|
|
|
- 萱获 缪
- 9 years ago
- Views:
Transcription
1
2
3
4
5
6
7
8
9
10
11 ο HOH O H A 1 1
12 A ο
13
14
15
16 Q C D t z = ρ z 1 1 z t D z z z t Qz = 1 2 z D z
17
18
19 2 2 Cl HCO SO CO 3 4 3
20
21 3 4 HCO SO F0. 005H SiO0. 032M T4 9 ( K + Na) Ca
22 C Ca
23 HCO 3 HCO + SO HCO SO 4 + Cl HCO 3 + Cl SO SO 4 + Cl + NH 4 HCO 3 SO 2 4 NO 3
24
25
26
27 1/2 S = K K K 15 3/ / K K ( 1-7) 15 15
28 HCO SO Cl NO Ca Na Mg K 3 4 3
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45 W T = 2 1 W
46
47
48
49
50 I Q = ds dt I t - θ t = s ( 2 2 ) I θ s t P + R E = 0 ( 2 6 ) P R R s
51 P E + R = P E R P = E 2 9 P = E ( P + P ) ( E + E ) = R 2 10 P = P + P E = P + E P E = R 2-11 P 800 E 485 R
52 P + R = E P R = E P + P = E + E 2-12 P = E 2-13
53
54
55 E R t L p = ( ) 2 15
56 E E M E = ϕ E 2-16 ϕ ϕ
57 ϕ j 2 E = 0. 22( e e ) u
58 E R H H n + a s = 2-23 L( 1+ β) H β = = C P T T 0 B 2-24 H e e e 0 R n + Ha Hs E = P T0 T L e e ( R n + H a + Hs) + LB( E2 s e 2) γ LE = / γ = e e 0s 2s T T s = k wρ u2 B k mρc 1 [ 1n( Z2 / Z1 )] u 2 Z2
59 Et F F Et = E t 2-30 F
60 E R = P ϕ 2-32 LP R E = P 1 e LP mm / a 2-33 E E th LP R = R L th LP p = ( mm / a) 2 34 R RP E = LP th L R ch R LP sh R 1 LP ( mm / a ) 2 35
61 E k c 2 = ( kg / m s) 2 36 x c c x k x Ex E x Ex + x x c x x t Ex c Ex Ex + x = x x 2-37 t
62 E c x + = x t c k x / x + c / t = x 2 c c = k x t x c c c c = k x 2 + k y 2 + k z t x ay z E = k c x T Q = λ w 2-42 Z 2 T Q w J / m s J / m s Z c t = 0
63 U c c c c = Dx 2 + Dy 2 + Dz x x y z c U c c c c + = Dx 2 + Dy 2 + Dz t x x y z U c x y z x
64 ϖ ρqu L Z ρqu ϖ P P = ρq Z Z P ρg ρqu ϖ L Z ρqu ϖ P 1 L ρg g qu ϖ = = L P 2-48 E = 1 ϖ qu 2-49 g ϖ ϖ u qu g = 1 ( / ) ( qu n L) i / q 2-50 F i ϖ ϖ 2-12, BC = 1, u(qu) / g ϖ ϖ u(qu) / g 0 ϖ ϖ u( qu) / g 0
65
66 P = R + E
67
68 P + P Pn P = Λ n
69 f1p1 + f2p2 + Λ + fnpn 1 P = = f + f + Λ + f f 1 2 n i= 1 f P i i P = f P F i i 2 54 i= 1
70 P j n PiWi i= 1 = = W n j W P n j i i i= 1 i 1 i= 2-56 W i j Wj Wi = n j j W i= 1 P N 1 P = Pj N j= 1 j 2-27
71
72
73 P = w m m P 2 59 w
74
75 df do = f 2 60 ϕ / z v = z k ( ϕ) ϕ z + k( ϕ ) 2-61 ϕ θ vz + = t z θ θ θ θ t = z D( ) z z k( ) 2 63
76 1 F( t) = st 2 + [ A2 + k( θi )] t df( t) 1 f( t) = = st 2 + [ A2 + k( θ i)] 2-65 dt 2 f k H s L p = s 2-66 L F = ( θ s θ i) L 2-67 ( θ s θ i ) S fp = k F f = df p dt 2 68 df = ks( 1 + ( θ s θ i ) S / F) 2-69 dt t F 1 k dt s F df s = ( θ θ ) / F ks t = F s( θs θi) 1 n + 1 ( θ s θ i) s 2-70 F k s + H k ( s + H )( θ θ ) 0 s o s i = 2 s t f = 2-71 θ θ 2t s i s i
77 βt f = fc + ( f0 fc) e 2 72 df = β( f f dt c ) n( f fc ) = βt + c t 1 t F = fdt = f t + f f 1 e β ( )( ) c β 0 c
78
79 Q QT R = F 1000Q M = 2 79 F α = R R P a 1 P
80
81
82
83
84
85
86 R ( t ) = idt i dt edt dt fdt ( 3-1) s t t t t t n s d 0 0 drs( t) = rs ( t) = i( t) f( t) ( 3-2) dt
87 t t t A B W(t) = W( 0 ) + f dt f dt R dt ( 3-3) ss dwt = 0 dt dwt = f A fb rss = 0 ( 3-4) dt rs s = fa f B ( 3-5) t t t 0 C 0 ( ) 0 g W = W + f dt r dt 3-6 t t t t 0 C 0 s s g 0 ( ) 0 W = W + f dt r dt r dt 3-7 dwt = f C rg = 0 dt
88 dw t = f C rss rg = 0 dt t t t t W t = W 0 + idt f Bdt rs sdt rs s t dt ( 3-8) dwt = 0 dt dwt = 0 = i rss fb rsa t dt
89 R = ( i f) 3-10 s t= n t= 0
90 dr dr 1 = 1 dp dp
91 Lm τ m = ( 3 12) V L m V
92 k = L 0 / V (3-13) n n I i( t) [ I i( t) ] ( ) i= = Φ i= 1
93 R = R = ( i t f t) ( 3-17) s n i s n i α = R P
94 2 v 1 2 v = mi 2R 3 ( 3-18) l1 l2 v = + ( 3-19) τ1 τ 2 ω = 1 = f τ Q = 1 ω1h t 1 = 1 t
95 ω1h 2 + ω 2h1 Q 1 = = t t ω1h3 + ω 2h 2 + ω 3h1 Q 3 = = t t ω 2h 3 + ω3h2 + ω4 h1 Q 4 = = t t ω3h3 + ω 4h2 + ω5h1 Q 5 = = t t ω5h3 + ω 5h2 Q 6 = = t t ω 5h3 Q 7 = = t t
96 h1 10Q Q1 = q 1 q 1 = 10 h 1 1 Q 2 h2 h h Q2 q 1 q 2 = 2 + q 1 q 10 2 = h /
97
98 Q t y = 1 F = = 20. 0( )
99
100 n 1 1 K e 1 t u( 0, t) = ( 3-20) k( n 1)! K n = c F I 1 α1 β1 K = c F I 2 α2 β2
101 9 6 GWF = [ 1 ( kk ) ]( 1+ KV GWS) SGW ( 3 22) 2 4 1
102 V = L (3 23) t
103 v = c RI ( 3 24) F 0 Q = dq = vdf
104 R = P E ( 3 25) R P E R n 1 Q 0 Q n Qi n ( 3 26) i= 1 Q n Q 0 Q n
105 n Ri fi R1 f1 + R 2 f2 + Λ + Rn fn i= 1 R0 = = ( 3 27) F F R = R + R i i+1 i 2
106 X R = = R 0 0 ( 3 28) X R = X α ( 3 29) 0 0 Cv = n ( Ki ) 2 1 n 1 ( 3 30 ) i= 1
107
108
109
110 2 12 Ki 1 i= K 1 Cv y = ( 3 31) 12 K = 100% = V Cr = ( 3 22) W 0 V Cr = ( 3 33) W V W C r 0
111
112 S Q = 0.278ϕ F ( 3-34 ) τ m n ϕ H 2 S = n τ τ v = c R I Q = F v K = F c R Q = K I K
113 F1 + F2 1 1 F = c = ( c1 + c 2) R = ( R1 + R2 ) K = ( K1 + K2) K 1 = F1 c1 R1 K 2 = F2 c2 R2 2 c = 1 n R 1 6 ( 3-35 ) c = 1 n R y ( 3-36 ) y = 2.5 n R n 0.10 R 3.0 3/2 Q = mb 2gH ( 3 37) 0
114 dw = Q dt dq Q = 1 k dt k t Q t = Q0e ( 3 38) 1 1 = α Q = Q e a t 0 k
115 R e = vd R e = vr R e 300 R e 300 v v U = U + U U U U U U 1 T U = U dt = 0 ( 3 40) T 0 U = U + U V = V + V W = W + W σ = U 2 σ σ u = V v w 2 = W 2 σu σ v σ w U V W * * *
116 σ u, U * i i i 0 i
117 Q L dl + Q t dt = 0 ( 3 41)
118 dl Q Q = dt t L dl = ω ( ω ) dt dl Q Q ω = = ( 3 42) dt t L Q F + = 0 L t Q F F Q = = L t Q t Q F Q ( 3 42 ) = ( )( ) t Q t ω = Q F dq VdF + FdV = = V + FdV df df df ω = V + FdV ( 3 43) df V = c Ri h R V = c hi dv c i dh c hi dh Vdh = = = 2 h 2 h h 2h ω = V + BhVdh = V + V 2hBdh 2 ω = 15. V ( 3 44)
119
120 u y u y u x u x u y 3 46c 0 u x
121 4 ρs ρ gd ω = ( 3 45) 3 ρ cd 2 1 ρ ρ s gd ω = ( 3 46) 18 ρ v
122 ρs ρ ω = gd ( 3 47) ρ 2 V d 2 2g 2 2 P rk d V x = 2g
123 2 2 V 3 rkd d rs r f g = ( ) rk d = 2fg r r V 2 ( ) A = s rk 2 fg( r r) d = AV 2 s d rs = rs ( AV ) = rsa V V c = R 6d 3 ( 3 50 ) h rs r h V c = d + 2 ( ) (3-51) d r d
124 2 V d h = ω ( 3-51) 5 5
125 v Riv S = ( 3-53) ω ω 0
126 2. 26 v S = ( 3-54) R ω 0 3 S = 0.07 v ( 3-55) ghω 3
127
128
129 1
130
131 ( m) ( c ) ( a ) m c a c a a a a a = = 2 2 b = 0 c = a = m n a a a ( 3-56) a
132
133
134
135
136
137
138
139
140
141 E = dρ (3-57) dh
142 2L 2L T = = C gh (3-59) 2L T = ( 3 61) n gh 2L λ = ( 3 62) n τ h = C al gh dos s α (3 63) ρ
143
144
145 W T = (3-66) Q 86400
146 V, V W 0 β W 0 Wv W0 W η = η = W W 0 0 c
147 (%) *
148
149
150
151
152
153 2 c = gh g
154
155 ϕ ϕ = A E A + A F E 100% ( 3-68) ϕ ϕ ϕ
156
157
158
159
160
161
162
163
164
165
166 c = ( 4 1 )
167
168 C = g λ 2πh s tanh ( 4 2 ) 2π λ gλ Cs = ( 4 3 ) 2π Cs = gh ( 4 4 ) C h g tanh 2 π h s λ λ π λ = gt 2 2π tanh 2 h ( 4 6) λ λ = gt 2 ( 4 7 ) 2π 4-3 c = 1.25 λ T = 0.8 λ g λ gt 2 2πc gt 2 2πλ 2πc 2π 2π g 2π g g k h r = r e (4 ) 8 0
169 r = a k = 2 π 0 u = rσ λ 2πh 2πk λ / k ( 4 2) tanh = λ λ c = gk ( 4 10)
170 x = x + PM = x + rsinθ ( 4 10) z = z PO = z rcosθ x = x + asinθs (4 ) 11 z = z - bcosθ
171
172 gλ C = π
173 C = gh ( 4 13) 1 1 h H 4l 2 ( 4 14)
174
175 c = gh
176 gh 1 h2 c 2 = gh 2 h1 h 2 c1 c2 c1dt sinθ1 = A B c2dt sinθ2 = A B sinθ1 c 2 = sinθ c 2 1 c2 sinθ = sin θ ( 4 15) 2 1 c1
177 f = k M E (4-16) R 2 f k M g M 2 = r p 1 x = 4 E x 2 2 ( 17) 2 k = k = g r r g E
178 f E k M E E = 2 ( 4 18) R f k M E = 2 ( 4 19) R
179 1 A F k M 1 k M 1 k Mr A = 2 2 = 2 3 ( R r) R r ( 4 20) 2 B F k M 1 k M 1 kmr B = 2 2 = 2 3 ( R + r) R R ( 4 21) 3 C F k M 1 k M 1 H( C) = 2 sin ϕ 2 = 0 L R ( 4 22) F k M 1 k Mr v( C) = 2 cos ϕ 0 = 3 L R 4 D F = 0 ( 4 23) ( 4-24) H( D) F = ( ) k M 1 V D 3 R 5 N F k M 1 k M 1 H( N) = 2 sin( θ + ϕ) 2 sin θ ( 4 25) L L 3 kmr = 3 sin2θ 2 R F k M 1 k M 1 V( N) = 2 cos( θ + ϕ) 2 cosθ L R = k Mr 2 3 ( 3cos θ 1) R 3 = k Mr (cos θ ) R 3 3 kmr 1 = 3 (cos 2θ + ) 2 R 3 3 ksr FH ( N) = 3 sin 2θ ( 4 26) 2 R 3 ksr 1 FV ( N) = 3 cos 2θ = ( 4 27) 2 R 3 3 F V M R = ( 4 28) FV S R
180
181
182
183
184 ω ϕ
185 τ ρ ρ D a = W 7. 6 ( 4 23) sinϕ ϕ W v0 = ( 4 33) sinϕ
186 π D h a vh = v0e ( 4 34)
187 mg P2 = ( 4 36) s P2 P1 h = ( 4 40) ρg
188
189
190
191
192
193
194
195
196 H N a = rszzdz ( 4 41) z 0 H N b = rsz ( H Z) dz ( 4 42) z 0 H N = Na + N b = rh SzdZ = rhv ( 4 43) z 0 1 P = EC ( 4 44) 2 g P = rh 2 T ( 4 45) 32π P = H T H T( / ) 32π
197 P = E = 10 = 500 ( 4-47) t 3 10
198 P = 1 2 mv + mgh ( 4-48) 2
199
200
201
202
203
204
205 ó
206
207
208
209
210 i f E = E p f i a E = Epexp (-k t) ( 5 3) i a E = C
211
212
213
214
215
216
217
218 ( H m a x H + ht) α = ( 5 9) P i
219
220
221 H n E = E ( 1 ) ( 5 10) o H o o d E = E e ( 5 11) 0 a k µ E 2 0 E = b ( 5 12) ( 1+ H)
222 V = k( I I 0 ) (5 13) V k H H 1 2 = I 0 ( 5 14) L
223 H = 2σ cos θ ( 5 17) r ρ g H = 10 6 ( ) H = ( ) ( 5 18) r d 0 Pc L P I = ( + ) = L V k P L c = L c L L ( 5 20) (5 ) 21
224 V k I I k P L c = ( 0 ) = I 0 ( 5 22) L V k P L c = I 0 ( 5 23) L P H c Pc = I H = ( 5 24) H I + 0 1
225 Q ka h h 1 2 = ( 5 25) L V = k h ( 5 L 29) V = k dh V ki ( 5 30) dl V k d Φ = ( 5 31) dl I = d Φ V dl
226 Q u = ( 5 32) nw γ k = k 0 ( 5 34) v
227
228 Q = ka dh ( 5 37) dr Q = k2π rh dh ( 5 38) dr R Q dr H = 2πk hdh r r 0 h0 Q k H 2 h 2 k H 2 h = π = ln( R / r ) lg( R / r ) 0 0 ( 5 39) ( 2H S0 ) S0 Q = k ( 5 40) lg( R / r ) 0 Q r Q r 2 2 h = ln 2 h = ln + h 0 ( 5 41) πl r + h πk r 0 0 Q( 1gR 1gr 0) 1gR 1gr 0 k = k = 0. 73Q ( 5 42) ( 2H S ) S ( 2H S ) S Q = k2π rm dh ( 5 43) dr H Q R dr dh = h km 0 2π r0 r Q km H h Q km H h 0 0 = 2π = (5 44 ) ln( R / r ) lg( R / r ) Q km S 0 = lg( R / r ) S 0 0 ( 5 45) Q R = lg ( 5 46) 2πkM r 0
229
230
231
232 P + R E = 0 ( 5 51) g 1 2
233 P R E Q = Q ( 5 52) a 0
234
235 y y 1 1 y y
236
237
238 6-1 ( )
239
240
241
242
243
244
245
246
247
248 6-4
249
250
251
252
253 V dc = Q(C 0 C) kcv 6 1 dt ( ) C = C Q 0 + C q k C Q C q i ( 0 + i ) ( 6 2 ) Q + q
254 Q q dw s = ( ) = 3 dt s 4 p w 2 ρ ρ ρ ρ D ( ) p ρs dw = 0 dt w = 4gD p ( ρ ρ ) s 3ρ p 1 1 ( 6 4 ) 2 4πr = = = ( / ) ( 6 5 ) 4 2 r πr
255
256
257 L = L k x u 0 exp( 1 / ) C = Cs ( Cs C k x u 0)exp( 2 / ) k1l0 + [exp( k1x / u) exp( k2x / u)] k1 k2 k L 1 0 D = D0 exp( k2x / u) 9exp( k1x / u) k1 k2 exp( k x u 2 / ) ( 6 6) u k Cs C k k 2 ( 0)( 2 1) xc = ln 1 ( 6 7 ) k 2 k1 k1 L0L1 k1l0 k1x0 Cc = Cs exp( ) k u 2 ( 6 8 ) k1l0 k1x0 Dc = exp( ) k u 2
258 =
259
260
261 C C Q + C = Q Q + Q p p E E P E ( 6 11) E = Ce Q p + QBi Qp Cp ( 6 13) i=1 C C Q C Q p p + E + E WS = + ( 6 14) Q 864. Q p C S P C Q C Q P P + E E r { 0 } = S ( 6 16) QP + QE Qi + k V Ce C0 exp t Qi kv V ( + ) E = ( 6 17) Qi + kv 1 exp V
262 Qi C C k V t { Q kv e 0 exp ( + ) } ( i + ) E = ( 6 18) Qi k + v t 1 exp ( )
263
ttian
T = l g = a - b a ϕ ϕ ϕ ϕ ϕ ϕ SiO SiO 65% SiO 52% SiO 52 45% SiO 45% 20 45% 10% 50% 95% 2.5 2.7 2.7 2.8 2.9 3.1 3.1 3.5 S0=1.94
m0 m = v2 1 c 2 F G m m 1 2 = 2 r m L T = 2 π ( m g 4 ) m m = 1 F AC F BC r F r F l r = sin sinl l F = h d G + S 2 = t v h = t 2 l = v 2 t t h = v = at v = gt t 1 l 1 a t g = t sin α 1 1 a = gsinα
-2 4 - cr 5 - 15 3 5 ph 6.5-8.5 () 450 mg/l 0.3 mg/l 0.1 mg/l 1.0 mg/l 1.0 mg/l () 0.002 mg/l 0.3 mg/l 250 mg/l 250 mg/l 1000 mg/l 1.0 mg/l 0.05 mg/l 0.05 mg/l 0.01 mg/l 0.001 mg/l 0.01 mg/l () 0.05 mg/l
P r = 1 + ecosθ 2 V = V + V 1 2 2V1V2 cosθ 2 2 = ( V V ) + 2V V ( 1 cos θ) 1 2 1 2 40000 V = 0. 5( / ) 24 60 60 λ m = 5100A = 0.51 Å 2 u e d s 3 1 e uud udd 3 2 3 e 1 3 e V = 2 9. 8 2000 = 198 V
Ζ # % & ( ) % + & ) / 0 0 1 0 2 3 ( ( # 4 & 5 & 4 2 2 ( 1 ) ). / 6 # ( 2 78 9 % + : ; ( ; < = % > ) / 4 % 1 & % 1 ) 8 (? Α >? Β? Χ Β Δ Ε ;> Φ Β >? = Β Χ? Α Γ Η 0 Γ > 0 0 Γ 0 Β Β Χ 5 Ι ϑ 0 Γ 1 ) & Ε 0 Α
! # % & ( & # ) +& & # ). / 0 ) + 1 0 2 & 4 56 7 8 5 0 9 7 # & : 6/ # ; 4 6 # # ; < 8 / # 7 & & = # < > 6 +? # Α # + + Β # Χ Χ Χ > Δ / < Ε + & 6 ; > > 6 & > < > # < & 6 & + : & = & < > 6+?. = & & ) & >&
W L Gates.Open Lecture The influences of the ocean on climate.scientific lecture at the 28th section of the ECWMO.WMO Bulletin. July1977168 169. WCP 1 WCRP2 WCAP 3 WCIP4 WCDP .. 1991 A Henderson-SellersP
Ps22Pdf
) ,,, :,,,,,,, ( CIP) /. :, 2001. 9 ISBN 7-5624-2368-7.......... TU311 CIP ( 2001) 061075 ( ) : : : : * : : 174 ( A ) : 400030 : ( 023) 65102378 65105781 : ( 023) 65103686 65105565 : http: / / www. cqup.
!! # % & ( )!!! # + %!!! &!!, # ( + #. ) % )/ # & /.
! # !! # % & ( )!!! # + %!!! &!!, # ( + #. ) % )/ # & /. #! % & & ( ) # (!! /! / + ) & %,/ #! )!! / & # 0 %#,,. /! &! /!! ) 0+(,, # & % ) 1 # & /. / & %! # # #! & & # # #. ).! & #. #,!! 2 34 56 7 86 9
untitled
arctan lim ln +. 6 ( + ). arctan arctan + ln 6 lim lim lim y y ( ln ) lim 6 6 ( + ) y + y dy. d y yd + dy ln d + dy y ln d d dy, dy ln d, y + y y dy dy ln y+ + d d y y ln ( + ) + dy d dy ln d dy + d 7.
Ρ Τ Π Υ 8 ). /0+ 1, 234) ς Ω! Ω! # Ω Ξ %& Π 8 Δ, + 8 ),. Ψ4) (. / 0+ 1, > + 1, / : ( 2 : / < Α : / %& %& Ζ Θ Π Π 4 Π Τ > [ [ Ζ ] ] %& Τ Τ Ζ Ζ Π
![Ρ Τ Π Υ 8 ). /0+ 1, 234) ς Ω! Ω! # Ω Ξ %& Π 8 Δ, + 8 ),. Ψ4) (. / 0+ 1, > + 1, / : ( 2 : / < Α : / %& %& Ζ Θ Π Π 4 Π Τ > [ [ Ζ ] ] %& Τ Τ Ζ Ζ Π](/thumbs/52/29426146.jpg "Ρ Τ Π Υ 8 ). /0+ 1, 234) ς Ω! Ω! # Ω Ξ %& Π 8 Δ, + 8 ),. Ψ4) (. / 0+ 1, > + 1, / : ( 2 : / < Α : / %& %& Ζ Θ Π Π 4 Π Τ > [ [ Ζ ] ] %& Τ Τ Ζ Ζ Π")
! # % & ( ) + (,. /0 +1, 234) % 5 / 0 6/ 7 7 & % 8 9 : / ; 34 : + 3. & < / = : / 0 5 /: = + % >+ ( 4 : 0, 7 : 0,? & % 5. / 0:? : / : 43 : 2 : Α : / 6 3 : ; Β?? : Α 0+ 1,4. Α? + & % ; 4 ( :. Α 6 4 : & %
untitled
f ( ) tan e, > = arcsin a = ae, a = tan e tan lim f ( ) = lim = lim =, arcsin + + + lim f = lim ae = a, y e ( ) =
untitled
4 y l y y y l,, (, ) ' ( ) ' ( ) y, y f ) ( () f f ( ) (l ) t l t lt l f ( t) f ( ) t l f ( ) d (l ) C f ( ) C, f ( ) (l ) L y dy yd π y L y cosθ, π θ : siθ, π yd dy L [ cosθ cosθ siθ siθ ] dθ π π π si
. () ; () ; (3) ; (4).. () : P.4 3.4; P. A (3). () : P. A (5)(6); B. (3) : P.33 A (9),. (4) : P. B 5, 7(). (5) : P.8 3.3; P ; P.89 A 7. (6) : P.
() * 3 6 6 3 9 4 3 5 8 6 : 3. () ; () ; (3) (); (4) ; ; (5) ; ; (6) ; (7) (); (8) (, ); (9) ; () ; * Email: [email protected] . () ; () ; (3) ; (4).. () : P.4 3.4; P. A (3). () : P. A (5)(6); B. (3) :
! Ν! Ν Ν & ] # Α. 7 Α ) Σ ),, Σ 87 ) Ψ ) +Ε 1)Ε Τ 7 4, <) < Ε : ), > 8 7
![! Ν! Ν Ν & ] # Α. 7 Α ) Σ ),, Σ 87 ) Ψ ) +Ε 1)Ε Τ 7 4, <) < Ε : ), > 8 7](/thumbs/48/24501715.jpg "! Ν! Ν Ν & ] # Α. 7 Α ) Σ ),, Σ 87 ) Ψ ) +Ε 1)Ε Τ 7 4, <) < Ε : ), > 8 7")
!! # & ( ) +,. )/ 0 1, 2 ) 3, 4 5. 6 7 87 + 5 1!! # : ;< = > < < ;?? Α Β Χ Β ;< Α? 6 Δ : Ε6 Χ < Χ Α < Α Α Χ? Φ > Α ;Γ ;Η Α ;?? Φ Ι 6 Ε Β ΕΒ Γ Γ > < ϑ ( = : ;Α < : Χ Κ Χ Γ? Ε Ι Χ Α Ε? Α Χ Α ; Γ ;
., /,, 0!, + & )!. + + (, &, & 1 & ) ) 2 2 ) 1! 2 2
! # &!! ) ( +, ., /,, 0!, + & )!. + + (, &, & 1 & ) ) 2 2 ) 1! 2 2 ! 2 2 & & 1 3! 3, 4 45!, 2! # 1 # ( &, 2 &, # 7 + 4 3 ) 8. 9 9 : ; 4 ), 1!! 4 4 &1 &,, 2! & 1 2 1! 1! 1 & 2, & 2 & < )4 )! /! 4 4 &! &,
3978 30866 4 3 43 [] 3 30 4. [] . . 98 .3 ( ) 06 99 85 84 94 06 3 0 3 9 3 0 4 9 4 88 4 05 5 09 5 8 5 96 6 9 6 97 6 05 7 7 03 7 07 8 07 8 06 8 8 9 9 95 9 0 05 0 06 30 0 .5 80 90 3 90 00 7 00 0 3
!!! #! )! ( %!! #!%! % + % & & ( )) % & & #! & )! ( %! ),,, )
! # % & # % ( ) & + + !!! #! )! ( %!! #!%! % + % & & ( )) % & & #! & )! ( %! ),,, ) 6 # / 0 1 + ) ( + 3 0 ( 1 1( ) ) ( 0 ) 4 ( ) 1 1 0 ( ( ) 1 / ) ( 1 ( 0 ) ) + ( ( 0 ) 0 0 ( / / ) ( ( ) ( 5 ( 0 + 0 +
4= 8 4 < 4 ϑ = 4 ϑ ; 4 4= = 8 : 4 < : 4 < Κ : 4 ϑ ; : = 4 4 : ;
! #! % & ( ) +!, + +!. / 0 /, 2 ) 3 4 5 6 7 8 8 8 9 : 9 ;< 9 = = = 4 ) > (/?08 4 ; ; 8 Β Χ 2 ΔΔ2 4 4 8 4 8 4 8 Ε Φ Α, 3Γ Η Ι 4 ϑ 8 4 ϑ 8 4 8 4 < 8 4 5 8 4 4
,!! #! > 1? = 4!! > = 5 4? 2 Α Α!.= = 54? Β. : 2>7 2 1 Χ! # % % ( ) +,. /0, , ) 7. 2
! # %!% # ( % ) + %, ). ) % %(/ / %/!! # %!! 0 1 234 5 6 2 7 8 )9!2: 5; 1? = 4!! > = 5 4? 2 Α 7 72 1 Α!.= = 54?2 72 1 Β. : 2>7 2 1 Χ! # % % ( ) +,.
/ Ν #, Ο / ( = Π 2Θ Ε2 Ρ Σ Π 2 Θ Ε Θ Ρ Π 2Θ ϑ2 Ρ Π 2 Θ ϑ2 Ρ Π 23 8 Ρ Π 2 Θϑ 2 Ρ Σ Σ Μ Π 2 Θ 3 Θ Ρ Κ2 Σ Π 2 Θ 3 Θ Ρ Κ Η Σ Π 2 ϑ Η 2 Ρ Π Ρ Π 2 ϑ Θ Κ Ρ Π
! # #! % & ( ) % # # +, % #. % ( # / ) % 0 1 + ) % 2 3 3 3 4 5 6 # 7 % 0 8 + % 8 + 9 ) 9 # % : ; + % 5! + )+)#. + + < ) ( # )# < # # % 0 < % + % + < + ) = ( 0 ) # + + # % )#!# +), (? ( # +) # + ( +. #!,
, ( 6 7 8! 9! (, 4 : : ; 0.<. = (>!? Α% ), Β 0< Χ 0< Χ 2 Δ Ε Φ( 7 Γ Β Δ Η7 (7 Ι + ) ϑ!, 4 0 / / 2 / / < 5 02
! # % & ( ) +, ) %,! # % & ( ( ) +,. / / 01 23 01 4, 0/ / 5 0 , ( 6 7 8! 9! (, 4 : : ; 0.!? Α% ), Β 0< Χ 0< Χ 2 Δ Ε Φ( 7 Γ Β Δ 5 3 3 5 3 1 Η7 (7 Ι + ) ϑ!, 4 0 / / 2 / 3 0 0 / < 5 02 Ν!.! %) / 0
1 s = v t + at, 0 v = v + at 0. 1 3.0 36 s = v t + at a 0 1 F ma a s v t at s 0 F f 1 a m = mgsin θ µ mg cosθ g sinθ µ cosmθ 1 s = v t + at 0 1 v tsinθ µ cosθ 0 1 1 1.0 5.0 9.8 50 0 04 3.. 5 57
8 9 8 Δ 9 = 1 Η Ι4 ϑ< Κ Λ 3ϑ 3 >1Ε Μ Ε 8 > = 8 9 =
!! % & ( & ),,., / 0 1. 0 0 3 4 0 5 3 6!! 7 8 9 8!! : ; < = > :? Α 4 8 9 < Β Β : Δ Ε Δ Α = 819 = Γ 8 9 8 Δ 9 = 1 Η Ι4 ϑ< Κ Λ 3ϑ 3 >1Ε 8 9 0 Μ Ε 8 > 9 8 9 = 8 9 = 819 8 9 =
1 2 / 3 1 A (2-1) (2-2) A4 6 A4 7 A4 8 A4 9 A ( () 4 A4, A4 7 ) 1 (2-1) (2-2) ()
(39mm E-Mail ( )( ), : : 1 1 ( ) 2 2 ( ) 29mm) WSK ( 1 2 / 3 1 A4 2 1 3 (2-1) 2-1 4 (2-2) 2-2 5 A4 6 A4 7 A4 8 A4 9 A4 10 11 ( () 4 A4, 5 6 7 8 A4 7 ) 1 (2-1) (2-2) () 1 2 (2-1) 3 (2-2) 4 5 6 7 (8 ) 9
Β 8 Α ) ; %! #?! > 8 8 Χ Δ Ε ΦΦ Ε Γ Δ Ε Η Η Ι Ε ϑ 8 9 :! 9 9 & ϑ Κ & ϑ Λ &! &!! 4!! Μ Α!! ϑ Β & Ν Λ Κ Λ Ο Λ 8! % & Π Θ Φ & Ρ Θ & Θ & Σ ΠΕ # & Θ Θ Σ Ε
! #!! % & ( ) +,. /. 0,(,, 2 4! 6! #!!! 8! &! % # & # &! 9 8 9 # : : : : :!! 9 8 9 # #! %! ; &! % + & + & < = 8 > 9 #!!? Α!#!9 Α 8 8!!! 8!%! 8! 8 Β 8 Α ) ; %! #?! > 8 8 Χ Δ Ε ΦΦ Ε Γ Δ Ε Η Η Ι Ε ϑ 8 9 :!
4 / ( / / 5 / / ( / 6 ( / / 7 1 2 / 3 ( 4 ( 2003 8 ( 2
: / ( 6 (2003 8 : ( 1 ( ( / / (,, ( ( - ( - (39mm 29mm 2 ( 1 2 3-6 3 6-24 6-48 12-24 8-12 WSK / WSK WSK 1 4 / ( / / 5 / / ( / 6 ( / / 7 1 2 / 3 ( 4 ( 2003 8 ( 2 9 5 ( 10 3 11 / (600 4 5 AA 710 AB 720 730
! * # + + *! # $ # #. 1 #! % &)# * 1 *! * ! % # * # * # + + +!!!! # * % # # + # + * & $ *! * # + * # % #& % &* # & # * %! + * # #!
!!#$! # $ %! ##&&!)*+, -##. -&/01& -#!!!% $! $!#!!!$23.1! 4)5 %#&&& &.1 %!!!! %! &!!!!!!!! /!!!!!!!!!!!!!!! %.&!!! #&!!!!0&&&& #!!!!!!!!!!!!!!! %.& /&&&/.! 0.!!!!!!!!!!!!!!! %1& 6 /. 1!!!!!!!!!!!!!!! %&
Ps22Pdf
Q CIP / / / 2880 Q Q Q Q Q QQ Q Q Q Q Q ec A c c A c c c Q a A A A c e A c a c c ea c c a A c c c a A c c c a A / A c c c c f a c c f a c c f a c f e A c f c f / c A c c a c c A e A c c e A c c ea c c
) Μ <Κ 1 > < # % & ( ) % > Χ < > Δ Χ < > < > / 7 ϑ Ν < Δ 7 ϑ Ν > < 8 ) %2 ): > < Ο Ε 4 Π : 2 Θ >? / Γ Ι) = =? Γ Α Ι Ρ ;2 < 7 Σ6 )> Ι= Η < Λ 2 % & 1 &
! # % & ( ) % + ),. / & 0 1 + 2. 3 ) +.! 4 5 2 2 & 5 0 67 1) 8 9 6.! :. ;. + 9 < = = = = / >? Α ) /= Β Χ Β Δ Ε Β Ε / Χ ΦΓ Χ Η Ι = = = / = = = Β < ( # % & ( ) % + ),. > (? Φ?? Γ? ) Μ
> # ) Β Χ Χ 7 Δ Ε Φ Γ 5 Η Γ + Ι + ϑ Κ 7 # + 7 Φ 0 Ε Φ # Ε + Φ, Κ + ( Λ # Γ Κ Γ # Κ Μ 0 Ν Ο Κ Ι Π, Ι Π Θ Κ Ι Π ; 4 # Ι Π Η Κ Ι Π. Ο Κ Ι ;. Ο Κ Ι Π 2 Η
1 )/ 2 & +! # % & ( ) +, + # # %. /& 0 4 # 5 6 7 8 9 6 : : : ; ; < = > < # ) Β Χ Χ 7 Δ Ε Φ Γ 5 Η Γ + Ι + ϑ Κ 7 # + 7 Φ 0 Ε Φ # Ε + Φ, Κ + ( Λ # Γ Κ Γ #
() (n = 4 3 )?( ) ( H) ( n) 0 H H,. , L. f(f < L 4 ),.? L L L L. + fl;. fl; 4 4 M f, = fm r (4) = E 3 - E h L L, 3 ; L L = P E S, ;, ;, ;,. 5 f. 4 u + v = f, u = 4v; 4v + v = f, v = 5 4 f..()80
# # 4 + % ( ) ( /! 3 (0 0 (012 0 # (,!./ %
#! # # %! # + 5 + # 4 + % ( ) ( /! 3 (0 0 (012 0 # (,!./ % ,9 989 + 8 9 % % % % # +6 # % 7, # (% ) ,,? % (, 8> % %9 % > %9 8 % = ΑΒ8 8 ) + 8 8 >. 4. ) % 8 # % =)= )
2 2 Λ ϑ Δ Χ Δ Ι> 5 Λ Λ Χ Δ 5 Β. Δ Ι > Ε!!Χ ϑ : Χ Ε ϑ! ϑ Β Β Β ϑ Χ Β! Β Χ 5 ϑ Λ ϑ % < Μ / 4 Ν < 7 :. /. Ο 9 4 < / = Π 7 4 Η 7 4 =
! # % # & ( ) % # ( +, & % # ) % # (. / ). 1 2 3 4! 5 6 4. 7 8 9 4 : 2 ; 4 < = = 2 >9 3? & 5 5 Α Α 1 Β ΧΔ Ε Α Φ 7 Γ 9Η 8 Δ Ι > Δ / ϑ Κ Α Χ Ε ϑ Λ ϑ 2 2 Λ ϑ Δ Χ Δ Ι> 5 Λ Λ Χ Δ 5 Β. Δ Ι > Ε!!Χ ϑ : Χ Ε ϑ!
8 9 < ; ; = < ; : < ;! 8 9 % ; ϑ 8 9 <; < 8 9 <! 89! Ε Χ ϑ! ϑ! ϑ < ϑ 8 9 : ϑ ϑ 89 9 ϑ ϑ! ϑ! < ϑ < = 8 9 Χ ϑ!! <! 8 9 ΧΧ ϑ! < < < < = 8 9 <! = 8 9 <! <
! # % ( ) ( +, +. ( / 0 1) ( 2 1 1 + ( 3 4 5 6 7! 89 : ; 8 < ; ; = 9 ; ; 8 < = 9! ; >? 8 = 9 < : ; 8 < ; ; = 9 8 9 = : : ; = 8 9 = < 8 < 9 Α 8 9 =; %Β Β ; ; Χ ; < ; = :; Δ Ε Γ Δ Γ Ι 8 9 < ; ; = < ; :
tbjx0164ZW.PDF
F = k Q Q r F = k Q = k Q r r Q Q = Fr k = C 0 5 C 9 0 5 Q 0 3 n = = 9 = 65. 0 e 6. 0 4 3 A B 7 7 9 6 C D 7 7 F = k q 7q = k 7q r r q + 7q = 4q F = k 4q 4q = k 6q r r F = 6 F 7 7q q = 3q s c = t s c =
B = F Il 1 = 1 1 φ φ φ B = k I r F Il F k I 2 = l r 2 10 = k 1 1-7 2 1 k = 2 10-7 2 B = ng Il. l U 1 2 mv = qu 2 v = 2qU m = 2 19 3 16. 10 13. 10 / 27 167. 10 5 = 5.0 10 /. r = m ν 1 qb r = m ν qb
UDC
CECS 102:2002 Technical specification for steed structure of light-eight Buildings ith gabled frames 2003 1 Technical specification for steed structure of light-eight Buildings ith gabled frames CECS102:2002
Π Ρ! #! % & #! (! )! + %!!. / 0% # 0 2 3 3 4 7 8 9 Δ5?? 5 9? Κ :5 5 7 < 7 Δ 7 9 :5? / + 0 5 6 6 7 : ; 7 < = >? : Α8 5 > :9 Β 5 Χ : = 8 + ΑΔ? 9 Β Ε 9 = 9? : ; : Α 5 9 7 3 5 > 5 Δ > Β Χ < :? 3 9? 5 Χ 9 Β
Ps22Pdf
( ) 158 :,,,,,, : 850 1168 1/ 32 : 12 : 311 1994 1 1 1998 11 2 : 5001 10000 ISBN 7 302 01340 3/ T B 15 : 13 00 ,,, 10 7, 2 80%, : 1 ;, :, :, ;, ;, 30%,,,, 20,,,, ,,,,,,!,,,! 1992 10 1 1 11 15 27 34 34
= Υ Ξ & 9 = ) %. Ο) Δ Υ Ψ &Ο. 05 3; Ι Ι + 4) &Υ ϑ% Ο ) Χ Υ &! 7) &Ξ) Ζ) 9 [ )!! Τ 9 = Δ Υ Δ Υ Ψ (
! # %! & (!! ) +, %. ( +/ 0 1 2 3. 4 5 6 78 9 9 +, : % % : < = % ;. % > &? 9! ) Α Β% Χ %/ 3. Δ 8 ( %.. + 2 ( Φ, % Γ Η. 6 Γ Φ, Ι Χ % / Γ 3 ϑκ 2 5 6 Χ8 9 9 Λ % 2 Χ & % ;. % 9 9 Μ3 Ν 1 Μ 3 Φ Λ 3 Φ ) Χ. 0
t H θ m [] Q Q q [] 9. 69kJ g q 32.0 620kJmol - 1 0. 500g mol C(s) 1 2 O (g) = CO(g) 2 SI n mol H 2H+H22H2+O2 [N2H41+O2g N2g+2H2O1] 1.2.3 qp qv [4] 1 C( ) O 2 (g) = CO(g), ( 2), 2 CO(g) CO (g),co(g) +
,..,.,,,,,.,,.,., ,.,,.,,.,, 1,,, ; 2,,,,.,,,,.,,.,,,.,.,.,,.,.,,,.,,,.,,,,.,.,,,, i
,..,.,,,,,.,,.,.,. 6 1,.,,.,,.,, 1,,, ;,,,,.,,,,.,,.,,,.,.,.,,.,.,,,.,,,.,,,,.,.,,,, i .,,,,.,,.,.,.,,.,,,., 1;,,,,,.,,,,.,,,.,.,,.,,.,,,.,,.,,.,.,.,,.,,.,..,.,,.,,,.,,,.,,,,,,.,,,,.,,????.,,,,,.,,,,.,
untitled
6 + a lim = 8, a =. a l. a a + a a a a lim = lim + = e, a a a e = 8 a= l ( 6,, ), 4 y+ z = 8. + y z = ( 6,, ) 4 y z 8 a ( 6,, ) + = = { } i j k 4,,, s = 6 = i+ j k. 4 ( ) ( y ) ( z ) + y z =. + =, () y
4 # = # 4 Γ = 4 0 = 4 = 4 = Η, 6 3 Ι ; 9 Β Δ : 8 9 Χ Χ ϑ 6 Κ Δ ) Χ 8 Λ 6 ;3 Ι 6 Χ Δ : Χ 9 Χ Χ ϑ 6 Κ
! # % & & ( ) +, %. % / 0 / 2 3! # 4 ) 567 68 5 9 9 : ; > >? 3 6 7 : 9 9 7 4! Α = 42 6Β 3 Χ = 42 3 6 3 3 = 42 : 0 3 3 = 42 Δ 3 Β : 0 3 Χ 3 = 42 Χ Β Χ 6 9 = 4 =, ( 9 6 9 75 3 6 7 +. / 9
:,,,, ( CIP ) /,. :, ISBN CIP ( 2001) : : 127, : : : ht t p: / / www. nwpup. com : :
:,,,, ( CIP ) /,. :, 2001. 8 ISBN 7 5612 1363 8............. 0342 CIP ( 2001) 027392 : : 127, : 710072 : 029-8493844 : ht t p: / / www. nwpup. com : : 787mm1 092mm : 19. 75 : 480 : 2001 8 1 2001 8 1 :
- 2-2- HCO HPO SO 3 4 4 P P R P QR Q R Q + - - H HCO3 - HCO 3 - - Cl HCO 3 HCO - 3 - HCO - 3 - HCO 3 - = 100 2- SO 4 - H 2 PO 4 HPO 4 - HPO 4 2- Na
2 23 (b) 4. (a) B X = µ 0I = (4π 10 7 )(1.5) X 2π(0.045) = 6.67 μt B Y = µ 0I = (4π 10 7 )(1.5) Y 2π(0.015) = 20 μt (b) B X = µ 0I = (4π 10 7 )(2) X 2
23 (b) 1 (p. 192) 1. (a) F (b) F 2. (a) C C B B B A (b) 2 (p. 196) 1. (a) T (b) F (c) T 2. (a) W 4. 3 (p. 205) 1. A A 2. (a) F (b) T 3. 4. (a) (b) Z 3. (a) 2 23 (b) 4. (a) B X = µ 0I = (4π 10 7 )(1.5)
1 2 1.1............................ 2 1.2............................... 3 1.3.................... 3 1.4 Maxwell.................... 3 1.5.......................... 4 1.6............................ 4
a b a = a ϕ λ ϕ λ ρ δ ρ δ ϕ λ M' J' x' = = m MJ x M' K' y' = = n MK y x' x = m 2-1 y' y = n 2 2 x + y = 1 2-2 2 2 x' y' 2 + 2 = 1 m n µ = ds ' ds 2 2 2 2 m + n = a + b 2-3 mnsinθ = ab 2-4 2 2 2 (
1 3 8 1 S 8 sini = n. ( 1` ) sinr n = c υ sini = n sinr 1 3 n 1 υ1 υ =, n1 =. υ υ 1 υ1 c / υ n 1 n1 = = = =. υ c / υ n n 1 1 1 sini n = n 1 =, sinr n1 60 n = 3 sin90 1 = =, sinc sinc n sin C = 1. n 3
! "! #!$$%!$$% &!!$$( # ) (
! " "!! " "!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! " #$$% #$$%!!% % & %!$ ( # ) #$$% *!!% ! "! #!$$%!$$% &!!$$( # ) ( " #$ %&!#& ( )*+,* -) " " "./012 )*+ 302 4056 7+1.6 0 3*8(*/.0-96 :*+/26) -+. 80;6
Microsoft Word - 24.doc
閱 讀 (1) 楊 桃 毒 素 過 去 有 許 多 糖 尿 病 或 高 血 壓 患 者 吃 過 新 鮮 楊 桃 或 楊 桃 汁 後, 出 現 嘔 吐 頭 暈 意 識 不 清 等 症 狀, 研 判 因 為 他 們 腎 功 能 不 佳, 無 法 排 出 楊 桃 中 不 明 的 中 毒 成 分 楊 桃 中 的 不 明 成 分 可 能 導 致 部 分 特 殊 體 質 的 尿 毒 症 病 患, 出 現 神 經
Cauchy Duhamel Cauchy Cauchy Poisson Cauchy 1. Cauchy Cauchy ( Duhamel ) u 1 (t, x) u tt c 2 u xx = f 1 (t, x) u 2 u tt c 2 u xx = f 2 (
Cauchy Duhamel Cauchy CauchyPoisson Cauchy 1. Cauchy Cauchy ( Duhamel) 1.1.......... u 1 (t, x) u tt c 2 u xx = f 1 (t, x) u 2 u tt c 2 u xx = f 2 (t, x) 1 C 1 C 2 u(t, x) = C 1 u 1 (t, x) + C 2 u 2 (t,
1-1 + 1 + + 2 + + 3 + 4 5 + 6 + 7 8 + 9 + 1-2 1 20000 20000 20000 20000 2 10000 30000 10000 30000 3 5000 5000 30000 4 10000 20000 10000 20000 5 3000 3000 20000 6 3000 3000 20000 7 5000 15000 8 5000 15000
x y 7 xy = 1 b c a b = x x = 1. 1 x + 17 + x 15 = 16 x + 17 x 15 + 17 15 x + 17 - x 15 = (x x ) ( ). x + 17 + x 15 x + y + 9 x + 4 y = 10 x + 9 y + 4 = 4xy. 9 4 ( x + ) + ( y + ) = 10 x y 9 ( x + )( ).
... II... III A A A.2...
ICS 13.200 X XX DZ DZ Specfcaton of desgn and constructon for landslde stablzaton - - - - 1 ... II... III 1... 1 2... 1 3... 1 4... 3 5... 4 6... 7 7... 12 8... 18 9... 24 10... 28 11... 32 12... 35 13...
υ υ υ υ υ υ υ υ υ υ υ υ υ υ υ è é é è υυ ν ε ε è α α α α α α α α α τ E h L. ν = λ = h p Ξ v k ν pe nµ Λ ν µ ν µ ε µ π ~ n p n np ~ π N Ξ + p n o o Λ Ξ Ξ SU 3
( ) (! +)! #! () % + + %, +,!#! # # % + +!
!! # % & & & &! # # % ( ) (! +)! #! () % + + %, +,!#! # # % + +! ! %!!.! /, ()!!# 0 12!# # 0 % 1 ( ) #3 % & & () (, 3)! #% % 4 % + +! (!, ), %, (!!) (! 3 )!, 1 4 ( ) % % + % %!%! # # !)! % &! % () (! %
1 ( CIP) /,. :, ISBN X753 CIP ( 2005 ) : : : : /
( 50474050) ( 20040533011) 1 ( CIP) /,. :, 2005. 12 ISBN 7-81105-260-1............. X753 CIP ( 2005 ) 151885 : : 410083 : 0731-8876770 : 0731-8710482 880 1230 1 /32 11. 25 277 2006 3 1 2006 3 1 ISBN 7-81105-260-1
第9章 排队论
9, 9. 9.. Nt () [, t] t Nt () { Nt ( ) t [, T]} t< t< t< t + N ( ( t+ ) i+ N( t) i, N( t) i,, N( t) i N + + N ( ( t ) i ( t ) i ) (9-) { Nt ( ) t [, T)} 9- t t + t, t,, t t t { Nt ( ) t [, T] } t< t,,
工程硕士网络辅导第一讲
< > < R R [ si t R si cos si cos si cos - sisi < si < si < < δ N δ { < δ δ > } www.tsighututor.com 6796 δ < < δ δ N δ { < < δ δ > b { < < b R} b] { b R} [ { > R} { R} } [ b { < b R} ] { b R} { R} X X Y
因 味 V 取 性 又 鸟 U 且 最 大 罗 海 惜 梅 理 春 并 贵 K a t h l ee n S c h w e r d t n er M f l e z S e b a s t i a n C A Fe rs e T 民 伊 ' 国 漳 尤 地 视 峰 州 至 周 期 甚 主 第 应
国 ' 东 极 也 直 前 增 东 道 台 商 才 R od e ric h P t ak 略 论 时 期 国 与 东 南 亚 的 窝 贸 易 * 冯 立 军 已 劳 痢 内 容 提 要 国 与 东 南 亚 的 窝 贸 易 始 于 元 代 代 大 规 模 开 展 的 功 效 被 广 为 颂 扬 了 国 国 内 市 场 窝 的 匮 乏 窝 补 虚 损 代 上 流 社 会 群 体 趋 之 若 鹜 食 窝
Persuasive Techniques (motorcycle helmet)
M O D E A T H E E L E M E N T S O F A N A R G U M E N T 1n t h l s t e s t i m o n y g iv e n b e f o r e t h e M a ry l a n d Se n a t e t h e s p e a ke r m a ke s a s t r o n g c l a i m a b o u t t
PowerPoint Presentation
1 1 2 3 4 2 2004 20044 2005 2006 5 2007 5 20085 20094 2010 4.. 20112116. 3 4 1 14 14 15 15 16 17 16 18 18 19 19 20 21 17 20 22 21 23 5 15 1 2 15 6 1.. 2 2 1 y = cc y = x y = x y =. x. n n 1 C = 0 C ( x
1 2 m v e 2 ö e m e m e m e m e e m m 1 1840 e m e m 2 v r Å Å Å 9999 10000 2 n λ = b( 2 2 ) n 2 Å 1 1854 1919 λ 1 1 1 2 2 λ = R ( Z H n ) 1 1 1 2 2 λ R H ( ) n f ni Z Z E ν = h mvr = n h 2π mvr = nh
% & :?8 & : 3 ; Λ 3 3 # % & ( ) + ) # ( ), ( ) ). ) / & /:. + ( ;< / 0 ( + / = > = =? 2 & /:. + ( ; < % >=? ) 2 5 > =? 2 Α 1 Β 1 + Α
# % & ( ) # +,. / 0 1 2 /0 1 0 3 4 # 5 7 8 / 9 # & : 9 ; & < 9 = = ;.5 : < 9 98 & : 9 %& : < 9 2. = & : > 7; 9 & # 3 2
C = C + C C = + + C C C C 1 2 3
C = C + C 1 2 3 1 1 1 1 + C = + + C C C C 1 2 3 17 Q = Q = Q C = Q U C 1 1 2 3 C 1 C 2 C 3 U = 1 1 1 U 1 U 2 U 3 = + + C C C 1 2 3 1) A B U A U B U U = AB A B AB G G R = R U = U U = 0 U = 4 B C BC CB C
微积分 授课讲义
2018 10 [email protected] 1201 / 18:00-20:20 213 14:00-17:00 I II Taylor : , n R n : x = (x 1, x 2,..., x n ) R; x, x y ; δ( ) ; ; ; ; ; ( ) ; ( / ) ; ; Ů(P 1,δ) P 1 U(P 0,δ) P 0 Ω P 1: 1.1 ( ). Ω
网络分析仪
5. A 5-, N l, N Hl 5- MKS l N r N H A/m l 5-a 5- CGS r 0.4πN H l (Oe) 5-b (CGS )Oe, MKS 0 (Oe) A/m 0.4π 5-. 5- N A A dψ d( BAN) e dt dt A N A B 5- db NA (5-) dt d ψ / dt. 5- H B B µh dh NAµ dt in H l N
Α 3 Α 2Η # # > # 8 6 5# Ι + ϑ Κ Ι Ι Ι Η Β Β Β Β Β Β ΔΕ Β Β Γ 8 < Φ Α Α # >, 0 Η Λ Μ Ν Ο Β 8 1 Β Π Θ 1 Π Β 0 Λ Μ 1 Ρ 0 Μ ϑ Σ ϑ Τ Ο Λ 8 ϑ
! # % & ( ) % + ( ), & ). % & /. % 0 1!! 2 3 4 5# 6 7 8 3 5 5 9 # 8 3 3 2 4 # 3 # # 3 # 3 # 3 # 3 # # # ( 3 # # 3 5 # # 8 3 6 # # # # # 8 5# :;< 6#! 6 =! 6 > > 3 2?0 1 4 3 4! 6 Α 3 Α 2Η4 3 3 2 4 # # >
koji-13.dvi
26 13 1, 2, 3, 4, 5, 6, 7 1 18 1. xy D D = {(x, y) y 2 x 4 y 2,y } x + y2 dxdy D 2 y O 4 x 2. xyz D D = {(x, y, z) x 1, y x 2, z 1, y+ z x} D 3. [, 1] [, 1] (, ) 2 f (1)