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4 Feb 2008: 1/1/2008 2008 Alexander Slocum 10-0. FUNdaMENTALS of DesignTopic 10Bearings. BearingsBearings are machine elements that allow compo-. nents to move with respect to each other. Bearings are used to support large skyscrapers to allow them to move

3 Jan 2008: 2 degree of freedom flutter solver mass=0.0486; polarI=4.05e-5; wh=12.0 wh=wh2.0pi wa=105.0 wa=wa2.0pi Kh=whwhmass Ka=wawapolarI ... i=1:100, A=masspolarI-crossIcrossI; B=0.5rhoVVSCla(crossIxacmass)-Kamass-KhpolarI; C=Kh(Ka-0.5rhoVVSClaxac); % %

5 Feb 2008: V = k (h/D)0.5. or Q =kA(h/D)0.5. or W= kA(hD)0.5. k is the discharge coefficient of theelement (which also reflects theunits of

6 Feb 2008: Affinities were calculated via fitting titration curves to the data. This resulted in the stoichiometry N, the binary equilibrium constant Ka, the dissociation constant Kd (which equals 1/Ka), the enthalpy

21 Oct 2008: ND#E77FC,=0FG=[=VS(NS=R6.I_4 MK]SI;GTMW3ZFSO]C3O]E;X0W2_KA%0LI7_O;%U9LT'PRRG_;NI#8IX4

26 Jun 2008: KA&&,#- 25,. @32 <-2<3-322<2 322. @3 2- 3<-332< -2< 32. FA661,6#$#.# ,, ,$1?

2 Jan 2008: Fund 4,780,166 – – 396,189 5,176,355 5,176,338The Genzyme Experimental Medicine Fund – – – – – –Li Ka Shing Fund for Oncology – 2,000,000 – 49,348 2,049,348

4 Feb 2008: by bearings, where the support points A and B are springs of constant KA and KB and the shaft diameter D1 changes to D2 a distance c from the origin (where ... The boundary conditions are δ(a) = -FA/KA and δ(b) = - FB/KB, hence the constants C1 and C2

8 Apr 2008: Plugging. into the equation we find c = kaa0kbka. for kb 6= ka.

4 Aug 2008: Curr Biol 2005, 1155::R514-R526. 40. Michie KA, Löwe J: DDyynnaammiicc ffiillaammeennttss ooff tthhee bbaacctteerriiaall ccyyttoosskkeelleettoonn.Annu Rev Biochem 2006, 7755::467-492.