由假设得到公式 j8 {3 _0 _ {1.We assume laminar flow and use Bernoulli's equation:(由假设得到的公式)' X `: m# O! p% K" W) Z
: _: W0 l. N- B" z公式 ! S! l7 V9 |' G, z/ ?1 G; H: v; v5 L! O$ A$ b
Where- J1 p% D- W0 e* j4 j+ ~
3 H: D' d9 j- Q y0 W+ w
符号解释 2 w9 F6 ^1 m% E ; {! u4 B) b( u) I. C/ S6 MAccording to the assumptions, at every junction we have (由于假设)9 `. o+ B, v! [4 O- }8 ]7 m6 h
1 C& s& M- y# \5 }6 V: H
公式 : K9 Q9 i1 |. \" u 9 e( i2 j" V! n8 o, _由原因得到公式 * e4 Q% O: o3 i# k4 P* K! {1 Y [2.Because our field is flat, we have公式, so the height of our source relative to our sprinklers does not affect the exit speed v2 (由原因得到的公式); 7 k, N0 ]4 j. r* H |" P& e0 v7 e1 m5 Z# B& y: i公式. G- L& t* k% d) ~& m* ~& p( Z2 J
9 V- c1 Q$ V$ e+ P* VSince the fluid is incompressible(由于液体是不可压缩的), we have* O" w b# @% x! W# I
i. w- i6 O6 k/ E公式 8 M( e) W8 `4 H% A. f6 m+ P R8 f* e9 {$ ?
Where% P, T9 k# p) W( J6 ~
5 f6 U8 Y1 s1 O6 Q- c$ _ n( D公式% ]# J: |6 O; u) i# D Z
& @ b3 A4 x; B8 W S用原来的公式推出公式 + C! m8 S+ }2 K+ ~% F% ]3.Plugging v1 into the equation for v2 ,we obtain (将公式1代入公式2中得到) ' f; Y ^- {" s5 u. g2 S3 @4 m+ ^6 i% K# B6 F9 W+ ~. K3 E
公式 2 Z- G" a' j# O1 l; m% X# j3 H+ _2 i, Z; `7 u
11.Putting these together(把公式放在一起), because of the law of conservation of energy, yields: 5 L* h4 v j7 b3 F- g6 l8 A8 A$ o x6 T p ]# h
公式8 I: p" ~4 t- d) i1 p2 V' F Y
% _1 @. R5 x% Z' v! G% l# Z12.Therefore, from (2),(3),(5), we have the ith junction(由前几个公式得)! ]' [- b6 {) J9 q( B+ `; S
' ?- {) M" c g* S9 N公式 / s" z+ `" r/ ]9 D4 `, b9 f# [/ ^5 u0 W& E
Putting (1)-(5) together, we can obtain pup at every junction . in fact, at the last junction, we have - f9 k2 O& o/ l - j& \8 C/ Z6 G% Z8 z0 @公式5 `& D0 f( `% b; G# l) p( ~
F3 N: J4 L2 h2 j5 E& K. Z WPutting these into (1) ,we get(把这些公式代入1中) 7 f1 i2 V' b( S( d 0 J- R+ i- E' T+ b- ~公式 ; x \2 D: h# E9 G/ ~4 X# n$ R+ h/ y, G+ h" z
Which means that the u' c/ Y8 S i# q9 u/ H) N
! h; G3 F0 l% z* P" F" \! [2 QCommonly, h is about, q. ]1 s/ s* \5 f" w, m" Y# M
* r0 g4 d" A: f/ u) h; e
From these equations, (从这个公式中我们知道)we know that ……… 9 I" O H+ P6 ]/ t4 f$ I' s' n ) t0 `+ m4 X( O " z# p! a2 t$ M9 V; C- Y2 r! L, V- k0 h. ]+ w; [" ~3 K% S
引出约束条件 2 n7 |7 c4 I y4.Using pressure and discharge data from Rain Bird 结果,7 b v$ f) g9 A5 W7 M, h
6 k3 _4 k2 S- }' W
We find the attenuation factor (得到衰减因子,常数,系数) to be . B, ? f) p. [, g$ J/ W5 N i! }3 `$ o4 G [
公式 $ H; y4 f4 v" C! f7 ?* f# b, I/ E7 q + V+ e& i6 i% s" u! v( _3 D计算结果& |7 G, v/ n3 d x9 f5 T. i. |
6.To find the new pressure ,we use the ( 0 0),which states that the volume of water flowing in equals the volume of water flowing out : (为了找到新值,我们用什么方程)% D; _7 `. d; f! Z( j( I9 b
8 Y) w/ g0 l A7 ]
公式 ' C! I- U8 b7 ]" n 5 b, A Z3 p& z# k; ^4 G5 p7 x' RWhere ' B4 p4 C! J+ K# E" J& x2 j$ A% D; Q& D- G
() is ;;! f) k; ]% S! l: ]
, `8 L6 V [* e0 r, p5 d9 _9 e' f7.Solving for VN we obtain (公式的解)3 H( q- g8 ~! v0 p
5 l* o: ?, x2 r
公式. z: y' C! B, k* D
1 Q- ^% S$ C' x8 b! BWhere n is the ….. u. U; G7 c3 N! ?5 r* Q# @: c# Y" I5 s4 ]
4 J" m1 {# }- I3 S H/ d# t , z8 U7 v5 \0 d: z9 t' o& b* u8.We have the following differential equations for speeds in the x- and y- directions: 8 B% ], b5 `0 r " h( O8 o7 H- }+ l3 |, T( D公式 2 ?$ p, a. D _- M: {7 {; w2 E* X' Z$ y+ J0 C
Whose solutions are (解)0 U0 c8 `+ _6 m/ _8 _
4 }7 _3 S W7 v7 K
公式 R. i2 \* \# i1 V D
9 C7 `- A, n% E; p. F7 P
9.We use the following initial conditions ( 使用初值 ) to determine the drag constant: ! j+ r p! h2 a7 q3 e 2 E: ~# `! K7 ]3 w; y公式 2 y1 w a8 ^' r/ k5 \4 J: F+ T . w9 _4 n* b J" q6 H* ]: y根据原有公式 " d+ p! S( r5 r4 c10.We apply the law of conservation of energy(根据能量守恒定律). The work done by the forces is / K$ z! f2 N! \& \1 F) r7 D2 O1 K; I7 t2 v8 k# P) H
公式 o( P* |1 l- D! M- _* l2 Q : S4 c# y! E& W/ Z7 l# y$ d7 pThe decrease in potential energy is (势能的减少); X# ?* }# n0 _* V
( n- F L! t8 s2 w
公式 - p: k4 R# }, V 5 t* w3 U* ]1 A ]The increase in kinetic energy is (动能的增加)4 M' s$ B# u$ ~. B
1 w( J! c& G1 L, V
公式 & l+ R4 ^9 {9 h& d3 G$ W1 e; Q3 w2 w) u& t: o
Drug acts directly against velocity, so the acceleration vector from drag can be found Newton's law F=ma as : (牛顿第二定律). k$ W% U0 k7 L; _, ~+ b
, u( F7 V; p# J$ N) Z" c
Where a is the acceleration vector and m is mass 7 D/ I4 i! Q, S: {& \+ ~" w( V3 s1 a% `+ W3 n2 N/ g" j* w
- g2 H+ ?: Z& r3 B9 d " w5 v' |+ i. H$ T) }( UUsing the Newton's Second Law, we have that F/m=a and/ d; d) n. y; Z, d
6 v9 z$ l$ k) M5 u. g7 ]3 W$ ]6 u
公式# A) n! g& b: t2 i6 @- L" d
6 G* ^* B& w3 V" b
So that * Q, E9 @' _: U8 F# C / c3 s1 b E2 L b/ s# h2 N0 }公式: C( { W5 y( W( c
# Q$ {& P" Y; RSetting the two expressions for t1/t2 equal and cross-multiplying gives& ^" g9 `5 z# g
4 f! H: E. p9 U0 U公式 8 j' {+ [( B# L0 N/ a) U9 b, O( j% ~1 Z( Q: I$ L3 `0 v
22.We approximate the binomial distribution of contenders with a normal distribution: ( e9 ?" Q1 @+ O+ v+ v ) y: N' z( o: [7 Y公式0 n& ]% X+ I" \8 p
" x! s! w1 W( R' [3 V5 TWhere x is the cumulative distribution function of the standard normal distribution. Clearing denominators and solving the resulting quadratic in B gives7 w2 D7 v: Q) W2 O% M2 Y/ W; h
" K5 l! E# R4 F5 m8 s* E3 j公式5 I9 Z$ _# y' E c) R
K+ d! L. Q8 H2 a- tAs an analytic approximation to . for k=1, we get B=c & d5 i+ N) x D+ E8 W- C" R6 ` ( T, ?3 [; b$ S. m + I+ t y/ R2 Q1 u% V( f
" ~8 O; S. r- o% C& F1 V& ]26.Integrating, (使结合)we get PVT=constant, where ( X) |& Y( g+ [) [! K 9 |( J1 h# L t8 c8 z: F5 q7 E公式 : ~4 N; G! a" i6 A7 P; M- u5 [0 F( p2 u9 u' b% w
The main composition of the air is nitrogen and oxygen, so i=5 and r=1.4, so 9 T6 c) L- X ^ : s" _3 @* S e f% H 7 A: Y+ c z+ P# w* A
! T2 Y5 F v. M+ Q& c# g; `- `
23.According to First Law of Thermodynamics, we get1 f5 L; N. u& k+ e2 y7 }
4 f$ ^9 z$ g( Q
公式 1 X% o) n3 ^$ b4 ] % c7 |1 Q# W# Q! B# H6 g0 MWhere ( ) . we also then have/ u6 q1 j1 D' J7 G% W
1 _2 D( b9 j9 A. S公式 ) c0 W) R" S1 p/ z4 x" j - H2 g }! X( UWhere P is the pressure of the gas and V is the volume. We put them into the Ideal Gas Internal Formula:: I( \0 K2 T$ c. N, t q5 u
$ v4 } B% e5 A# ?9 Z公式! B+ l5 V# V" z. `0 l
6 P1 L$ b1 H& [6 P8 |3 E0 G
Where9 f/ i5 `4 D$ S& u$ h& M
% ]- o( L! n' b6 A9 p2 {; Y 9 L; w% b; X8 q
# j! b$ y a' e' `8 }1 P/ U, m+ ~
对公式变形7 H1 r+ m+ N7 I; b2 c
13.Define A=nlw to be the ( )(定义); rearranging (1) produces (将公式变形得到); |7 w) B; W) `* q
; d9 Q1 `: f$ y! B- ~' K" |公式6 Z: ?1 f3 h7 v/ X6 A
! }; S, G( M3 M
We maximize E for each layer, subject to the constraint (2). The calculations are easier if we minimize 1/E.(为了得到最大值,求他倒数的最小值) Neglecting constant factors (忽略常数), we minimize9 ~2 w( c+ {# _+ n/ u
6 X- ~. R) V# D# O& n/ a! @公式7 q4 ` Q' q3 u' b0 e3 K# i
9 j5 j8 B+ G. k" L5 W; v
使服从约束条件8 c4 a& v' a5 K# p
14.Subject to the constraint (使服从约束条件) # p$ H) }. G- d% _: @/ e# A/ K: x% Z0 N3 O" a( `" U( Y
公式 7 K6 }6 B: q8 M7 H/ \ R" `. `& a ; c( G( D# j8 f+ o7 O6 |; \8 a$ J+ o- |Where B is constant defined in (2). However, as long as we are obeying this constraint, we can write (根据约束条件我们得到) 9 ?# S7 a4 n K3 I" B& a+ g1 A- A8 }8 |) o+ Z O- l
公式 ! {6 ^" i5 K# D: E( z+ P* z A( b
And thus f depends only on h , the function f is minimized at (求最小值)* J& B5 w7 g6 D
) A u' s% S; Q) n+ n
公式 # t3 [3 t. F, y- s$ P. r( z$ R) c* u% W* |, A& s( }: `$ |
At this value of h, the constraint reduces to 3 I8 T. G0 Y+ u" Z& B" o0 s; Q9 ?; t! a" w; A' e/ P7 F/ |
公式9 I+ _) |7 `5 _. r1 ?# y. s* f
( N# s9 T/ ?$ v3 y) E结果说明 ' [5 [1 P ~3 ?15.This implies(暗示) that the harmonic mean of l and w should be 4 Z3 z! Z5 X) k0 h' a) }" K5 ~ _- x6 v# B
公式$ g. a& i+ v }+ q$ {. d u
; F; F2 f x+ G; @So , in the optimal situation. ………% m. U" T3 r+ ]! s
8 [" y/ s' u. X1 W6 c/ U, t- K5 ]
5.This value shows very little loss due to friction.(结果说明) The escape speed with friction is 6 \+ S: F1 e& o9 F+ T# S( A2 i ( D* j6 O# k; d- U- H% G1 a- s公式 , g+ _) w$ ]; \. ` * p: q( p9 t {- Y4 }16. We use a similar process to find the position of the droplet, resulting in 1 |3 }+ J" U6 I2 q3 p4 D + a9 d1 c- ~2 k& L公式. n3 U$ E* ~! J9 S! q8 V
9 ?$ m% Q6 }' m: N L* HWith t=0.0001 s, error from the approximation is virtually zero.8 m! H) J7 l j
; c1 R; n6 Q9 o* J( `
# m! [5 r9 t( v' ^2 w
# q2 g2 s2 k/ f2 L( o
17.We calculated its trajectory(轨道) using " F) l7 D8 R4 C+ {9 q6 w" R; ]: i$ W
公式0 }6 Q% p$ e6 b" H7 ^
& w; W3 H* O( D4 o8 \# z2 D( Z
18.For that case, using the same expansion for e as above,& G& e" n0 f. M1 O
: e/ v& U& V; Z! N$ H4 h/ t公式; n7 K* F4 H( l! J
' N6 W0 k+ o2 E- E" ~: i9 r" i4 ^
19.Solving for t and equating it to the earlier expression for t, we get- X* l: o; W1 j* C. V4 R
0 `) F4 s0 b' h7 K) z" o
公式! s; n+ P" m- k0 a( M7 ]# k
$ Z& r' c; D% N" b
20.Recalling that in this equality only n is a function of f, we substitute for n and solve for f. the result is . f+ G5 q4 Z$ M0 e& \* i2 J( [5 o+ ~/ t4 k0 y
公式; Y! ?6 }& U2 U
1 {. t1 ~7 J! R+ N! B! b8 aAs v=…, this equation becomes singular (单数的).$ J) q) T2 h, N- R) F
" G" H/ ?& `& I3 F, {+ t/ \
2 P$ G6 ]/ T8 f3 Y" I; K, k6 t$ L5 m+ u& m2 j; }7 e- {
由语句得到公式 K4 Y- {- h. S' J
21.The revenue generated by the flight is $ k* }7 b3 k8 ]1 b4 p c( X! E( E" q z; p' I" {公式 , g2 v# b% l' r2 g5 t # j1 j0 X. E. S 2 z) o+ j, P* C1 ]1 n2 [9 D$ h7 \8 L* R# T( j+ g6 {
24.Then we have7 v4 R) k$ {4 c' y3 x s
) D E9 [8 ~. v( p4 T公式 ' L/ g' x# k0 C9 w) q, }* e1 v! i' e ~: P- S, v
We differentiate the ideal-gas state equation/ M8 I6 N* ^& V+ l B& k" g
& ~3 K: Y4 G" r# T& j) m. S
公式 & t6 A3 e# u, F: O# V# h7 Y9 p 0 X: G1 F" D' a) f1 uGetting " d, `# `0 ?% c2 z9 ~4 x . U, z1 {& X+ N3 R8 \公式 * C I! N& D! _' M( q" { ( a6 e9 Q0 Y% s8 D0 K8 S$ ~25.We eliminate dT from the last two equations to get (排除因素得到) 3 R& x$ O( a7 k7 w1 e# B 6 h, [2 a- D. D: c( x/ q公式2 [& N; m* k" C9 W( S' ]9 J) Y9 R
! [1 o; e W2 Y) z, @ Q0 @4 U
A0 t! O3 }' K6 d 7 }/ ~4 a7 V% H' \22.We fist examine the path that the motorcycle follows. Taking the air resistance into account, we get two differential equations7 k4 \- L+ o W. q
0 e+ v- S. @! H' I& r
公式6 N. T4 n# i q* o9 G
1 i# C( L4 Z- v* ~- e& O8 u& ^Where P is the relative pressure. We must first find the speed v1 of water at our source: (找初值) x4 f3 [) J# y4 F