Theorem 2.7 and (2.18).

U. U. U. =. -. =. cos. cos. sin. sin. sin. ... U. in (cos, cos, sin) with the place. of (cos, cos, sin)).

φ, bv = b cos θ , andgk = g sin θ cos φ with b and g the magnitudes of the Voigt andFaraday effects. ... These can be calculatedfrom the Fresnel equations [23],. tan ϕh = 2 cos ϑi.

i cos(γπκ2. ) sin(γπκ. 2)). , (36). where γ, b are model-dependent constants, while H(κ) is a model-dependent product ofgamma-functions. ... A(x) = cos(γπvx2. )eγvx log x+. k=1 z2k1(vx)2k1. , (37). where zk is proportional to ζk in a

For example, Fig. 1 shows two block designs with v = 15, b = 7 and. ... Thus current is defined on( B) (B ) and voltage is defined on T B.

The nonconvex minimization problem theninvolves eigen-strains. E1 = 0.0113m m 0.0102n n and E2 = 0.0102m m 0.0113n n,for m = (cos(π/3), sin(π/3)) and ... n = ( sin(π/3), cos(π/3)), and the material tensor Cdefined for cubic anisotropy by.

2.0. 1.5. 1.0. Ene. rgy. (eV. ). (a). (b). d=. 900. ... 2.0. 1.5. 1.0. Ene. rgy. (eV. ). (a). (b)(b). d=. 900.

such that sin φ, cos φ 6= 0) we apply twice the identity. ... dv, (3.20). where g(v) = 2πi12. (w2v2) cos φwvsin φ. We have.

trjt = r. jt. (rjt )2 1. (rjt )2 2rjt cos(ϑ. ... jTt). (rjt )2 2rjt cos(ϑ. jt ξ. jTt) 1. (23). (where we have suppressed the dependence on z to ease notation).By observing that the right hand side of (22)

Kd2=110; % kd for MBP-BRC4-HumRadA18. max. 0. A. AASF bb. =. dedK. dedF. d. b. =. )3/cos()3(23. )3/cos()3(22. 1. 2. ϑ. ϑ. ... Fsb=(2.sqrt(d.2-3.e).cos(theta./3)-d)./(3.Kd12.sqrt(d.2-3.e).cos(theta./3)-d);. semilogx(BRC3,Fsb,'Color','red');. S13.

Indeed, writing. f(x) = [f]. n=1. (an cos(nx) bn sin(nx). ),. ... we have. Ckh (f′) =. n=1. ncoth(nkh)(an cos(nx) bn sin(nx). ),.

For (DBC) another straightforward, although a little longer computation gives thatbifurcation values are of the form α = 1/|cos b|, where b solves. ... Cb tan b = 0. (16). We denote by b0 (π/2,π) the smallest positive solution, set α0 = 1/|cos b0|,

Λk =. {. 2a(0) 4(M1)/2. j=1 a(j) cos kj if M is odd. ... TN )jl =2. π. π. k=0. Λk|Λk|. cos kj cos klk M.

b. b bτ xzzzp. τ =. τ =. shea. r st. ... gz λκηu2). εgzb. y,. η := cos (ψ(y) ϕ(x) ϕ0) ,. in which. Kx = Kxact/pass = 2sec2φ(. 1(. 1 cos2 φ/cos2 δ)1/2. ).

E. F. C. D. X 0.1.0.2.0.1 cos (45) 0.2 cos (45) 0.1 cos (22.5) 0.2 cos (22.5). ... sin. cos. sin cos. sin cos. sin cos. cos. sin. 2.

STS. STS. LO. pFsITS. OTS. Cos. ITS. OTSpFs. Cos. LOV3aV3a. V1V1. ... temporal sulcus, ITS: inferior temporal sulcus, OTS: occipitotemporal. sulcus, CoS: collateral sulcus.

the following metric. ds2 = 4dudv r+(u). 2ν2(1 cos(u). ) dx2 1 ν4. ... ν2(1 cos(u). ) dxdy r(u). 2ν2(1 cos(u). ) dy2. (1 ν2).

1. cos 2ϕϕul) = 0. (4.34). Besides the obvious SO(n 1) symmetry these equations are invariant under the following formaltransformation. ... n , (4.37). one arrives at the system (4.34) for φ,ul with the obvious replacement of cos ϕ, sin ϕ, tan ϕ