KENZOU Karman) x
|
|
- るるみ かいじ
- 5 years ago
- Views:
Transcription
1 KENZO Karman) x / x ( ) ============================ 1
2 R e 3 R e x y u v ū, v u v 1 u = ū + u, v = v + v x u y v v u = ū + u, v = v (1.1) F ig.56 u ū y v = v ū ū u O t O x τ r τ l τ t τ r = τ l + τ t τ l Fig.56 x ρv u x ρv (ū + u ) 1 t t ū u v t ρv (ū + u )dt = ρū t = ρ t t t v dt + ρ t t u v dt u v dt ρ u v (1.) ū = 1 t t udt =, 1 t t u dt =, 1 t t v dt = (1.3) 1.τ r τ r = ρ u v (1.4) Reynolds stress u v τ τ = τ + τ r = µ dū dt ρ u v = ρν dū dt ρ u v (1.5) 1 7 N-S Q Q Fig.57 τ r
3 ρν(dū/dt) ρ u v τ F ig.57 H-P H-P u u max ū Boussinesq τ = ρ u v = ρɛ dū (1.6) 3 τ τ = ρν dū dt ρ u v = ρ(ν + ɛ) dū (1.7) ɛ ν 1.1 u v Prandtl 4 5 mixing length ρu v ( ) dū τ r = ρ u v = ρ l (1.8) l mixing length l u v = l dū (1.9) Boussinesq 4 Ludwig Prandtl ) 5 η η = (1/3)mρv l l
4 F ig.58 y l m(dū/) l ū(y + l m) ū(y) ū(y l m) l m(dū/) ū(y + l m). = ū(y) = ū(y l m). = x u(y) u(y) dū + l m u(y) dū l m τ r τ r x x y ū u(y) l m l m ū(y + l m ) =. u(y) dū + l m ū(y) = u(y) ū(y l m ) =. u(y) dū l m (1.1) y y + l m y y l m y dū ū(y + l m ) ū(y) = l m ū(y l m ) u(y) dū = l m (1.11) y ū(y) u ū ū = 1 { l m dū } + dū l m = l m dū (1.1) y y v ū v = ( ) u dū = ( ) l m (1.13) u v y u v y u v c u v = u v (1.14) u v = c ū v (1.15) 1.8 τ r τ r = ρ u v = ρc ū v ρc lm dū ( ) dū = ρ l (1.16) 1.8l c l m l τ = ρɛ(dū/) ɛ ɛ = l dū/ ɛ 7 l m Prandtl l 4
5 1. (P89) l.4 l =.4y l y l = κy (1.17) κ κ = dū = 1 τr κy ρ (1.18) 1.3 τ r R e < 5 viscous sublayer 8 δ 7 τ w τ w = µ du = µu y (y δ ) (1.19) ρ τ w ρ = ν u y (1.) τ w /ρ [ML T L 3 /M 1 L 3 ] 1/ = [LT 1 ] τ w /ρ = u u y = δ u = u δ R δ u δ = u δ = R δ (1.1) u ν F ig.59 y δ u u δ x 8 5
6 τ r = τ w 9 τ r ( ) dū τ r = κ ρy dū = 1 τr κy ρ (1.) τ r u 1. dū = u 1 κ y ū = 1 u κ ln y + const 1 ū ln y + const =.33 κ u κ log 1 y + const (1.3) ū u y = 5.75 log u (1.4) ν 1 log law 7 τ τ = µ dū = τ w = const τ w dū = τ w µ = ( ) τw 1 ρ ν = u ν u ū = u ν y + const y = ū = const = ū = u ν y ū = u y u ν wall cordinate y + = yu /ν [L T 1 /L T 1 ] u + = ū u = u y ν = y+ 1.4 / ū/u max u ( y ) 1/n ( = = 1 r 1/n (1.5) u max a a) 9 1 h u/u = log 1 y/h by Nikuradse) 6
7 a y (r n R e = n = 7 u = u max (y/a) 1/7 (1.6) 1/7 11 R e = 1 5 τ w ū/u max =.8.8 (1.7) ( ) 1/4 τ w =.ρu ν max (1.8) u max a 8 cm 6cm/sec 1 1 ν =.13cm /sec R e = ūd/ν6 /.13 = < R e < 1 5 1/7 ( ) 1/4 u max = ū/.8 = 6/.8 = 75.cm/sec, τ w =.5ρu ν max =.83kg/m (1.9) u max a Re = L/ν = / 1 du/ du/ = free stream main stream 1 Boundary layer 1 F ig.6 y x.99 δ(x 1) x 1 x 99% y δ 11 R e 1 7 n = 1/8, 1/
8 laminar boundary layerturbulent boundary layer transition region R c = t + u x + v y = 1 ( p ) ρ x + ν u x + u y v t + u v x + v v y = 1 ( p ) ρ y + ν v x + v y x + v y = (13.1) x y δ L x u, x L, v y x L v L δ, v x L δ, y δ, u x L, v y Lδ = L L δ u y δ (13.) (1) u () v (3) u (4) v [ ] [ u ν x + u y ν L + L ] L δ [ ] [ v ν x + v δ y ν L L + ] L L δ u x + v y ( L + δ ) L δ = L + L u v x + v v y δ ( L + δ ) L L = δ L L + δ L L (13.3) (13.4) (13.5) (13.6) δ L ( δ L 1) (a) (b) (a) (b), (c) (d) (c) (d) 13.1 t + u x + v y = 1 p ρ x + ν u y p y = x + v y = (13.7) 13 8
9 ( ρ ū ū ) ū + v = ρ p x y x + τ y τ = µ ū y ρu v p y = ū x + v y = (13.8) 9 L b ρ µ δ δ 13. u x L, ν u y ν δ (13.9) δ L ν δ δ νl (13.1) 15 3 / y y= k y = u = v = u x + v y = 1 ρ p x + ν u x = 1 p ρ x + ν u x u y = 1 p ρν x = 1 p µ x y / y = (1/µ)( p/ x)y + const const = k (... y = / y = k) y = u = u = 1 p µ x y + ky (13.11) x y x δ x t = 14 ) δ = 5.r νl 9
10 y u(y) y = δ u y =, y=δ y = y=δ y x p x = p x y=δ u x = 1 p y=δ ρ x y=δ.. p. = ρu x x = ρ y=δ x (13.1) x x (d/dx = ) x dp/dx = 13.1 ( 1 x + p ) = p + 1 ρ ρ = const = p + 1 ρ (13.13) const δ u 99% y δ =.99 δ (1) δ displacement thickness δ F ig.61 y δ u x = δ u Fig δ
11 Q δ = Q δ = 1 ( u) (13.14) y > δ u δ () θ momentum thickness θ θ = 1 u( u) (13.15) ρ ρ θ = uρ( u) ρ( u) Fig.61 u ρθ ρθ (3) θ energy thickness θ θ = 1 3 u( u ) (13.16) H shape factor δ θ H=.59 H=1.4 H = δ θ (13.17) adverse pressure gradient H θ H H H=3.5 H= /7 δ θ 1/7 1.6 δ = 1 θ = 1 u = (y/δ) 1/7 { ( y ) } 1/7 ( u) = 1 = δ δ 8 ( y ) { 1/7 ( y ) } 1/7 u( u) = 1 = 7 δ δ 7 δ 11
12 13..3 ( ) 18 u x + v y = 1 ρ 13.3 y = y = δ p x + ν u y (13.18) x + v y = (13.19) x = y = y = δ { u x + v } { } 1 p = y ρ x + ν u y v ( δ y = = ) x y = x + { u x + v } = y ν u y = 13.) 1 p ρ x = ν y δ v y = v y=δ (13.) u x u y { u x } x d dx δ + = ν y ν y=δ y = 1 y= ρ τ w u x (13.1) ( u x x d ) = 1 dx ρ τ w (13.) u x x d dx = x (u ) (u) ud x dx + d dx = d [u(u )] ( u) x dx 13.) d dx u( u) + d dx δ θ 13.3 d dx ( θ) + δ d dx = τ w ρ dθ d + θ dx dx + δ d dx = τ w ρ ( + δ θ dθ dx + ( u) = 1 ρ τ w (13.3) dθ dx + ) ( + δ θ d θ dx = τ w ρ ) θ d dx = τ w ρ (13.4) 1
13 x = const d/dx = 13.3 τ w = d dx ρu( u) (13.5) 3 (δ y)y u = δ (13.6) δ x dθ dx + ) ( + δ θ d θ dx = τ w ρ x θ θ = 1 (δ y)y = δ u( u) = 1 { 1 τ w ( ) τ w = µ = µ y y= y dθ dx = τ w ρ 15 dθ dx = τ w ρ (13.7) (δ y)y δ { (δ y)y δ { } = 3 δ 8 15 δ = 15 δ } (δ y)y δ } (δ y)y δ = µ δ (δ y) y= = µ δ dδ dx = 1 ( ) µ ρ 1 dδ δ 15 dx =... δdδ = 15ν dx δ = 3 νx + const ν, ν = µ/ρ δ x = δ = const = x 3νx δ(x) δ(x) = 33 δ u = a + by + cy τ a =, b = /δ, c = /δ 13.5 τ = ρ d dx y = u = = a y = δ u = = a + bδ + cδ y = δ y = = b + cδ u = [(y/δ) (y/δ) ] u( u) = ρ d ( ) dx 15 δ = dδ ρ 15 dx ( ) du τ = µ = µ y= d [ { ( y ) ( y ) }] = µ δ δ y= δ 13
14 dδ ρ 15 dx = µ δ... δdδ = 15 ν dx δdδ = 15 ν x dx νx δ = 5.48 τ τ = µ δ =.73ρ ν x /7 τ ( 1/4 τ =.5ρ ν δ(x)) { ) dθ τ = ( dx + + δ θ d θ dx = 7 { dδ 7 dx δ d dx } ρ = } ρ { 7 dδ 7 dx + ( + 7 ) δ } d ρ dx τ ( ) 1/4 ν τ =.5ρ (13.8) δ(x) d/dx = 7 dδ 7 dx =.5ρ ( ) 1/4 ν 7 δ(x) 7 δ 1/4 dδ =.5 x ( ν ) 1/4.. ( ν ) 1/5. δ(x) =.371 x 1/5 (13.9) x 4/5 x 13.3 p/ x = / t = u x + v y = ν u y, 9 δ νx δ x, y ξ, η 19 ξ = x, η = (y/) = y δ νx 19 η = y/δ y/ x + v y = (13.3) 14
15 ψ = νxf(η) (13.31) f(η) x, y ξ, η ϕ(ξ, η) ϕ(ξ, η) x ϕ(ξ, η) y = ϕ ξ ϕ(ξ, η) y = ξ x + ϕ η η x = ϕ ξ + η x ϕ νx η = ϕ ξ ξ y + ϕ η η y = 1 ( ) ( ) ϕ = 1 y y 4 νx u, v ϕ η x = ξ η x η y = 1 νx η ϕ y y = 1 4 νx η (13.3) u = ψ y = 1 ψ νx η = 1 f (13.33) v = ψ ( x = ξ η ) ψ = 1 ν x η x (f η f) ( x = ξ η x u y = 8νx f η ) f (η) = η 4x f, f y = 4 νx f (13.34) ff + f = (13.35) η y = u = v =, y = u = y = u =, v = η = f =, f = y = u = η = f = f(η) η f(η) = A + A 1 1! η + A! η + A 3 3! η3 (13.36)... f (η) = A 1 + A η + A 3! η + f() = A =, f () = A 1 = f(η) = A! η + A 3 3! η3 + A n n! ηn +, f (η) = A η + A 3! η + + A n f (η) = A + A 3 η ( ) A A 3 + A 4 η + + A 5 η +! η (n 1)! ηn 1 + A n (n )! ηn +, f (η) = A 3 + A 4 η + A 5! η + + A n (n 3)! ηn 3 ( 4A A 3 + A 6 3! ) η 3 + = A 3 =, A 4 =, A + A 5 =, 4A A 3 + A 6 = ( A 6 = ), A 7 = 15
16 f(η) f(η) = A! η A = A 1/3 5! η5 + 11A3 8! { (A 1/3 η) 3! η 8 375A4 η ! (A1/3 η) 5 5! + 11(A1/3 η) 8 8! [ = A 1/3 Y Y Y 8 ] Y +! 5! 8! 11! } 375(A1/3 η) ! ( Y = A 1/3 η) = A 1/3 F (Y ) (13.37) A ( ) f lim f Y = lim = A /3 η lim Y Y η F (Y ) Y η = f = A /3 lim Y F (Y ) = A = lim Y F (Y ) 3/ A = (13.38) f (η) = f () = A = τ w τ w = µ ( ) = y y= 4 νx f () = 4 ( ν ) 1/ ρ νx f () =.6641 x (13.39) δ xν δ(x) = 5. (13.4) xν xν δ (x) = 1.7, θ =.664 (13.41) ( ) 8 OK (!?) 9 16
Untitled
II 14 14-7-8 8/4 II (http://www.damp.tottori-u.ac.jp/~ooshida/edu/fluid/) [ (3.4)] Navier Stokes [ 6/ ] Navier Stokes 3 [ ] Reynolds [ (4.6), (45.8)] [ p.186] Navier Stokes I 1 balance law t (ρv i )+ j
More information(5) 75 (a) (b) ( 1 ) v ( 1 ) E E 1 v (a) ( 1 ) x E E (b) (a) (b)
(5) 74 Re, bondar laer (Prandtl) Re z ω z = x (5) 75 (a) (b) ( 1 ) v ( 1 ) E E 1 v (a) ( 1 ) x E E (b) (a) (b) (5) 76 l V x ) 1/ 1 ( 1 1 1 δ δ = x Re x p V x t V l l (1-1) 1/ 1 δ δ δ δ = x Re p V x t V
More informationII ( ) (7/31) II ( [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Re
II 29 7 29-7-27 ( ) (7/31) II (http://www.damp.tottori-u.ac.jp/~ooshida/edu/fluid/) [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Reynolds [ (4.6), (45.8)] [ p.186] Navier Stokes I Euler Navier
More information微分積分 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. このサンプルページの内容は, 初版 1 刷発行時のものです.
微分積分 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. ttp://www.morikita.co.jp/books/mid/00571 このサンプルページの内容は, 初版 1 刷発行時のものです. i ii 014 10 iii [note] 1 3 iv 4 5 3 6 4 x 0 sin x x 1 5 6 z = f(x, y) 1 y = f(x)
More informationTOP URL 1
TOP URL http://amonphys.web.fc.com/ 3.............................. 3.............................. 4.3 4................... 5.4........................ 6.5........................ 8.6...........................7
More information2.2 h h l L h L = l cot h (1) (1) L l L l l = L tan h (2) (2) L l 2 l 3 h 2.3 a h a h (a, h)
1 16 10 5 1 2 2.1 a a a 1 1 1 2.2 h h l L h L = l cot h (1) (1) L l L l l = L tan h (2) (2) L l 2 l 3 h 2.3 a h a h (a, h) 4 2 3 4 2 5 2.4 x y (x,y) l a x = l cot h cos a, (3) y = l cot h sin a (4) h a
More information9. 05 L x P(x) P(0) P(x) u(x) u(x) (0 < = x < = L) P(x) E(x) A(x) P(L) f ( d EA du ) = 0 (9.) dx dx u(0) = 0 (9.2) E(L)A(L) du (L) = f (9.3) dx (9.) P
9 (Finite Element Method; FEM) 9. 9. P(0) P(x) u(x) (a) P(L) f P(0) P(x) (b) 9. P(L) 9. 05 L x P(x) P(0) P(x) u(x) u(x) (0 < = x < = L) P(x) E(x) A(x) P(L) f ( d EA du ) = 0 (9.) dx dx u(0) = 0 (9.2) E(L)A(L)
More informationD v D F v/d F v D F η v D (3.2) (a) F=0 (b) v=const. D F v Newtonian fluid σ ė σ = ηė (2.2) ė kl σ ij = D ijkl ė kl D ijkl (2.14) ė ij (3.3) µ η visco
post glacial rebound 3.1 Viscosity and Newtonian fluid f i = kx i σ ij e kl ideal fluid (1.9) irreversible process e ij u k strain rate tensor (3.1) v i u i / t e ij v F 23 D v D F v/d F v D F η v D (3.2)
More information1 (Berry,1975) 2-6 p (S πr 2 )p πr 2 p 2πRγ p p = 2γ R (2.5).1-1 : : : : ( ).2 α, β α, β () X S = X X α X β (.1) 1 2
2005 9/8-11 2 2.2 ( 2-5) γ ( ) γ cos θ 2πr πρhr 2 g h = 2γ cos θ ρgr (2.1) γ = ρgrh (2.2) 2 cos θ θ cos θ = 1 (2.2) γ = 1 ρgrh (2.) 2 2. p p ρgh p ( ) p p = p ρgh (2.) h p p = 2γ r 1 1 (Berry,1975) 2-6
More informationII A A441 : October 02, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka )
II 214-1 : October 2, 214 Version : 1.1 Kawahira, Tomoki TA (Kondo, Hirotaka ) http://www.math.nagoya-u.ac.jp/~kawahira/courses/14w-biseki.html pdf 1 2 1 9 1 16 1 23 1 3 11 6 11 13 11 2 11 27 12 4 12 11
More information: , 2.0, 3.0, 2.0, (%) ( 2.
2017 1 2 1.1...................................... 2 1.2......................................... 4 1.3........................................... 10 1.4................................. 14 1.5..........................................
More informationJKR Point loading of an elastic half-space 2 3 Pressure applied to a circular region Boussinesq, n =
JKR 17 9 15 1 Point loading of an elastic half-space Pressure applied to a circular region 4.1 Boussinesq, n = 1.............................. 4. Hertz, n = 1.................................. 6 4 Hertz
More information( ; ) C. H. Scholz, The Mechanics of Earthquakes and Faulting : - ( ) σ = σ t sin 2π(r a) λ dσ d(r a) =
1 9 8 1 1 1 ; 1 11 16 C. H. Scholz, The Mechanics of Earthquakes and Faulting 1. 1.1 1.1.1 : - σ = σ t sin πr a λ dσ dr a = E a = π λ σ πr a t cos λ 1 r a/λ 1 cos 1 E: σ t = Eλ πa a λ E/π γ : λ/ 3 γ =
More informationQMI13a.dvi
I (2013 (MAEDA, Atsutaka) 25 10 15 [ I I [] ( ) 0. (a) (b) Plank Compton de Broglie Bohr 1. (a) Einstein- de Broglie (b) (c) 1 (d) 2. Schrödinger (a) Schrödinger (b) Schrödinger (c) (d) 3. (a) (b) (c)
More informationII No.01 [n/2] [1]H n (x) H n (x) = ( 1) r n! r!(n 2r)! (2x)n 2r. r=0 [2]H n (x) n,, H n ( x) = ( 1) n H n (x). [3] H n (x) = ( 1) n dn x2 e dx n e x2
II No.1 [n/] [1]H n x) H n x) = 1) r n! r!n r)! x)n r r= []H n x) n,, H n x) = 1) n H n x) [3] H n x) = 1) n dn x e dx n e x [4] H n+1 x) = xh n x) nh n 1 x) ) d dx x H n x) = H n+1 x) d dx H nx) = nh
More information201711grade1ouyou.pdf
2017 11 26 1 2 52 3 12 13 22 23 32 33 42 3 5 3 4 90 5 6 A 1 2 Web Web 3 4 1 2... 5 6 7 7 44 8 9 1 2 3 1 p p >2 2 A 1 2 0.6 0.4 0.52... (a) 0.6 0.4...... B 1 2 0.8-0.2 0.52..... (b) 0.6 0.52.... 1 A B 2
More informationTOP URL 1
TOP URL http://amonphys.web.fc2.com/ 1 6 3 6.1................................ 3 6.2.............................. 4 6.3................................ 5 6.4.......................... 6 6.5......................
More informationKENZOU
KENZOU 2008 8 9 5 1 2 3 4 2 5 6 2 6.1......................................... 2 6.2......................................... 2 6.3......................................... 4 7 5 8 6 8.1.................................................
More information医系の統計入門第 2 版 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. このサンプルページの内容は, 第 2 版 1 刷発行時のものです.
医系の統計入門第 2 版 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. http://www.morikita.co.jp/books/mid/009192 このサンプルページの内容は, 第 2 版 1 刷発行時のものです. i 2 t 1. 2. 3 2 3. 6 4. 7 5. n 2 ν 6. 2 7. 2003 ii 2 2013 10 iii 1987
More informationVenkatram and Wyngaard, Lectures on Air Pollution Modeling, m km 6.2 Stull, An Introduction to Boundary Layer Meteorology,
65 6 6.1 No.4 1982 1 1981 J. C. Kaimal 1993 1994 Turbulence and Diffusion in the Atmosphere : Lectures in Environmental Sciences, by A. K. Blackadar, Springer, 1998 An Introduction to Boundary Layer Meteorology,
More informationtomocci ,. :,,,, Lie,,,, Einstein, Newton. 1 M n C. s, M p. M f, p d ds f = dxµ p ds µ f p, X p = X µ µ p = dxµ ds µ p. µ, X µ.,. p,. T M p.
tomocci 18 7 5...,. :,,,, Lie,,,, Einstein, Newton. 1 M n C. s, M p. M f, p d ds f = dxµ p ds µ f p, X p = X µ µ p = dxµ ds µ p. µ, X µ.,. p,. T M p. M F (M), X(F (M)).. T M p e i = e µ i µ. a a = a i
More informationBethe-Bloch Bethe-Bloch (stopping range) Bethe-Bloch FNAL (Fermi National Accelerator Laboratory) - (SciBooNE ) SciBooNE Bethe-Bloch FNAL - (SciBooNE
21 2 27 Bethe-Bloch Bethe-Bloch (stopping range) Bethe-Bloch FNAL (Fermi National Accelerator Laboratory) - (SciBooNE ) SciBooNE Bethe-Bloch FNAL - (SciBooNE ) Bethe-Bloch 1 0.1..............................
More informationDVIOUT
A. A. A-- [ ] f(x) x = f 00 (x) f 0 () =0 f 00 () > 0= f(x) x = f 00 () < 0= f(x) x = A--2 [ ] f(x) D f 00 (x) > 0= y = f(x) f 00 (x) < 0= y = f(x) P (, f()) f 00 () =0 A--3 [ ] y = f(x) [, b] x = f (y)
More informationTOP URL 1
TOP URL http://amonphys.web.fc2.com/ 1 30 3 30.1.............. 3 30.2........................... 4 30.3...................... 5 30.4........................ 6 30.5.................................. 8 30.6...............................
More information9 2 1 f(x, y) = xy sin x cos y x y cos y y x sin x d (x, y) = y cos y (x sin x) = y cos y(sin x + x cos x) x dx d (x, y) = x sin x (y cos y) = x sin x
2009 9 6 16 7 1 7.1 1 1 1 9 2 1 f(x, y) = xy sin x cos y x y cos y y x sin x d (x, y) = y cos y (x sin x) = y cos y(sin x + x cos x) x dx d (x, y) = x sin x (y cos y) = x sin x(cos y y sin y) y dy 1 sin
More information( ) ( )
20 21 2 8 1 2 2 3 21 3 22 3 23 4 24 5 25 5 26 6 27 8 28 ( ) 9 3 10 31 10 32 ( ) 12 4 13 41 0 13 42 14 43 0 15 44 17 5 18 6 18 1 1 2 2 1 2 1 0 2 0 3 0 4 0 2 2 21 t (x(t) y(t)) 2 x(t) y(t) γ(t) (x(t) y(t))
More informationx E E E e i ω = t + ikx 0 k λ λ 2π k 2π/λ k ω/v v n v c/n k = nω c c ω/2π λ k 2πn/λ 2π/(λ/n) κ n n κ N n iκ k = Nω c iωt + inωx c iωt + i( n+ iκ ) ωx
x E E E e i ω t + ikx k λ λ π k π/λ k ω/v v n v c/n k nω c c ω/π λ k πn/λ π/(λ/n) κ n n κ N n iκ k Nω c iωt + inωx c iωt + i( n+ iκ ) ωx c κω x c iω ( t nx c) E E e E e E e e κ e ωκx/c e iω(t nx/c) I I
More information(3) (2),,. ( 20) ( s200103) 0.7 x C,, x 2 + y 2 + ax = 0 a.. D,. D, y C, C (x, y) (y 0) C m. (2) D y = y(x) (x ± y 0), (x, y) D, m, m = 1., D. (x 2 y
[ ] 7 0.1 2 2 + y = t sin t IC ( 9) ( s090101) 0.2 y = d2 y 2, y = x 3 y + y 2 = 0 (2) y + 2y 3y = e 2x 0.3 1 ( y ) = f x C u = y x ( 15) ( s150102) [ ] y/x du x = Cexp f(u) u (2) x y = xey/x ( 16) ( s160101)
More informationi 18 2H 2 + O 2 2H 2 + ( ) 3K
i 18 2H 2 + O 2 2H 2 + ( ) 3K ii 1 1 1.1.................................. 1 1.2........................................ 3 1.3......................................... 3 1.4....................................
More informationall.dvi
72 9 Hooke,,,. Hooke. 9.1 Hooke 1 Hooke. 1, 1 Hooke. σ, ε, Young. σ ε (9.1), Young. τ γ G τ Gγ (9.2) X 1, X 2. Poisson, Poisson ν. ν ε 22 (9.) ε 11 F F X 2 X 1 9.1: Poisson 9.1. Hooke 7 Young Poisson G
More information64 3 g=9.85 m/s 2 g=9.791 m/s 2 36, km ( ) 1 () 2 () m/s : : a) b) kg/m kg/m k
63 3 Section 3.1 g 3.1 3.1: : 64 3 g=9.85 m/s 2 g=9.791 m/s 2 36, km ( ) 1 () 2 () 3 9.8 m/s 2 3.2 3.2: : a) b) 5 15 4 1 1. 1 3 14. 1 3 kg/m 3 2 3.3 1 3 5.8 1 3 kg/m 3 3 2.65 1 3 kg/m 3 4 6 m 3.1. 65 5
More information5. [1 ] 1 [], u(x, t) t c u(x, t) x (5.3) ξ x + ct, η x ct (5.4),u(x, t) ξ, η u(ξ, η), ξ t,, ( u(ξ,η) ξ η u(x, t) t ) u(x, t) { ( u(ξ, η) c t ξ ξ { (
5 5.1 [ ] ) d f(t) + a d f(t) + bf(t) : f(t) 1 dt dt ) u(x, t) c u(x, t) : u(x, t) t x : ( ) ) 1 : y + ay, : y + ay + by : ( ) 1 ) : y + ay, : yy + ay 3 ( ): ( ) ) : y + ay, : y + ay b [],,, [ ] au xx
More informationNote.tex 2008/09/19( )
1 20 9 19 2 1 5 1.1........................ 5 1.2............................. 8 2 9 2.1............................. 9 2.2.............................. 10 3 13 3.1.............................. 13 3.2..................................
More information2000年度『数学展望 I』講義録
2000 I I IV I II 2000 I I IV I-IV. i ii 3.10 (http://www.math.nagoya-u.ac.jp/ kanai/) 2000 A....1 B....4 C....10 D....13 E....17 Brouwer A....21 B....26 C....33 D....39 E. Sperner...45 F....48 A....53
More information2011de.dvi
211 ( 4 2 1. 3 1.1............................... 3 1.2 1- -......................... 13 1.3 2-1 -................... 19 1.4 3- -......................... 29 2. 37 2.1................................ 37
More informationI
I 6 4 10 1 1 1.1............... 1 1................ 1 1.3.................... 1.4............... 1.4.1.............. 1.4................. 1.4.3........... 3 1.4.4.. 3 1.5.......... 3 1.5.1..............
More information2007 5 iii 1 1 1.1.................... 1 2 5 2.1 (shear stress) (shear strain)...... 5 2.1.1...................... 6 2.1.2.................... 6 2.2....................... 7 2.2.1........................
More informationV(x) m e V 0 cos x π x π V(x) = x < π, x > π V 0 (i) x = 0 (V(x) V 0 (1 x 2 /2)) n n d 2 f dξ 2ξ d f 2 dξ + 2n f = 0 H n (ξ) (ii) H
199 1 1 199 1 1. Vx) m e V cos x π x π Vx) = x < π, x > π V i) x = Vx) V 1 x /)) n n d f dξ ξ d f dξ + n f = H n ξ) ii) H n ξ) = 1) n expξ ) dn dξ n exp ξ )) H n ξ)h m ξ) exp ξ )dξ = π n n!δ n,m x = Vx)
More information9 1. (Ti:Al 2 O 3 ) (DCM) (Cr:Al 2 O 3 ) (Cr:BeAl 2 O 4 ) Ĥ0 ψ n (r) ω n Schrödinger Ĥ 0 ψ n (r) = ω n ψ n (r), (1) ω i ψ (r, t) = [Ĥ0 + Ĥint (
9 1. (Ti:Al 2 O 3 ) (DCM) (Cr:Al 2 O 3 ) (Cr:BeAl 2 O 4 ) 2. 2.1 Ĥ ψ n (r) ω n Schrödinger Ĥ ψ n (r) = ω n ψ n (r), (1) ω i ψ (r, t) = [Ĥ + Ĥint (t)] ψ (r, t), (2) Ĥ int (t) = eˆxe cos ωt ˆdE cos ωt, (3)
More informationN/m f x x L dl U 1 du = T ds pdv + fdl (2.1)
23 2 2.1 10 5 6 N/m 2 2.1.1 f x x L dl U 1 du = T ds pdv + fdl (2.1) 24 2 dv = 0 dl ( ) U f = T L p,t ( ) S L p,t (2.2) 2 ( ) ( ) S f = L T p,t p,l (2.3) ( ) U f = L p,t + T ( ) f T p,l (2.4) 1 f e ( U/
More informationmeiji_resume_1.PDF
β β β (q 1,q,..., q n ; p 1, p,..., p n ) H(q 1,q,..., q n ; p 1, p,..., p n ) Hψ = εψ ε k = k +1/ ε k = k(k 1) (x, y, z; p x, p y, p z ) (r; p r ), (θ; p θ ), (ϕ; p ϕ ) ε k = 1/ k p i dq i E total = E
More information/ Christopher Essex Radiation and the Violation of Bilinearity in the Thermodynamics of Irreversible Processes, Planet.Space Sci.32 (1984) 1035 Radiat
/ Christopher Essex Radiation and the Violation of Bilinearity in the Thermodynamics of Irreversible Processes, Planet.Space Sci.32 (1984) 1035 Radiation and the Continuing Failure of the Bilinear Formalism,
More information211 kotaro@math.titech.ac.jp 1 R *1 n n R n *2 R n = {(x 1,..., x n ) x 1,..., x n R}. R R 2 R 3 R n R n R n D D R n *3 ) (x 1,..., x n ) f(x 1,..., x n ) f D *4 n 2 n = 1 ( ) 1 f D R n f : D R 1.1. (x,
More informationt = h x z z = h z = t (x, z) (v x (x, z, t), v z (x, z, t)) ρ v x x + v z z = 0 (1) 2-2. (v x, v z ) φ(x, z, t) v x = φ x, v z
I 1 m 2 l k 2 x = 0 x 1 x 1 2 x 2 g x x 2 x 1 m k m 1-1. L x 1, x 2, ẋ 1, ẋ 2 ẋ 1 x = 0 1-2. 2 Q = x 1 + x 2 2 q = x 2 x 1 l L Q, q, Q, q M = 2m µ = m 2 1-3. Q q 1-4. 2 x 2 = h 1 x 1 t = 0 2 1 t x 1 (t)
More information構造と連続体の力学基礎
II 37 Wabash Avenue Bridge, Illinois 州 Winnipeg にある歩道橋 Esplanade Riel 橋6 6 斜張橋である必要は多分無いと思われる すぐ横に道路用桁橋有り しかも塔基部のレストランは 8 年には営業していなかった 9 9. 9.. () 97 [3] [5] k 9. m w(t) f (t) = f (t) + mg k w(t) Newton
More information8 300 mm 2.50 m/s L/s ( ) 1.13 kg/m MPa 240 C 5.00mm 120 kpa ( ) kg/s c p = 1.02kJ/kgK, R = 287J/kgK kPa, 17.0 C 118 C 870m 3 R = 287J
26 1 22 10 1 2 3 4 5 6 30.0 cm 1.59 kg 110kPa, 42.1 C, 18.0m/s 107kPa c p =1.02kJ/kgK 278J/kgK 30.0 C, 250kPa (c p = 1.02kJ/kgK, R = 287J/kgK) 18.0 C m/s 16.9 C 320kPa 270 m/s C c p = 1.02kJ/kgK, R = 292J/kgK
More informationD = [a, b] [c, d] D ij P ij (ξ ij, η ij ) f S(f,, {P ij }) S(f,, {P ij }) = = k m i=1 j=1 m n f(ξ ij, η ij )(x i x i 1 )(y j y j 1 ) = i=1 j
6 6.. [, b] [, d] ij P ij ξ ij, η ij f Sf,, {P ij } Sf,, {P ij } k m i j m fξ ij, η ij i i j j i j i m i j k i i j j m i i j j k i i j j kb d {P ij } lim Sf,, {P ij} kb d f, k [, b] [, d] f, d kb d 6..
More information.2 ρ dv dt = ρk grad p + 3 η grad (divv) + η 2 v.3 divh = 0, rote + c H t = 0 dive = ρ, H = 0, E = ρ, roth c E t = c ρv E + H c t = 0 H c E t = c ρv T
NHK 204 2 0 203 2 24 ( ) 7 00 7 50 203 2 25 ( ) 7 00 7 50 203 2 26 ( ) 7 00 7 50 203 2 27 ( ) 7 00 7 50 I. ( ν R n 2 ) m 2 n m, R = e 2 8πε 0 hca B =.09737 0 7 m ( ν = ) λ a B = 4πε 0ħ 2 m e e 2 = 5.2977
More information数学の基礎訓練I
I 9 6 13 1 1 1.1............... 1 1................ 1 1.3.................... 1.4............... 1.4.1.............. 1.4................. 3 1.4.3........... 3 1.4.4.. 3 1.5.......... 3 1.5.1..............
More informationb3e2003.dvi
15 II 5 5.1 (1) p, q p = (x + 2y, xy, 1), q = (x 2 + 3y 2, xyz, ) (i) p rotq (ii) p gradq D (2) a, b rot(a b) div [11, p.75] (3) (i) f f grad f = 1 2 grad( f 2) (ii) f f gradf 1 2 grad ( f 2) rotf 5.2
More information( ) sin 1 x, cos 1 x, tan 1 x sin x, cos x, tan x, arcsin x, arccos x, arctan x. π 2 sin 1 x π 2, 0 cos 1 x π, π 2 < tan 1 x < π 2 1 (1) (
6 20 ( ) sin, cos, tan sin, cos, tan, arcsin, arccos, arctan. π 2 sin π 2, 0 cos π, π 2 < tan < π 2 () ( 2 2 lim 2 ( 2 ) ) 2 = 3 sin (2) lim 5 0 = 2 2 0 0 2 2 3 3 4 5 5 2 5 6 3 5 7 4 5 8 4 9 3 4 a 3 b
More informationGauss Gauss ɛ 0 E ds = Q (1) xy σ (x, y, z) (2) a ρ(x, y, z) = x 2 + y 2 (r, θ, φ) (1) xy A Gauss ɛ 0 E ds = ɛ 0 EA Q = ρa ɛ 0 EA = ρea E = (ρ/ɛ 0 )e
7 -a 7 -a February 4, 2007 1. 2. 3. 4. 1. 2. 3. 1 Gauss Gauss ɛ 0 E ds = Q (1) xy σ (x, y, z) (2) a ρ(x, y, z) = x 2 + y 2 (r, θ, φ) (1) xy A Gauss ɛ 0 E ds = ɛ 0 EA Q = ρa ɛ 0 EA = ρea E = (ρ/ɛ 0 )e z
More information,. Black-Scholes u t t, x c u 0 t, x x u t t, x c u t, x x u t t, x + σ x u t, x + rx ut, x rux, t 0 x x,,.,. Step 3, 7,,, Step 6., Step 4,. Step 5,,.
9 α ν β Ξ ξ Γ γ o δ Π π ε ρ ζ Σ σ η τ Θ θ Υ υ ι Φ φ κ χ Λ λ Ψ ψ µ Ω ω Def, Prop, Th, Lem, Note, Remark, Ex,, Proof, R, N, Q, C [a, b {x R : a x b} : a, b {x R : a < x < b} : [a, b {x R : a x < b} : a,
More information7. y fx, z gy z gfx dz dx dz dy dy dx. g f a g bf a b fa 7., chain ule Ω, D R n, R m a Ω, f : Ω R m, g : D R l, fω D, b fa, f a g b g f a g f a g bf a
9 203 6 7 WWW http://www.math.meiji.ac.jp/~mk/lectue/tahensuu-203/ 2 8 8 7. 7 7. y fx, z gy z gfx dz dx dz dy dy dx. g f a g bf a b fa 7., chain ule Ω, D R n, R m a Ω, f : Ω R m, g : D R l, fω D, b fa,
More informationK E N Z OU
K E N Z OU 11 1 1 1.1..................................... 1.1.1............................ 1.1..................................................................................... 4 1.........................................
More information70の法則
70 70 1 / 27 70 1 2 3 4 5 6 2 / 27 70 70 70 X r % = 70 2 r r r 10 72 70 72 70 : 1, 2, 5, 7, 10, 14, 35, 70 72 : 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72 3 / 27 r = 10 70 r = 10 70 1 : X, X 10 = ( X + X
More informationNo δs δs = r + δr r = δr (3) δs δs = r r = δr + u(r + δr, t) u(r, t) (4) δr = (δx, δy, δz) u i (r + δr, t) u i (r, t) = u i x j δx j (5) δs 2
No.2 1 2 2 δs δs = r + δr r = δr (3) δs δs = r r = δr + u(r + δr, t) u(r, t) (4) δr = (δx, δy, δz) u i (r + δr, t) u i (r, t) = u i δx j (5) δs 2 = δx i δx i + 2 u i δx i δx j = δs 2 + 2s ij δx i δx j
More informationD xy D (x, y) z = f(x, y) f D (2 ) (x, y, z) f R z = 1 x 2 y 2 {(x, y); x 2 +y 2 1} x 2 +y 2 +z 2 = 1 1 z (x, y) R 2 z = x 2 y
5 5. 2 D xy D (x, y z = f(x, y f D (2 (x, y, z f R 2 5.. z = x 2 y 2 {(x, y; x 2 +y 2 } x 2 +y 2 +z 2 = z 5.2. (x, y R 2 z = x 2 y + 3 (2,,, (, 3,, 3 (,, 5.3 (. (3 ( (a, b, c A : (x, y, z P : (x, y, x
More informationTOP URL 1
TOP URL http://amonphys.web.fc.com/ 1 19 3 19.1................... 3 19.............................. 4 19.3............................... 6 19.4.............................. 8 19.5.............................
More informationn ξ n,i, i = 1,, n S n ξ n,i n 0 R 1,.. σ 1 σ i .10.14.15 0 1 0 1 1 3.14 3.18 3.19 3.14 3.14,. ii 1 1 1.1..................................... 1 1............................... 3 1.3.........................
More information1 No.1 5 C 1 I III F 1 F 2 F 1 F 2 2 Φ 2 (t) = Φ 1 (t) Φ 1 (t t). = Φ 1(t) t = ( 1.5e 0.5t 2.4e 4t 2e 10t ) τ < 0 t > τ Φ 2 (t) < 0 lim t Φ 2 (t) = 0
1 No.1 5 C 1 I III F 1 F 2 F 1 F 2 2 Φ 2 (t) = Φ 1 (t) Φ 1 (t t). = Φ 1(t) t = ( 1.5e 0.5t 2.4e 4t 2e 10t ) τ < 0 t > τ Φ 2 (t) < 0 lim t Φ 2 (t) = 0 0 < t < τ I II 0 No.2 2 C x y x y > 0 x 0 x > b a dx
More information6kg 1.1m 1.m.1m.1 l λ ϵ λ l + λ l l l dl dl + dλ ϵ dλ dl dl + dλ dl dl 3 1. JIS 1 6kg 1% 66kg 1 13 σ a1 σ m σ a1 σ m σ m σ a1 f f σ a1 σ a1 σ m f 4
35-8585 7 8 1 I I 1 1.1 6kg 1m P σ σ P 1 l l λ λ l 1.m 1 6kg 1.1m 1.m.1m.1 l λ ϵ λ l + λ l l l dl dl + dλ ϵ dλ dl dl + dλ dl dl 3 1. JIS 1 6kg 1% 66kg 1 13 σ a1 σ m σ a1 σ m σ m σ a1 f f σ a1 σ a1 σ m
More informationH 0 H = H 0 + V (t), V (t) = gµ B S α qb e e iωt i t Ψ(t) = [H 0 + V (t)]ψ(t) Φ(t) Ψ(t) = e ih0t Φ(t) H 0 e ih0t Φ(t) + ie ih0t t Φ(t) = [
3 3. 3.. H H = H + V (t), V (t) = gµ B α B e e iωt i t Ψ(t) = [H + V (t)]ψ(t) Φ(t) Ψ(t) = e iht Φ(t) H e iht Φ(t) + ie iht t Φ(t) = [H + V (t)]e iht Φ(t) Φ(t) i t Φ(t) = V H(t)Φ(t), V H (t) = e iht V (t)e
More informationRadiation from moving charges#1 Liénard-Wiechert potential Yuji Chinone 1 Maxwell Maxwell MKS E (x, t) + B (x, t) t = 0 (1) B (x, t) = 0 (2) B (x, t)
Radiation from moving harges# Liénard-Wiehert potential Yuji Chinone Maxwell Maxwell MKS E x, t + B x, t = B x, t = B x, t E x, t = µ j x, t 3 E x, t = ε ρ x, t 4 ε µ ε µ = E B ρ j A x, t φ x, t A x, t
More informationuntitled
8- My + Cy + Ky = f () t 8. C f () t ( t) = Ψq( t) () t = Ψq () t () t = Ψq () t = ( q q ) ; = [ ] y y y q Ψ φ φ φ = ( ϕ, ϕ, ϕ,3 ) 8. ψ Ψ MΨq + Ψ CΨq + Ψ KΨq = Ψ f ( t) Ψ MΨ = I; Ψ CΨ = C; Ψ KΨ = Λ; q
More information,,,17,,, ( ),, E Q [S T F t ] < S t, t [, T ],,,,,,,,
14 5 1 ,,,17,,,194 1 4 ( ),, E Q [S T F t ] < S t, t [, T ],,,,,,,, 1 4 1.1........................................ 4 5.1........................................ 5.........................................
More information1 1.1 Excel Excel Excel log 1, log 2, log 3,, log 10 e = ln 10 log cm 1mm 1 10 =0.1mm = f(x) f(x) = n
1 1.1 Excel Excel Excel log 1, log, log,, log e.7188188 ln log 1. 5cm 1mm 1 0.1mm 0.1 4 4 1 4.1 fx) fx) n0 f n) 0) x n n! n + 1 R n+1 x) fx) f0) + f 0) 1! x + f 0)! x + + f n) 0) x n + R n+1 x) n! 1 .
More informationohpr.dvi
2003/12/04 TASK PAF A. Fukuyama et al., Comp. Phys. Rep. 4(1986) 137 A. Fukuyama et al., Nucl. Fusion 26(1986) 151 TASK/WM MHD ψ θ ϕ ψ θ e 1 = ψ, e 2 = θ, e 3 = ϕ ϕ E = E 1 e 1 + E 2 e 2 + E 3 e 3 J :
More information7 π L int = gψ(x)ψ(x)φ(x) + (7.4) [ ] p ψ N = n (7.5) π (π +,π 0,π ) ψ (σ, σ, σ )ψ ( A) σ τ ( L int = gψψφ g N τ ) N π * ) (7.6) π π = (π, π, π ) π ±
7 7. ( ) SU() SU() 9 ( MeV) p 98.8 π + π 0 n 99.57 9.57 97.4 497.70 δm m 0.4%.% 0.% 0.8% π 9.57 4.96 Σ + Σ 0 Σ 89.6 9.46 K + K 0 49.67 (7.) p p = αp + βn, n n = γp + δn (7.a) [ ] p ψ ψ = Uψ, U = n [ α
More information50 2 I SI MKSA r q r q F F = 1 qq 4πε 0 r r 2 r r r r (2.2 ε 0 = 1 c 2 µ 0 c = m/s q 2.1 r q' F r = 0 µ 0 = 4π 10 7 N/A 2 k = 1/(4πε 0 qq
49 2 I II 2.1 3 e e = 1.602 10 19 A s (2.1 50 2 I SI MKSA 2.1.1 r q r q F F = 1 qq 4πε 0 r r 2 r r r r (2.2 ε 0 = 1 c 2 µ 0 c = 3 10 8 m/s q 2.1 r q' F r = 0 µ 0 = 4π 10 7 N/A 2 k = 1/(4πε 0 qq F = k r
More informationI A A441 : April 15, 2013 Version : 1.1 I Kawahira, Tomoki TA (Shigehiro, Yoshida )
I013 00-1 : April 15, 013 Version : 1.1 I Kawahira, Tomoki TA (Shigehiro, Yoshida) http://www.math.nagoya-u.ac.jp/~kawahira/courses/13s-tenbou.html pdf * 4 15 4 5 13 e πi = 1 5 0 5 7 3 4 6 3 6 10 6 17
More informationW u = u(x, t) u tt = a 2 u xx, a > 0 (1) D := {(x, t) : 0 x l, t 0} u (0, t) = 0, u (l, t) = 0, t 0 (2)
3 215 4 27 1 1 u u(x, t) u tt a 2 u xx, a > (1) D : {(x, t) : x, t } u (, t), u (, t), t (2) u(x, ) f(x), u(x, ) t 2, x (3) u(x, t) X(x)T (t) u (1) 1 T (t) a 2 T (t) X (x) X(x) α (2) T (t) αa 2 T (t) (4)
More informationI, II 1, A = A 4 : 6 = max{ A, } A A 10 10%
1 2006.4.17. A 3-312 tel: 092-726-4774, e-mail: hara@math.kyushu-u.ac.jp, http://www.math.kyushu-u.ac.jp/ hara/lectures/lectures-j.html Office hours: B A I ɛ-δ ɛ-δ 1. 2. A 1. 1. 2. 3. 4. 5. 2. ɛ-δ 1. ɛ-n
More information. ev=,604k m 3 Debye ɛ 0 kt e λ D = n e n e Ze 4 ln Λ ν ei = 5.6π / ɛ 0 m/ e kt e /3 ν ei v e H + +e H ev Saha x x = 3/ πme kt g i g e n
003...............................3 Debye................. 3.4................ 3 3 3 3. Larmor Cyclotron... 3 3................ 4 3.3.......... 4 3.3............ 4 3.3...... 4 3.3.3............ 5 3.4.........
More information<4D F736F F D B B83578B6594BB2D834A836F815B82D082C88C60202E646F63>
基礎からの冷凍空調 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. http://www.morikita.co.jp/books/mid/067311 このサンプルページの内容は, 初版 1 刷発行当時のものです. http://www.morikita.co.jp/support. 03-3817-5670FAX 03-3815-8199 i () () Q&A
More information2 1 κ c(t) = (x(t), y(t)) ( ) det(c (t), c x (t)) = det (t) x (t) y (t) y = x (t)y (t) x (t)y (t), (t) c (t) = (x (t)) 2 + (y (t)) 2. c (t) =
1 1 1.1 I R 1.1.1 c : I R 2 (i) c C (ii) t I c (t) (0, 0) c (t) c(i) c c(t) 1.1.2 (1) (2) (3) (1) r > 0 c : R R 2 : t (r cos t, r sin t) (2) C f : I R c : I R 2 : t (t, f(t)) (3) y = x c : R R 2 : t (t,
More information, 1.,,,.,., (Lin, 1955).,.,.,.,. f, 2,. main.tex 2011/08/13( )
81 4 2 4.1, 1.,,,.,., (Lin, 1955).,.,.,.,. f, 2,. 82 4.2. ζ t + V (ζ + βy) = 0 (4.2.1), V = 0 (4.2.2). (4.2.1), (3.3.66) R 1 Φ / Z, Γ., F 1 ( 3.2 ). 7,., ( )., (4.2.1) 500 hpa., 500 hpa (4.2.1) 1949,.,
More informationuntitled
20010916 22;1017;23;20020108;15;20; 1 N = {1, 2, } Z + = {0, 1, 2, } Z = {0, ±1, ±2, } Q = { p p Z, q N} R = { lim a q n n a n Q, n N; sup a n < } R + = {x R x 0} n = {a + b 1 a, b R} u, v 1 R 2 2 R 3
More information総研大恒星進化概要.dvi
The Structure and Evolution of Stars I. Basic Equations. M r r =4πr2 ρ () P r = GM rρ. r 2 (2) r: M r : P and ρ: G: M r Lagrange r = M r 4πr 2 rho ( ) P = GM r M r 4πr. 4 (2 ) s(ρ, P ) s(ρ, P ) r L r T
More information4 2016 3 8 2.,. 2. Arakawa Jacobin., 2 Adams-Bashforth. Re = 80, 90, 100.. h l, h/l, Kármán, h/l 0.28,, h/l.., (2010), 46.2., t = 100 t = 2000 46.2 < Re 46.5. 1 1 4 2 6 2.1............................
More information/02/18
3 09/0/8 i III,,,, III,?,,,,,,,,,,,,,,,,,,,,?,?,,,,,,,,,,,,,,!!!,? 3,,,, ii,,,!,,,, OK! :!,,,, :!,,,,,, 3:!,, 4:!,,,, 5:!,,! 7:!,,,,, 8:!,! 9:!,,,,,,,,, ( ),, :, ( ), ( ), 6:!,,, :... : 3 ( )... iii,,
More informationMicrosoft Word - 11問題表紙(選択).docx
A B A.70g/cm 3 B.74g/cm 3 B C 70at% %A C B at% 80at% %B 350 C γ δ y=00 x-y ρ l S ρ C p k C p ρ C p T ρ l t l S S ξ S t = ( k T ) ξ ( ) S = ( k T) ( ) t y ξ S ξ / t S v T T / t = v T / y 00 x v S dy dx
More informationContents 1 Jeans (
Contents 1 Jeans 2 1.1....................................... 2 1.2................................. 2 1.3............................... 3 2 3 2.1 ( )................................ 4 2.2 WKB........................
More informationSeptember 25, ( ) pv = nrt (T = t( )) T: ( : (K)) : : ( ) e.g. ( ) ( ): 1
September 25, 2017 1 1.1 1.2 p = nr = 273.15 + t : : K : 1.3 1.3.1 : e.g. 1.3.2 : 1 intensive variable e.g. extensive variable e.g. 1.3.3 Equation of State e.g. p = nr X = A 2 2.1 2.1.1 Quantity of Heat
More informationgr09.dvi
.1, θ, ϕ d = A, t dt + B, t dtd + C, t d + D, t dθ +in θdϕ.1.1 t { = f1,t t = f,t { D, t = B, t =.1. t A, tdt e φ,t dt, C, td e λ,t d.1.3,t, t d = e φ,t dt + e λ,t d + dθ +in θdϕ.1.4 { = f1,t t = f,t {
More information24 I ( ) 1. R 3 (i) C : x 2 + y 2 1 = 0 (ii) C : y = ± 1 x 2 ( 1 x 1) (iii) C : x = cos t, y = sin t (0 t 2π) 1.1. γ : [a, b] R n ; t γ(t) = (x
24 I 1.1.. ( ) 1. R 3 (i) C : x 2 + y 2 1 = 0 (ii) C : y = ± 1 x 2 ( 1 x 1) (iii) C : x = cos t, y = sin t (0 t 2π) 1.1. γ : [a, b] R n ; t γ(t) = (x 1 (t), x 2 (t),, x n (t)) ( ) ( ), γ : (i) x 1 (t),
More information19 σ = P/A o σ B Maximum tensile strength σ % 0.2% proof stress σ EL Elastic limit Work hardening coefficient failure necking σ PL Proportional
19 σ = P/A o σ B Maximum tensile strength σ 0. 0.% 0.% proof stress σ EL Elastic limit Work hardening coefficient failure necking σ PL Proportional limit ε p = 0.% ε e = σ 0. /E plastic strain ε = ε e
More information平成12年度
1 1-1 (1) 150[ml] 500[ml/] Cerebral Ventricle Brain 1-1 2 ( ) 1-1 1-2 0.20.5[mm] 13 14[mm] 1-2 3 ( ) (2) 4 2-1 (cerebral ventricle) (peritoneum) R O p O Cerebral Ventricle Valve Brain R o R i P i Peritoneum
More informationユニセフ表紙_CS6_三.indd
16 179 97 101 94 121 70 36 30,552 1,042 100 700 61 32 110 41 15 16 13 35 13 7 3,173 41 1 4,700 77 97 81 47 25 26 24 40 22 14 39,208 952 25 5,290 71 73 x 99 185 9 3 3 3 8 2 1 79 0 d 1 226 167 175 159 133
More informationf : R R f(x, y) = x + y axy f = 0, x + y axy = 0 y 直線 x+y+a=0 に漸近し 原点で交叉する美しい形をしている x +y axy=0 X+Y+a=0 o x t x = at 1 + t, y = at (a > 0) 1 + t f(x, y
017 8 10 f : R R f(x) = x n + x n 1 + 1, f(x) = sin 1, log x x n m :f : R n R m z = f(x, y) R R R R, R R R n R m R n R m R n R m f : R R f (x) = lim h 0 f(x + h) f(x) h f : R n R m m n M Jacobi( ) m n
More information() x + y + y + x dy dx = 0 () dy + xy = x dx y + x y ( 5) ( s55906) 0.7. (). 5 (). ( 6) ( s6590) 0.8 m n. 0.9 n n A. ( 6) ( s6590) f A (λ) = det(a λi)
0. A A = 4 IC () det A () A () x + y + z = x y z X Y Z = A x y z ( 5) ( s5590) 0. a + b + c b c () a a + b + c c a b a + b + c 0 a b c () a 0 c b b c 0 a c b a 0 0. A A = 7 5 4 5 0 ( 5) ( s5590) () A ()
More informationI A A441 : April 21, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka ) Google
I4 - : April, 4 Version :. Kwhir, Tomoki TA (Kondo, Hirotk) Google http://www.mth.ngoy-u.c.jp/~kwhir/courses/4s-biseki.html pdf 4 4 4 4 8 e 5 5 9 etc. 5 6 6 6 9 n etc. 6 6 6 3 6 3 7 7 etc 7 4 7 7 8 5 59
More information: 2005 ( ρ t +dv j =0 r m m r = e E( r +e r B( r T 208 T = d E j 207 ρ t = = = e t δ( r r (t e r r δ( r r (t e r ( r δ( r r (t dv j =
72 Maxwell. Maxwell e r ( =,,N Maxwell rot E + B t = 0 rot H D t = j dv D = ρ dv B = 0 D = ɛ 0 E H = μ 0 B ρ( r = j( r = N e δ( r r = N e r δ( r r = : 2005 ( 2006.8.22 73 207 ρ t +dv j =0 r m m r = e E(
More information8 i, III,,,, III,, :!,,,, :!,,,,, 4:!,,,,,,!,,,, OK! 5:!,,,,,,,,,, OK 6:!, 0, 3:!,,,,! 7:!,,,,,, ii,,,,,, ( ),, :, ( ), ( ), :... : 3 ( )...,, () : ( )..., :,,, ( ), (,,, ),, (ϵ δ ), ( ), (ˆ ˆ;),,,,,,!,,,,.,,
More informationsec13.dvi
13 13.1 O r F R = m d 2 r dt 2 m r m = F = m r M M d2 R dt 2 = m d 2 r dt 2 = F = F (13.1) F O L = r p = m r ṙ dl dt = m ṙ ṙ + m r r = r (m r ) = r F N. (13.2) N N = R F 13.2 O ˆn ω L O r u u = ω r 1 1:
More informationJuly 28, H H 0 H int = H H 0 H int = H int (x)d 3 x Schrödinger Picture Ψ(t) S =e iht Ψ H O S Heisenberg Picture Ψ H O H (t) =e iht O S e i
July 8, 4. H H H int H H H int H int (x)d 3 x Schrödinger Picture Ψ(t) S e iht Ψ H O S Heisenberg Picture Ψ H O H (t) e iht O S e iht Interaction Picture Ψ(t) D e iht Ψ(t) S O D (t) e iht O S e ih t (Dirac
More information.5 z = a + b + c n.6 = a sin t y = b cos t dy d a e e b e + e c e e e + e 3 s36 3 a + y = a, b > b 3 s363.7 y = + 3 y = + 3 s364.8 cos a 3 s365.9 y =,
[ ] IC. r, θ r, θ π, y y = 3 3 = r cos θ r sin θ D D = {, y ; y }, y D r, θ ep y yddy D D 9 s96. d y dt + 3dy + y = cos t dt t = y = e π + e π +. t = π y =.9 s6.3 d y d + dy d + y = y =, dy d = 3 a, b
More informationv er.1/ c /(21)
12 -- 1 1 2009 1 17 1-1 1-2 1-3 1-4 2 2 2 1-5 1 1-6 1 1-7 1-1 1-2 1-3 1-4 1-5 1-6 1-7 c 2011 1/(21) 12 -- 1 -- 1 1--1 1--1--1 1 2009 1 n n α { n } α α { n } lim n = α, n α n n ε n > N n α < ε N {1, 1,
More informationma22-9 u ( v w) = u v w sin θê = v w sin θ u cos φ = = 2.3 ( a b) ( c d) = ( a c)( b d) ( a d)( b c) ( a b) ( c d) = (a 2 b 3 a 3 b 2 )(c 2 d 3 c 3 d
A 2. x F (t) =f sin ωt x(0) = ẋ(0) = 0 ω θ sin θ θ 3! θ3 v = f mω cos ωt x = f mω (t sin ωt) ω t 0 = f ( cos ωt) mω x ma2-2 t ω x f (t mω ω (ωt ) 6 (ωt)3 = f 6m ωt3 2.2 u ( v w) = v ( w u) = w ( u v) ma22-9
More information