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7 Oct 2008: tn. g(t) dt. If b(t) = ε cos ϑ(t) then the integral in(2.4) is of the form. ... The first step is to rewriteb(t) in the form. b(t) = 12 [cos(ω̃1t) cos(ω̃2t)],whereω̃1 = ω1 ω2 and ω̃2 = ω1 ω2.

7 Dec 2023: r3; (12b). _c % 32r3ð1þB cos 2cÞ%a. rsin c; (12c). where we recall that B = (a2 % 1)/(a2 1) A (0,1) is the shapefactor of the spheroid. ... Fig. 3A). Otherwise, it will escape due tothe fact that :r p cos c.

1 Dec 2023: tan δ = (Ws 1) sin Φ(Ws 1) cos Φ Q, (2.18). ... Fx(δ, n, Bi) Bifx(|δ|) sgn(δ) Fz(δ, n, Bi) Bifz(|δ|) sgn(cos δ), (3.6a,b)for some functions fx and fz.

25 Nov 2013: v2 =g sin2 θ. cos θ. where cos θ = d/. ... the particle and deduce that the particleleaves the surface of the sphere when cos θ = 2/3.

1 Dec 2023: u= eσ cos(2πy). I0(σ)1, v=0. (3.2a,b). Given this fixed, known velocity the concentration simply evolves via the advection–diffusion equation (2.7). ... u(x, y, t)= eσ cos(2πy)Rc′(x,y,t) 1. 0eσ cos(2πs)Rc. ′(x,s,t) ds1. (4.4).

12 Oct 2020: S =. α β. β α. ββ α.  ,are given by λk = α 2 i β cos kx, and wk = (im sin kmx)Mm=1, where x = πh =. πM1. So, all ... λk(A) =λk(C). λk(B)=. 1 12 i µ cos kx. 1 12 i µ cos kx |λk(A)| = 1, k = 1.M.

25 Feb 2008: Show that. θ̇2 =6gb. a2 3b2(cos θ cos α) (1). where b is the length OX, θ is the angle between OXand the vertical and α is the maximum value ... Combining these results,. 12 (. 13 a. 2 b2)θ̇2 gb cos θ = E/M.

26 Jun 2024: n(x, t) = A(k) cos (!t k x) B(k) sin (!t k x) (3.6). ... We’ll focus on the probability distribution of some scalar field (x) which, in the cos-.

26 Jun 2024: which diverges at = 0 and =. However, the numerator 1 cos has a zero when = 0 and the gauge connection is fine there. ... cos = eg/4L. So far, our discussion has been classical. Now we invoke some.

8 Apr 2008: In polar coordinates z = reiθ, Ar2r(B cos θC sin θ)D =0. ... Under the inversion 1/z ρeiφ = 1/reiθ this leads to Aρ(B cos φC sin φ)Dρ2 = 0which is also a circle.

11 Nov 2020: x(t) = X R sin(!B(t t0)) and y(t) = Y R cos(!B(t t0)). ... AN=. g. 4r. 1 cos. sin(1.25). The resulting magnetic field is.

26 Feb 2008: constant k:. mẍ = kx. so x = A sin(. k/m t) B cos(. ... mẍ = kx cẋ F cos t. The general solution consists of two parts: (a) the particular integral; and (b) the com-.

18 May 2022: uU0 sin b cos h. pr0r. 2. þX1n2. sin nb cos nhpn. ... Pane, and B. Nelson, “Soft microma-chines with programmable motility and morphology,” Nat.

21 Oct 2022: x′ = x cos ωt y sin ωt. y′ = y cos ωtx sin ωtz′ = z (2.18). ... 12ml2θ̇2 mgl cos θ (2.39). From which we may derive Lagrange’s equations using the coordinate θ directly.

2 Mar 2004: Why?) The solution is. T =. A0 2G(1 β)1r cos θ r < a,. ... The solution is. Φ(r, θ) ={ (. 1 3 ln(r/a))r cos θ/90 r < a.

25 Jan 2007: pair abmeans add or subtract the same expression with a and b interchanged. ... cos θ =gabU. aV b. |U a||V b|. (103). If the two vectors are timelike, and both are future directed, the rapidity θbetween them is given by.

10 Sep 2021: e1 = cos χ [cos θ (cos φex sin φey ) sin θ ez ] sin χ [ sin φex cos φey], (1). ... the helix is R = sin(ψ )/(π N ) and the pitch is p = 2 cos(ψ )/N.

9 May 2023: taking the opportunity to remind ourselves of this. In Electromagnetism, the magnetic field obeys r B = 0 which means that it can be. ... magnetic field B in terms of the current density. B(x) =µ04.

10 Sep 2021: helical turn and b = cos ψ. ... ez Fp = ζ T1 LT / cos ψ. 0[U (βT 1)b(R sin ψ U b)] d s. =

7 Nov 2017: b β(b b‖) , (43). respectively, where. P (θ ) = (1 U0)b b‖ cos θ (1 U0)(b b‖) cos2 θ, (44). ... uγ̇ th =. sin2 θ 2 b‖b. (cos θ q)2, (45).