Posts filled under #mosfet

Do I need eyepro when I'm

Do I need eyepro when I'm at home aswell? New video is up! Go check it out while I'm at Tjrnan for some pewpew. FINALLY --------------------------------------------------------- Follow these awesome peeps: @airsoft_locos @mr_hitman_nld @tacticalsnoopy @taurus_rumble @snu7.inf05 --------------------------------------------------------- My sponsor: @redwolfairsoft --------------------------------------------------------- #redwolf #redwolfairsoft #airsoft #gg #mosfet #firehawk #tjrnan #sweden #multicamo #aeg #worldairsoft #featureairsoft #airsofterphoto #doairsoft #rifle #instagram #pewpew #airsoftglobus #milsim #airsoftobsessed #bbwarz #airsoftinternational #airsoftphotography #airsofter #usa #milsim #milsimnation #airsofteurope #airsoftworldwide #warzoneinc #featureairsoft

Our Class AB amplifier. D

Our Class AB amplifier. Designed completely over 2 years we are nearing its release. We don't resell. We create and build from conception! Stay engaged for amazing specs and details. We won't build it unless its unique and spectacular. Chassis mods are in the works and excited to start demonstrations. #audiobliss # analog #soundengineering #hauteaudio #mosfet #linear #amplifier #audiophile #muskoka #gravenhurst #bracebridgeaudio #bracebridge #torontolife #portcarling #handmade #madeincanada #igaudio

  ,     -  -2017  ,

, - -2017 , . . "" , , . , , , , . , . . /Our partners: @dz_grozny /Skydive/The biggest drop zone and wind tunnel in Russia; @fps_chr /Federation of practical shooting of the Chechen Republic; @tacticalrock /The shop of professional equipment and uniforms #comouflage #mosfet #tjarnan #aeg #worldairsoft #bbwarz #milsimnation #warzoneinc #fighterjet #troopthanks #dailycombat #firearmsdaily #gunsofinstagram #gunnut #gunlife #pewpewprofessional #gunsofig #dailyguns #dailygundose #dailygunporn #gunfantastics #pew #c130 #556 # # # #hummerh1 #grappling #mmfighter @world_of_armies @war.rusarmy @spetsnaz.alfa @military.picture @soldier.pic @qmgraphics @alfa.operators

#Repost @itcsf_russia

#Repost @itcsf_russia . /Our partners: @dz_grozny t/Skydive/The biggest drop zone and wind tunnel in Russia; @fps_chr /Federation of practical shooting of the Chechen Republic; @tacticalrock /The shop of professional equipment and uniforms #comouflage #mosfet #tjarnan #aeg #worldairsoft #bbwarz #milsimnation #warzoneinc #fighterjet #troopthanks #dailycombat #firearmsdaily #gunsofinstagram #gunnut #gunlife #pewpewprofessional #gunsofig #dailyguns #dailygundose #dailygunporn #gunfantastics #pew #c130 #556 # # # #hummerh1 #grappling @world_of_armies @war.rusarmy @spetsnaz.alfa @military.picture @soldier.pic @qmgraphics @alfa.operators

#Repost @airsoft_locos (@

#Repost @airsoft_locos (@get_repost) #Repost @klockar Oldie but goldie... my DMR build is going to get a mosfet install and maybe I should try one of those new motors from @actionsportgames to install... Anyone out there tried them yet? Allready have the CNC line in two of my guns and just love them! / Klockar #dmr #dmrbuild #paintjob #camo #sniper #scope #bipod #mosfet #silencer #rail #airsofter #airsoftgun #callsignklockar #klockar #airsoftsweden #airsoftare #forest #outdoor #backcountry #moss #techjob #airsoftfamily

An extract on #mosfet

We can identify the space of linear operators on a vector space V, defined over the field F, with the space V V {\displaystyle V\otimes V^{*}} , where v h = ( w h ( w ) v ) {\displaystyle v\otimes h=\left(w\mapsto h(w)v\right)} . We also have a canonical bilinear function t : V V F {\displaystyle t\colon V\times V^{*}\to F} that consists of applying an element w* of V* to an element v of V to get an element of F, in symbols t ( v , w ) := w ( v ) F {\displaystyle t\left(v,w^{*}\right):=w^{*}(v)\in F} . This induces a linear function on the tensor product (by its universal property) t : V V F {\displaystyle t\colon V\otimes V^{*}\to F} , which, as it turns out, when that tensor product is viewed as the space of operators, is equal to the trace. This also clarifies why tr ( A B ) = tr ( B A ) {\displaystyle \operatorname {tr} (AB)=\operatorname {tr} (BA)} and why tr ( A B ) tr ( A ) tr ( B ) {\displaystyle \operatorname {tr} (AB)\neq \operatorname {tr} (A)\operatorname {tr} (B)} , as composition of operators (multiplication of matrices) and trace can be interpreted as the same pairing. Viewing End ( V ) V V {\displaystyle \operatorname {End} (V)\cong V\otimes V^{*}} , one may interpret the composition map End ( V ) End ( V ) End ( V ) {\displaystyle \operatorname {End} (V)\times \operatorname {End} (V)\to \operatorname {End} (V)} as ( V V ) ( V V ) ( V V ) {\displaystyle (V\otimes V^{*})\times (V\otimes V^{*})\to (V\otimes V^{*})} coming from the pairing V V F {\displaystyle V^{*}\times V\to F} on the middle terms. Taking the trace of the product then comes from pairing on the outer terms, while taking the product in the opposite order and then taking the trace just switches which pairing is applied first. On the other hand, taking the trace of A and the trace of B corresponds to applying the pairing on the left terms and on the right terms (rather than on inner and outer), and is thus different. In coordinates, this corresponds to indexes: multiplication is given by ( A B ) i k = j a i j b j k {\displaystyle \textstyle {(AB)_{ik}=\sum _{j}a_{ij}b_{jk}}} , so tr ( A B ) = i j a i j b j i {\displaystyle \textstyle {\operatorname {tr} (AB)=\sum _{ij}a_{ij}b_{ji}}} and tr ( B A ) = i j b i j a j i {\displaystyle \textstyle {\operatorname {tr} (BA)=\sum _{ij}b_{ij}a_{ji}}} which is the same, while tr ( A ) tr ( B ) = i a i i j b j j {\displaystyle \textstyle {\operatorname {tr} (A)\cdot \operatorname {tr} (B)=\sum _{i}a_{ii}\cdot \sum _{j}b_{jj}}} , which is different. For V {\displaystyle V} finite-dimensional, with basis { e i } {\displaystyle \left\{e_{i}\right\}} and dual basis { e i } {\displaystyle \left\{e^{i}\right\}} , then e i e j {\displaystyle e_{i}\otimes e^{j}} is the ij-entry of the matrix of the operator with respect to that basis. Any operator A {\displaystyle A} is therefore a sum of the form A = a i j e i e j {\displaystyle A=a_{ij}e_{i}\otimes e^{j}} . With t {\displaystyle t} defined as above, t ( A ) = a i j t ( e i e j ) {\displaystyle t(A)=a_{ij}t\left(e_{i}\otimes e^{j}\right)} . The latter, however, is just the Kronecker delta, being 1 if i = j and 0 otherwise. This shows that t ( A ) {\displaystyle t(A)} is simply the sum of the coefficients along the diagonal. This method, however, makes coordinate invariance an immediate consequence of the definition.

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