Q = va = kia (1.2) 1.2 ( ) 2 ( 1.2) 1.2(a) (1.2) k = Q/iA = Q L/h A (1.3) 1.2(b) t 1 t 2 h 1 h 2 a
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- みひな さくいし
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1 (Darcy) v(cm/s) (1.1) v = ki (1.1) v k i 1.1 h ( )L i = h/l 1.1 t 1 h(cm) (t 2 t 1 ) 1.1 A Q(cm 3 /s)
2 Q = va = kia (1.2) 1.2 ( ) 2 ( 1.2) 1.2(a) (1.2) k = Q/iA = Q L/h A (1.3) 1.2(b) t 1 t 2 h 1 h 2 a
3 1.2 3 (a) (b) 1.2 a dt dh dq dq = adh dq dq = kaidt adh = kaidt = ka(h/l)dt h2 1 1 dh = ka h al dt h 1 1 h 1 t2 dh = ka dt al t 1
4 4 1 [log e h] h 2 h 1 = ka 1 al [t]t 2 t 1 (log e h 2 log e h 1 ) = ka 1 al (t 2 t 1 ) (log e h 1 log e h 2 ) = ka 1 al (t 2 t 1 ) k = k = h 1 log e = ka (t 2 t 1 ) h 2 al al A(t 2 t 1 ) log h 1 e (1.4) h aL A(t 2 t 1 ) log h 1 10 (1.5) h ( 10 cm 12.7 cm) 5 B ( θ S r k r ) B 0.95 ( 0.9 ) B B
5 1.3 5 (1) cm ( 28 cm 16 cm 9850 cm 3 ) ( 23,000 cm 3 ) 1.0 cm 12.0 cm 26.2 cm 50 kpa 1.3 (1.6) k s = QL Ath = 0.26 Q th (1.6)
6 6 第 1 章 飽和土の透水 ここに Q は流量 A は供試体の断面積 (19.6 cm2 ) h は水位差 L は供試体の高さ (5.1 cm) t は計測時間である 変水位透水試験の飽和透水係数は 次の式 (1.7) より求めることができる ks = 2.3aL h h1 log10 = log10 At h2 t h2 (1.7) ここで a はアクリル管の断面積 (13.85 cm2 ) A は供試体の断面積 (19.6 cm2 ) L は 供試体の高さ (5.1 cm) h1 は計測開始時の水位差 h2 は計測終了時の水位差 t は計測 時間である チャンバー内の供試体寸法は 式 (1.6) 式 (1.7) にも出てきたように 直径 5.0 cm 高さ 5.1 cm で体積は 100 cm3 である (写真-1.1) 供試体の組立は カラムにゴム製の O リングを取り付け回転させるだけで 試料受器に固定できるように工夫してあり 通水断 面を容易に確保できる 写真 1.1 チャンバー内の供試体 測定手順は 供試体をセットし底面から浸透させ チャンバー内の所定の水位まで給水 する この水位と二重管ビューレット内の水位を一致させ 真空脱気を 1 時間行い 背圧 P (圧力が低いと計測が難しいので kpa) をかけて供試体内に流入した水量 V を 計測し 式 (1.8) から飽和度を換算する Sr = 1 P0 V 1 H P Vv ここで H はヘンリー係数 P0 は大気圧 Vv は供試体の間隙部分の体積である (1.8)
7 1.3 7 B (29.4 kpa 9.81 kpa B ) B (2) 1.4 B B B B B 1.4 B 1.5 B ( ) 1.0 cm
8 Re (Re < 1) B B 3 B B B
9 B 0.98 (3) a) Re (laminar flow) (turbulent flow) (Reynolds) (Reynold s number) (Critical Reynold s number) R e = vd ν (1.9)
10 10 1 v (m/s) D (m) ν (m 2 /s) R e 2000 R e 1 R e = ρ wvd µ (1.10) v (cm/s) D (D 50 D 60 )(cm) µ (g/s cm) (1.1) m 0.5 < m < 1 v = ki m (1.11) b) B (A.W. Skempton) u = B { σ 3 + A ( σ 1 + σ 3 )} (1.12) A B A A f > 0 A f < 0 A( σ 1 σ 3 ) B B < 1 B = 1 K 0 V 0 e 0 n 0 V 0 n 0 σ v0 ( σ = hγ w + σ v0 h γ w ) u
11 V w K 0 C w V w = n 0 V 0 C w u (1.13) ( σ = hγ w + σ v0 ) σ v ε v ε v = C s σ v (1.14) C s K 0 V s V s = V 0 C s σ v (1.15) (1.13) (1.15) n 0 V 0 C w u = V 0 C s σ v (1.16) σ v = σ v + u (1.16) σ v u = n 0 C w /C s σ v (1.17) (1.17) C w /C s C w /C s (1.17) (1.18) u = σ v (1.18) (1.18) B B = u σ v = 1 (1.19) (1.20) B = u σ v < 1 (1.20)
12 12 1 (back pressure) c) 1.1 C K( ) H (1.21) S r S r = 1 [ ] 1 H P 0 1 P 0 P 1 H S r0 (1.21) H 1.1 P 0 (98.1 kn/m 2 ) P S r0 P 0 (V a0 + HV w0 ) = (P 0 + P )(V a1 + HV w1 ) (1.22) V a0 V a1 P V w0 V w1 P
13 V V a0 V a1 = V V w1 V w0 = V (1.23) (1.22) (1.23) V a1 V w1 V (H 1)(P 0 + P ) + P (V a0 + HV w0 ) = 0 (1.24) S r0 S r0 = 1 H V (P 0 + P ) V v P (1.25) (1.25) V v V P (1.21) (1.25) ( (1.8)) S r = 1 1 H P 0 V P V v (1.26) 1.4 (v = ki) k v k h 1.7 x v x ab (1.27) (1.28) ( v x x, z + z ) = v x (x, z) ( v x (x, z) + 1 ) v x 2 z z v x z (1.27) z z (1.28)
14 cd x (1.29) (1.30) ( v x x + x, z + z ) = v x (x, z) + v x 2 x x + 1 v x z (1.29) 2 z ( v x (x, z) + v x x x + 1 ) v x 2 z z z (1.30) 1.8(a) x q x (1.28) (1.30) ( q x = v x (x, z) + 1 ) ( v x 2 z z z v x (x, z) + v x x x + 1 ) v x 2 z z z = v x x z (1.31) x 1.8(b) q z = v z x z (1.32) z (1.31) (1.32) 0 q x + q z = v x x x z + v z z x z = v x x + v z z = 0 (1.33)
15 (a) (b) 1.8 v x k h ( ) h v x = k h i = k h x i h x i = h x, i = h z, v x = k h h x, v z = k v h z (1.34) (1.34) (1.33) k h = k v 2 h k h x 2 + k 2 h v z 2 = 0 (1.35) 2 h x h z 2 = 0 (1.36) (1.36)
16 ( ) H v ( α+ β = 90 ) H h H =
17 H x z(x z ) h h x = z = 0 (1.37) x z 1.9 (1.38) (1.39) v x H = cos α, z h v x = k h x, v x v = cos β, v z v H = sin α (1.38) v h z = k v z (1.39) = sin β (1.40) (1.39) (1.40) cos β = k h v h x, sin β = k v v h z (1.41) (1.38) (1.41) cos(α + β) = cos α cos β sin α sin β = x H = k ( ) h h x + Hv x z z ( k v ) ( h z ) ( k x H v ) h z (1.42) (1.42) (1.37) 0 y = cos x 0 π/2 α + β = ( ) 1 Q Q = kiat(cm 3 ) ( 1.11) a h 8 1 8( 1 8 ) N d
18 N f ( 4(1 4 ) ) N d N f 1 q q = kia = k h N d 1 a a 1 = k h N d (1.43) h/n d (1/a) i = h L = h 1 a 1 a N d a 1 A Q (1.43)
19 N f Q = k h N f N d (1.44) k = cm/s 1 m Q = k h N f N d = = 0.75 cm3 /s (1.45) h 27.0 m 19.5 m 7.5 m=750 cm cm 1 m 100 cm 2 v = ki = k( h/l) ( ) ( ) 3 (1.46) (p/ρ w ) z v 2 /2g p ρ w + z + v2 2g = const. (1.46) p ρ w + z = const. (1.47) (1.47) 1.12 A B h = =7.5 m El.18.0 m A 3.0 m B
20 m + = A p + z = H h ρ w N d p + ( 3.0) = ρ w 8 1 p = m (1.48) ρ w H z (7.5/8) 1 ( 1) B ( 6) p + ( 6.0) = ρ w 8 6 p = 9.38 m ρ w El.0.0 m A H = 27.0 m z = 15.0 m p = ρ w 8 1 p = m ρ w
21 N d N f ( ) 1 % Lambe N f = 4 N d = 12 (N f /N d = 0.333) k = m/s Q m 3 /s Lambe N f N d N f /N d Q( 10 6 m 3 /s) (%) (1.48) ( ) 1.14(a) (b) (a) (b) N f = 4 N d = m II 1 ( ) p + ( 1.5) = ρ w p = 4.71 m p = 4.71 t/m 2 = 46.2 kn/m 2 ρ w
22
23 (kn/m 2 ) II , III III 1 ( ) p + ( 1.5) = ρ w 14 2 p = 6.64 m p = 6.64 t/m 2 = 65.2 kn/m 2 ρ w
24 24 1 II x z R ( ) 1/R 1.15 k h k v 9
25 R = kv 1 = k h 9 = 1 3 (1.49) x ( ) R 1.15 III (a) (b) 1.14(b) 1.7 (Muskat) 1.16 Muskat D d 1 d 2 Q k h 1.10 (1.45) 1.10 k = cm/s 1 m 1.16 Muskat Q = k h N f N d = = 0.75 cm3 /s
26 D = 18.0 m d 1 = 9.0 m d 2 = 0 m d 1 /D = 9.0/18.0 = 0.5 d 2 /d 1 = Q/kh = 0.5 Q Q = 0.5 kh = = 0.75 cm 3 /s Q/kh N f /N d d α α (1.44) α = N d /N f k = Q h N d N f = Q h α (1.50) N d N f α
27 (Dupuit) (a) (b) (c) (a) o R abc 1.18(b) 1.18(b) (c) n q = kd dh dx (1.51) D = 2 tan θx θ = 45 D = 2x x = nr h = H x = R h = H/2 (1.52)
28 28 1 (1.51) (1.52) dh = q 2xk dx h = q 2k ln x + C C = q ln nr + H 2k H 2 = q 2k ln R q ln nr + H 2k H 2 = q 2k ln R nr k = q H ln 1 n (1.53) (1.50) (1.53) α ln n nr q = πrv = πrk dh dr (1.54)
29 (1.54) (1.52) dh = q πrk dr h = q πk ln r + C C = q ln nr + H πk H 2 = q πk ln R + q ln nr πk H 2 = q πk ln R nr k = q H ln 1 2 n π (1.55) α ln n 1 2/π (1.53) (1.55) α n 1.20 α 1.20
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降雨 地下水汚染 蒸発 揚水量 河川 地盤掘削 ダム 涵養 斜面崩壊 地すべり 漏水 地下水の塩水化 シールドトンネル 浸透圧 井戸地盤沈下 浸透量 浸透破壊湧水 流動 地下水に関する問題 トンネル掘削湧水, 周辺地下水低下 吸着水 地下水面 重力水 毛管水 不飽和領域 土粒子 地下水 飽和領域 土中水の存在形態 重力水は雨水, 地表面の貯留水, 流水などが地下に浸透し, 重力の作用により, 地下水面に向かって移動する水である
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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
1 (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
(1.2) T D = 0 T = D = 30 kn 1.2 (1.4) 2F W = 0 F = W/2 = 300 kn/2 = 150 kn 1.3 (1.9) R = W 1 + W 2 = = 1100 N. (1.9) W 2 b W 1 a = 0
1 1 1.1 1.) T D = T = D = kn 1. 1.4) F W = F = W/ = kn/ = 15 kn 1. 1.9) R = W 1 + W = 6 + 5 = 11 N. 1.9) W b W 1 a = a = W /W 1 )b = 5/6) = 5 cm 1.4 AB AC P 1, P x, y x, y y x 1.4.) P sin 6 + P 1 sin 45
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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
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65 8. K 8 8 7 8 K 6 7 8 K 6 M Q σ (6.4) M O ρ dθ D N d N 1 P Q B C (1 + ε)d M N N h 2 h 1 ( ) B (+) M 8.1: σ = E ρ (E, 1/ρ ) (8.1) 66 σ σ (8.1) σ = 0 0 σd = 0 (8.2) (8.2) (8.1) E ρ d = 0... d = 0 (8.3)
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II Karel Švadlenka * [1] 1.1* 5 23 m d2 x dt 2 = cdx kx + mg dt. c, g, k, m 1.2* u = au + bv v = cu + dv v u a, b, c, d R
![II Karel Švadlenka * [1] 1.1* 5 23 m d2 x dt 2 = cdx kx + mg dt. c, g, k, m 1.2* u = au + bv v = cu + dv v u a, b, c, d R](/thumbs/95/125540821.jpg "II Karel Švadlenka * [1] 1.1* 5 23 m d2 x dt 2 = cdx kx + mg dt. c, g, k, m 1.2* u = au + bv v = cu + dv v u a, b, c, d R")
II Karel Švadlenka 2018 5 26 * [1] 1.1* 5 23 m d2 x dt 2 = cdx kx + mg dt. c, g, k, m 1.2* 5 23 1 u = au + bv v = cu + dv v u a, b, c, d R 1.3 14 14 60% 1.4 5 23 a, b R a 2 4b < 0 λ 2 + aλ + b = 0 λ =
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Microsoft PowerPoint - 水と土の科学④
降雨 地下水汚染 蒸発 揚水量 河川 地盤掘削 ダム 涵養 斜面崩壊 地すべり 漏水 地下水の塩水化 シールドトンネル 浸透圧 井戸地盤沈下 浸透量 浸透破壊湧水 流動 地下水に関する問題 トンネル掘削湧水, 周辺地下水低下 吸着水 地下水面 重力水 毛管水 不飽和領域 土粒子 地下水 飽和領域 土中水の存在形態 重力水は雨水, 地表面の貯留水, 流水などが地下に浸透し, 重力の作用により, 地下水面に向かって移動する水である
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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
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
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38 5 Cauchy.,,,,., σ.,, 3,,. 5.1 Cauchy (a) (b) (a) (b) 5.1: 5.1. Cauchy 39 F Q Newton F F F Q F Q 5.2: n n ds df n ( 5.1). df n n df(n) df n, t n. t n = df n (5.1) ds 40 5 Cauchy t l n mds df n 5.3: t
c y /2 ddy = = 2π sin θ /2 dθd /2 [ ] 2π cos θ d = log 2 + a 2 d = log 2 + a 2 = log 2 + a a 2 d d + 2 = l
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c 28. 2, y 2, θ = cos θ y = sin θ 2 3, y, 3, θ, ϕ = sin θ cos ϕ 3 y = sin θ sin ϕ 4 = cos θ 5.2 2 e, e y 2 e, e θ e = cos θ e sin θ e θ 6 e y = sin θ e + cos θ e θ 7.3 sgn sgn = = { = + > 2 < 8.4 a b 2
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断面積 (A) を使わずに, 間隙率を使う透水係数の算定 図に示したような 本の孔を掘って, 上流側から食塩を投入した 食塩を投入してから,7 時間後に下流側に食塩が到達したことが分かった この地盤の透水係数を求めよ 地盤の間隙比は e=0.77, 水位差は 0 cmであった なお, この方法はトレーサ法の中の食塩法と呼ばれている Nacl 計測器 0 cm 0.0 m 断面積 (A) を使わずに,
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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
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I-2 (100 ) (1) y(x) y dy dx y d2 y dx 2 (a) y + 2y 3y = 9e 2x (b) x 2 y 6y = 5x 4 (2) Bernoulli B n (n = 0, 1, 2,...) x e x 1 = n=0 B 0 B 1 B 2 (3) co
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量子力学 問題
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9 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 (
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(Compton Scattering) Beaming 1 exp [i (k x ωt)] k λ k = 2π/λ ω = 2πν k = ω/c k x ωt ( ω ) k α c, k k x ωt η αβ k α x β diag( + ++) x β = (ct, x) O O x
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Compton Scattering Beaming exp [i k x ωt] k λ k π/λ ω πν k ω/c k x ωt ω k α c, k k x ωt η αβ k α x β diag + ++ x β ct, x O O x O O v k α k α β, γ k γ k βk, k γ k + βk k γ k k, k γ k + βk 3 k k 4 k 3 k
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)
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知能科学：ニューラルネットワーク
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知能科学：ニューラルネットワーク
2 3 4 (Neural Network) (Deep Learning) (Deep Learning) ( x x = ax + b x x x ? x x x w σ b = σ(wx + b) x w b w b .2.8.6 σ(x) = + e x.4.2 -.2 - -5 5 x w x2 w2 σ x3 w3 b = σ(w x + w 2 x 2 + w 3 x 3 + b) x,
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x A Aω ẋ ẋ 2 + ω 2 x 2 = ω 2 A 2. (ẋ, ωx) ζ ẋ + iωx ζ ζ dζ = ẍ + iωẋ = ẍ + iω(ζ iωx) dt dζ dt iωζ = ẍ + ω2 x (2.1) ζ ζ = Aωe iωt = Aω cos ωt + iaω sin
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1 I 1.1 ± e = = - = C C MKSA [m], [Kg] [s] [A] 1C 1A 1 MKSA 1C 1C +q q +q q 1
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9 7 A = A x x + A y y + A, B = B x x + B y y + B, C = C x x + C y y + C..6 x y A B C = A x x + A y y + A B x B y B C x C y C { B = A x x + A y y + A y B B x x B } B C y C y + x B y C x C C x C y B = A
ω 0 m(ẍ + γẋ + ω0x) 2 = ee (2.118) e iωt x = e 1 m ω0 2 E(ω). (2.119) ω2 iωγ Z N P(ω) = χ(ω)e = exzn (2.120) ϵ = ϵ 0 (1 + χ) ϵ(ω) ϵ 0 = 1 +
2.6 2.6.1 ω 0 m(ẍ + γẋ + ω0x) 2 = ee (2.118) e iωt x = e 1 m ω0 2 E(ω). (2.119) ω2 iωγ Z N P(ω) = χ(ω)e = exzn (2.120) ϵ = ϵ 0 (1 + χ) ϵ(ω) ϵ 0 = 1 + Ne2 m j f j ω 2 j ω2 iωγ j (2.121) Z ω ω j γ j f j
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3 3.1 ( 1 m d2 x(t dt 2 = kx(t k = (3.1 d 2 x dt 2 = ω2 x, ω = x(t = 0, ẋ(0 = v 0 k m (3.2 x = v 0 ω sin ωt (ẋ = v 0 cos ωt (3.3 E = 1 2 mẋ2 + 1 2 kx2 = 1 2 mv2 0 cos 2 ωt + 1 2 k v2 0 ω 2 sin2 ωt = 1
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4 4 ) a + b = i + 6 b c = 6i j ) a = 0 b = c = 0 ) â = i + j 0 ˆb = 4) a b = b c = j + ) cos α = cos β = 6) a ˆb = b ĉ = 0 7) a b = 6i j b c = i + 6j + 8) a b a b = 6i j 4 b c b c 9) a b = 4 a b) c = 7
ma22-9 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
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Mｏｔｔ散乱によるParity対称性の破れを検証
Mott Parity P2 Mott target Mott Parity Parity Γ = 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 t P P ),,, ( 3 2 1 0 1 γ γ γ γ γ γ ν ν µ µ = = Γ 1 : : : Γ P P P P x x P ν ν µ µ vector axial vector ν ν µ µ γ γ Γ ν γ
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Chap11.dvi
. () x 3 + dx () (x )(x ) dx + sin x sin x( + cos x) dx () x 3 3 x + + 3 x + 3 x x + x 3 + dx 3 x + dx 6 x x x + dx + 3 log x + 6 log x x + + 3 rctn ( ) dx x + 3 4 ( x 3 ) + C x () t x t tn x dx x. t x
橡博論表紙.PDF
Study on Retaining Wall Design For Circular Deep Shaft Undergoing Lateral Pressure During Construction 2003 3 Study on Retaining Wall Design For Circular Deep Shaft Undergoing Lateral Pressure During Construction
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5. [. ] z = f(, y) () z = 3 4 y + y + 3y () z = y (3) z = sin( y) (4) z = cos y (5) z = 4y (6) z = tan y (7) z = log( + y ) (8) z = tan y + + y ( ) () z = 3 8y + y z y = 4 + + 6y () z = y z y = (3) z =
18 2 F 12 r 2 r 1 (3) Coulomb km Coulomb M = kg F G = ( ) ( ) ( ) 2 = [N]. Coulomb
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r 1 r 2 r 1 r 2 2 Coulomb Gauss Coulomb 2.1 Coulomb 1 2 r 1 r 2 1 2 F 12 2 1 F 21 F 12 = F 21 = 1 4πε 0 1 2 r 1 r 2 2 r 1 r 2 r 1 r 2 (2.1) Coulomb ε 0 = 107 4πc 2 =8.854 187 817 10 12 C 2 N 1 m 2 (2.2)
m(ẍ + γẋ + ω 0 x) = ee (2.118) e iωt P(ω) = χ(ω)e = ex = e2 E(ω) m ω0 2 ω2 iωγ (2.119) Z N ϵ(ω) ϵ 0 = 1 + Ne2 m j f j ω 2 j ω2 iωγ j (2.120)
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1 1.1 [ 1] velocity [/s] 8 4 (1) MKS? (2) MKS? 1.2 [ 2] (1) (42.195k) k 2 (2) (3) k/hr [ 3] t = 0
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50 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
.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
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x () g(x) = f(t) dt f(x), F (x) 3x () g(x) g (x) f(x), F (x) (3) h(x) = x 3x tf(t) dt.9 = {(x, y) ; x, y, x + y } f(x, y) = xy( x y). h (x) f(x), F (x
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液晶の物理１：連続体理論（弾性，粘性）
The Physics of Liquid Crystals P. G. de Gennes and J. Prost (Oxford University Press, 1993) Liquid crystals are beautiful and mysterious; I am fond of them for both reasons. My hope is that some readers
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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
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6 2 2 x y x y t P P = P t P = I P P P ( ) ( ) ,, ( ) ( ) cos θ sin θ cos θ sin θ, sin θ cos θ sin θ cos θ y x θ x θ P
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H 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) = [
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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
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