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- えつみ おおばま
- 9 years ago
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20 7 1 22 7 1 1 2 3 7 8 9 10 11 13 14 15 17 18 19 21 22 - 1 - - 2 - - 3 - - 4 - 50 200 50 200-5 - 50 200 50 200 50 200 - 6 - - 7 - () - 8 - (XY) - 9 - 112-10 - - 11 - - 12 - - 13 - - 14 - - 15 - - 16 -
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19 1 19 19 3 8 1 19 1 61 2 479 1965 64 1237 148 1272 58 183 X 1 X 2 12 2 15 A B 5 18 B 29 X 1 12 10 31 A 1 58 Y B 14 1 25 3 31 1 5 5 15 Y B 1 232 Y B 1 4235 14 11 8 5350 2409 X 1 15 10 10 B Y Y 2 X 1 X
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15 10 16 42 55 55 56 60 62 199310 1995 134 10 8 15 1 13 1311 a s d f 141412 2 g h j 376104 3 104102 232 4 5 51 30 53 27 36 6 Y 7 8 9 10 8686 86 11 1310 15 12 Z 13 14 15 16 102193 23 1712 60 27 17 18 Z
7 27 7.1........................................ 27 7.2.......................................... 28 1 ( a 3 = 3 = 3 a a > 0(a a a a < 0(a a a -1 1 6
26 11 5 1 ( 2 2 2 3 5 3.1...................................... 5 3.2....................................... 5 3.3....................................... 6 3.4....................................... 7
応力とひずみ.ppt
in [email protected] 2 3 4 5 x 2 6 Continuum) 7 8 9 F F 10 F L L F L 1 L F L F L F 11 F L F F L F L L L 1 L 2 12 F L F! A A! S! = F S 13 F L L F F n = F " cos# F t = F " sin# S $ = S cos# S S
ac b 0 r = r a 0 b 0 y 0 cy 0 ac b 0 f(, y) = a + by + cy ac b = 0 1 ac b = 0 z = f(, y) f(, y) 1 a, b, c 0 a 0 f(, y) = a ( ( + b ) ) a y ac b + a y
01 4 17 1.. y f(, y) = a + by + cy + p + qy + r a, b, c 0 y b b 1 z = f(, y) z = a + by + cy z = p + qy + r (, y) z = p + qy + r 1 y = + + 1 y = y = + 1 6 + + 1 ( = + 1 ) + 7 4 16 y y y + = O O O y = y
( ) x y f(x, y) = ax
013 4 16 5 54 (03-5465-7040) [email protected] hp://lecure.ecc.u-okyo.ac.jp/~nkiyono/inde.hml 1.. y f(, y) = a + by + cy + p + qy + r a, b, c 0 y b b 1 z = f(, y) z = a + by + cy z = p + qy
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変 位 変位とは 物体中のある点が変形後に 別の点に異動したときの位置の変化で あり ベクトル量である 変位には 物体の変形の他に剛体運動 剛体変位 が含まれている 剛体変位 P(x, y, z) 平行移動と回転 P! (x + u, y + v, z + w) Q(x + d x, y + dy,
変 位 変位とは 物体中のある点が変形後に 別の点に異動したときの位置の変化で あり ベクトル量である 変位には 物体の変形の他に剛体運動 剛体変位 が含まれている 剛体変位 P(x, y, z) 平行移動と回転 P! (x + u, y + v, z + w) Q(x + d x, y + dy, z + dz) Q! (x + d x + u + du, y + dy + v + dv, z +
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1 ( 12 11 44 7 20 10 10 1 1 ( ( 2 10 46 11 10 10 5 8 3 2 6 9 47 2 3 48 4 2 2 ( 97 12 ) 97 12 -Spencer modulus moduli (modulus of elasticity) modulus (le) module modulus module 4 b θ a q φ p 1: 3 (le) module
42,955 27,585 6,696 53,020 44,577 2,198 43,625 25,917 16,007 12,213 12,062 11,788 0 20,000 40,000 60,000 80,000 100,000 120,000 140,000
Disclosure 2005 TOMAKOMAI SHINKIN BANK 11 350,000 300,000 250,000 200,000 150,000 100,000 50,000 24,281 21,469 22,693 66,792 67,769 73,165 216,099 222,407 228,456 0 7 42,955 27,585 6,696 53,020 44,577
p q p q p q p q p q p q p q p q p q x y p q t u r s p q p p q p q p q p p p q q p p p q P Q [] p, q P Q [] P Q P Q [ p q] P Q Q P [ q p] p q imply / m
P P N p() N : p() N : p() N 3,4,5, L N : N : N p() N : p() N : p() N p() N p() p( ) N : p() k N : p(k) p( k ) k p(k) k k p( k ) k k k 5 k 5 N : p() p() p( ) p q p q p q p q p q p q p q p q p q x y p q
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96 7 1m =2 10 7 N 1A 7.1 7.2 a C (1) I (2) A C I A A a A a A A a C C C 7.2: C A C A = = µ 0 2π (1) A C 7.2 AC C A 3 3 µ0 I 2 = 2πa. (2) A C C 7.2 A A
7 Lorentz 7.1 Ampère I 1 I 2 I 2 I 1 L I 1 I 2 21 12 L r 21 = 12 = µ 0 2π I 1 I 2 r L. (7.1) 7.1 µ 0 =4π 10 7 N A 2 (7.2) magnetic permiability I 1 I 2 I 1 I 2 12 21 12 21 7.1: 1m 95 96 7 1m =2 10 7 N
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数学Ⅲ立体アプローチ.pdf
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dy + P (x)y = Q(x) (1) dx dy dx = P (x)y + Q(x) P (x), Q(x) dy y dx Q(x) 0 homogeneous dy dx = P (x)y 1 y dy = P (x) dx log y = P (x) dx + C y = C exp
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Microsoft Word - 計算力学2007有限要素法.doc
95 2 x y yz = zx = yz = zx = { } T = { x y z xy } () {} T { } T = { x y z xy } = u u x y u z u x x y z y + u y (2) x u x u y x y x y z xy E( ) = ( + )( 2) 2 2( ) x y z xy (3) E x y z z = z = (3) z x y
Image Browser Ver3.5 Manual. Zeiss Image Browser Ver3.5 p2. p3. p5. p7. p9. p13. p17. p19. p23 Appendix 1. p25 Appendix 2. p27 LSM Image Browser http:
Image Browser Ver3.5 Manual 200505av10 Image Browser Ver3.5 Manual. Zeiss Image Browser Ver3.5 p2. p3. p5. p7. p9. p13. p17. p19. p23 Appendix 1. p25 Appendix 2. p27 LSM Image Browser http://www.zeiss.de/imagebrowser
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2011 I 2 II III 17, 18, 19 7 7 1 2 2 2 1 2 1 1 1.1.............................. 2 1.2 : 1.................... 4 1.2.1 2............................... 5 1.3 : 2.................... 5 1.3.1 2.....................................
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n =3, 200 2 10 1 1 n =3, 2 n 3 x n + y n = z n x, y, z 3 a, b b = aq q a b a b b a b a a b a, b a 0 b 0 a, b 2 a, b (a, b) =1a b 1 x 2 + y 2 = z 2, (x, y) =1, x 0 (mod 2) (1.1) x =2ab, y = a 2 b 2, z =
B line of mgnetic induction AB MN ds df (7.1) (7.3) (8.1) df = µ 0 ds, df = ds B = B ds 2π A B P P O s s Q PQ R QP AB θ 0 <θ<π
8 Biot-Svt Ampèe Biot-Svt 8.1 Biot-Svt 8.1.1 Ampèe B B B = µ 0 2π. (8.1) B N df B ds A M 8.1: Ampèe 107 108 8 0 B line of mgnetic induction 8.1 8.1 AB MN ds df (7.1) (7.3) (8.1) df = µ 0 ds, df = ds B
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Microsoft Word - K-ピタゴラス数.doc
- ピタゴラス数の代数と幾何学 津山工業高等専門学校 菅原孝慈 ( 情報工学科 年 ) 野山由貴 ( 情報工学科 年 ) 草地弘幸 ( 電子制御工学科 年 ) もくじ * 第 章ピタゴラス数の幾何学 * 第 章ピタゴラス数の代数学 * 第 3 章代数的極小元の幾何学の考察 * 第 章ピタゴラス数の幾何学的研究の動機 交点に注目すると, つの曲線が直交しているようにみえる. これらは本当に直交しているのだろうか.
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座標変換におけるテンソル成分の変換行列
座標変換におけるテンソル成分の変換行列 座標変換におけるテンソル成分の変換関係は 次元数によらず階数によって定義される変換行列で整理することができる 位置ベクトルの変換行列を D としてそれを示そう D の行列式を ( = D ) とするとき 鏡映や回映といった pseudo rotation に対しては = -1 である が問題になる基底は 対称操作に含まれる pseudo rotation に依存する
1 1 2 1 3 1 4 2 4.1 AKB............................................... 2 4.2......................................... 6 4.3...........................
24 3 28 : 1 1 2 1 3 1 4 2 4.1 AKB............................................... 2 4.2......................................... 6 4.3............................................. 9 5 9 5.1.........................................
Microsoft Word - 触ってみよう、Maximaに2.doc
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m dv = mg + kv2 dt m dv dt = mg k v v m dv dt = mg + kv2 α = mg k v = α 1 e rt 1 + e rt m dv dt = mg + kv2 dv mg + kv 2 = dt m dv α 2 + v 2 = k m dt d
m v = mg + kv m v = mg k v v m v = mg + kv α = mg k v = α e rt + e rt m v = mg + kv v mg + kv = m v α + v = k m v (v α (v + α = k m ˆ ( v α ˆ αk v = m v + α ln v α v + α = αk m t + C v α v + α = e αk m
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4.6: 3 sin 5 sin θ θ t θ 2t θ 4t : sin ωt ω sin θ θ ωt sin ωt 1 ω ω [rad/sec] 1 [sec] ω[rad] [rad/sec] 5.3 ω [rad/sec] 5.7: 2t 4t sin 2t sin 4t
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システムの概要
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: LES Energy RANS Kawai & Lele JCP (2010) Wavenumber Accuracy LES, DNS RANS 1970 s Efficiency
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10. : (NMR) ( 1 H) MRI ρ H (x,y,z) NMR (Nuclear Magnetic Resonance) spectrometry: NMR NMR s( B ) m m = µ 0 IA = γ J (1) γ: :Planck constant J: Ĵ 2 = J(J +1),Ĵz = J J: (J = 1 2 for 1 H) I m A 173/197 10.1
.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
a n a n ( ) (1) a m a n = a m+n (2) (a m ) n = a mn (3) (ab) n = a n b n (4) a m a n = a m n ( m > n ) m n 4 ( ) 552
3 3.0 a n a n ( ) () a m a n = a m+n () (a m ) n = a mn (3) (ab) n = a n b n (4) a m a n = a m n ( m > n ) m n 4 ( ) 55 3. (n ) a n n a n a n 3 4 = 8 8 3 ( 3) 4 = 8 3 8 ( ) ( ) 3 = 8 8 ( ) 3 n n 4 n n
120 9 I I 1 I 2 I 1 I 2 ( a) ( b) ( c ) I I 2 I 1 I ( d) ( e) ( f ) 9.1: Ampère (c) (d) (e) S I 1 I 2 B ds = µ 0 ( I 1 I 2 ) I 1 I 2 B ds =0. I 1 I 2
9 E B 9.1 9.1.1 Ampère Ampère Ampère s law B S µ 0 B ds = µ 0 j ds (9.1) S rot B = µ 0 j (9.2) S Ampère Biot-Savart oulomb Gauss Ampère rot B 0 Ampère µ 0 9.1 (a) (b) I B ds = µ 0 I. I 1 I 2 B ds = µ 0
