# Register now to access 7 million high quality study materials (What's Course Hero?) Course Hero is the premier provider of high quality online educational resources. With millions of study documents, online tutors, digital flashcards and free courseware, Course Hero is helping students learn more efficiently and effectively. Whether you're interested in exploring new subjects or mastering key topics for your next exam, Course Hero has the tools you need to achieve your goals.

100 Pages

### dif9-page4

Course: MATH 203, Spring 2009
School: Stony Brook University
Rating:

Word Count: 416

#### Document Preview

2. Lemma If two functions f,g e M coincide on a neighborhood U of p e M ------------------------- and D is a derivation at p then D(f) = D(g). Proof. ------------------------- Sublemma. If x e R n and B(x,e) is a fixed open ball about it, then there n exists a smooth function h:R R such that h(y) is equal to 1 if y e ---------------------------------------- ----------L B(x,e/2), positive if y e B(x,e) and...

Register Now

#### Unformatted Document Excerpt

Coursehero >> New York >> Stony Brook University >> MATH 203

Course Hero has millions of student submitted documents similar to the one
below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

Course Hero has millions of student submitted documents similar to the one below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.
2. Lemma If two functions f,g e M coincide on a neighborhood U of p e M ------------------------- and D is a derivation at p then D(f) = D(g). Proof. ------------------------- Sublemma. If x e R n and B(x,e) is a fixed open ball about it, then there n exists a smooth function h:R R such that h(y) is equal to 1 if y e ---------------------------------------- ----------L B(x,e/2), positive if y e B(x,e) and zero if y m B(x,e). Define the function h of the real variable t by the formula &e-(1-t2)-1if t e (-1,1) h (t) = { 0 70 otherwise. It is a good exercise to prove that h is a smooth function on R. Set 0 h (y) := h (4NyN /e), let c denote the characteristic function of the ball 1 0 B(x,3e/4), and define the function h as follows 2 0 i h (y) := 2 jnc(z)h1(y-z)dz. R If we put h(y) = h (y)/h (x) then we get a desirable function. 2 2 Now let us prove the lemma. Using the construction above we can define a smooth function h on M which is zero outside U and such that h(p) = 1. In this case h(f-g) is the constant function 0 on M. Thus we have 0 = D(0) = D(h(f-g)) = D(h) (f(p)-g(p)) + h(p) D(f-g) = D(f) - D(g). 44444 Remarks. ----------------------------------- i) The sublemma shows that the mapping i:M ----------L F*(M) above is indeed an inclusion. If p \$ q are distinct points of M, then there is a smooth function h on M such that [i(p)](h) = h(p) = 1 \$ [i(q)](h) = h(q) = 0. ii) We can extend a derivation D at a point p on functions f defined only in a neighborhood U of p by taking a smooth function h on M such that h is zero outside U and constant 1 in a neighborhood of p and putting D(f) := ~ D(f), where ( ~ f(x)h(x) f(x) = { 0 9 for for xeU x m U. By lemma 2 this extension of D is correctly defined. Lemma 3. Let f:B ----------L ------------------------- R be a smooth function defined on an open ball B C Rn around the origin. Then there exist smooth functions g 1< i<n on B such that i n f(x) = f(0) + S x g (x) for x = (x ,...,x ) e B ii 1 n i=1 and df g (0) = (0). i dx i -------------------- 4
Find millions of documents on Course Hero - Study Guides, Lecture Notes, Reference Materials, Practice Exams and more. Course Hero has millions of course specific materials providing students with the best way to expand their education.

Below is a small sample set of documents:

University of Phoenix - COM 155 - 155
I chose to read paragraphs four and five for the final assignment. While reading the twoparagraphs I thought that paragraph four was more effective of the two. I feel thatparagraph four was written better and it flowed nicely. Paragraph five was a bitc
University of Phoenix - COM 155 - 155
I chose to read paragraphs four and five for the final assignment. While reading the twoparagraphs i felt that paragraph four was more effective of the two. I feel that paragraph four waswritten better and it flowed nicely. Paragraph five was a bit conf
University of Phoenix - COM 155 - 155
University of Phoenix - COM 155 - 155
University of Phoenix - COM 155 - 155
3. Assignment: Paragraphs and Topic SentencesWrite two paragraphs about how interest rates affect our purchasing decisions.Identify the topic sentence in each by underlining it.Post your paper as an attachment.Attach the Certificate of Originality sep
University of Phoenix - COM 155 - 155
Axia College MaterialAppendix EFueling UpMotorists often complain about rising gas prices. Some motorists purchase fuel efficient vehicles andparticipate in trip reduction plans, such as carpooling and using alternative transportation. Other driverst
University of Phoenix - COM 155 - 155
WageIndustry111862085210997144761378719452166671523439888131622079319284134811678911702114513335137771256701331229191417802997725166303088344315957217163346128219316916062652762221333209416796351851769015193751651
University of Phoenix - COM 155 - 155
American-StylePit BarbequeChina:Business Plan:(Microsoft clip art, n.d.)(Microsoft clip art, n.d.)Team A: Joseph Cooney, Gene Janes, Joel Ramsey, Elizabeth Stevens, TinaSugliaUniversity of Phoenix OnlineGBM/389 Strategic Topics in Global Business
University of Phoenix - COM 155 - 155
Axia College MaterialAppendix CBudgets MatrixDirections: Using the matrix, define each of the budgets listed and briefly describe its uses.BudgetSales budgetDefinitionIs a budget which contains all ofthe sales prediction for thecompany in a month
University of Phoenix - COM 155 - 155
Hello Classmates and Mr. Bull:My name is Tina Suglia; I was born in Mesa, Arizona. I moved around a lot as a kid andnow live on the East Coast. Which I enjoy the 4 seasons anyway, in Arizona its just plainhot. But I miss it out there sometimes to. I ha
University of Phoenix - COM 155 - 155
First Position for a Senior Accountant located in Cambridge, MA. The educationalrequirements for this job position are to have a Bachelors Degree in Accounting orFinance, with 4 to 6 years experience financial analyst or accounting experience, have agr
University of Phoenix - COM 155 - 155
Based on what you know about accounting, what role do you see it playing inbusiness operations?The role that accounting plays in the business operation is several important things, likebilling is one part that several people are assigned to type out in
Penn State - CH E - 210
8.53 (contd)bgbg1 kJkJJ 16.2 b301gb61g + b170gb27g= 813 kJ mol.molmol 10 JH = H v + C pl 52 113 H m + C ps 25 52= 42.1kJmol3Required heat transfer: Q = H = nH =2.05 mol 813 kJ.s1 kWmol1 kJ s= 167 kW8.54Basis: 100 kg wet film a
Penn State - CH E - 210
8.54 (contd)T&amp;ET&amp;Ed. T f 2 = 34 C Ta1 = 506 C , T f 2 = 36 C Ta1 = 552 Ce.8.55In an adiabatic system, when a liquid evaporates, the temperature of the remaining condensedphase drops. In this problem, the heat transferred from the air goes to (1) va
Penn State - CH E - 210
8.56 (contd)References: Solids (0.01C), H 2 O (l, 0.01oC)SubstanceminmoutHH outinbgH Obv g20080062.85125.7200571.4428.6209.6419.12676H 2 O , 1.6 barm12696.2m1m ( kg h ) H H 2 O from steam tables475.4SolidsH 2O l2E.B. Q = H =m
Penn State - CH E - 210
8.57 (contd)bgbgbC 3 H 6 O v , 329 K C 3 H 6 O l , 329 K C 3 H 6 O l , 303 Kzg303bg C dT = 520.6 + b2.3gb26g = 580.4 kJ kgb2 6831gkg 580.4 kJ = 7.93 10 kJ h= H = nH =H = H v 329 K +pl329Qcc.6hkgOverall process energy balanceReferenc
Penn State - CH E - 210
8.58 (contd)Energy balance on 1st effect:b gbg b gb g b gb= 2534 kg H Obv g hg b gb gH = 0 n 2 2654 + n1 360 + n5 605 2738 5000 1131 = 0.n5n1 = 3182n2 =1818c.2Mass balance on 2nd effect: 3182 = n 3 + n 4 (1)bH = 0gbn gb2610g + bn gb252g + b
Penn State - CH E - 210
8.59 a.b0.035gb5000g = 583 kg hSalt balance: x L 7 n L 7 = x L1 n L1 n L1 =0.30Fresh water produced: n L 7 n L1 = 5000 583 = 4417 kg fresh water hb. Final result given in Part (d).c.Salt balance on i th effect:b g bx gn Li x Li = n Li +1L i +1
Penn State - CH E - 210
8.60 a.dC i = dC ipvplb g dC i= 20 cal (mol C) ; Cvpvvb R 10 2calg mol C = 8 cal (mol C)b.n0 (mol N2)n0 (mol N2)o3.00 L@ 93 C, 1 atmn2 [mol A(v)]85oC, P(atm)n1 (mol A(l)n3 [mol A(l)]o85oC, P(atm)0.70 mL, 93 Cn0 =3.00 Ln1 =70.0 m
Penn State - CH E - 210
8.61 (a) i)FG mIJHV KExpt 1 =b4.4553 3.2551gkg = 0.600 kg bSGg2.000 LliquidLbliquid= 0.600gii) Expt 2 Mass of gas = 3.2571 3.2551 kg = 0.0020 kg = 2.0 gMoles of gas =b763 500gmm Hg2.000 L 273 K363 K1 mol= 0.0232 mol22.4 liters STPbg7
Penn State - CH E - 210
8.61 (contd)Plot p (log scale) vs. 1 T (linear scale); straight line fit yields3770ln p =+ 17.28 or p = 3196 107 exp 3770 T.TKbbggbg1 17.28 ln 760== 2.824 10 3 K 1 Tb = 354 KA v Tb3770Partvi) p = 760 mm Hg vii)H v= 3770 K H v = 3770 K
Penn State - CH E - 210
8.61 (contd)bgN 2 510 K : H N 2 (510K) - H N 2 ( 262 K) = H N 2 (237 o C) - H N 2 ( 11o C)Table B.8B= [6.24 - (-1.05)] kJ / mol = 7.286 kJ / mol = 7286 J / molbgbgA(v, 262K): H = C pl Tb 262 + Hv 359 K +z262TbC pv dTPart (a) results for Tb
Penn State - CH E - 210
8.62 (contd)zbg6.01 kJ26H = H m 0 C + C p dT =1 mol 10 3 g 2.17 kJ18.02 g 1 kg + kg CmolA026 C= 390 kJ kgTable B.1bbggbb50 0g CH 2 O l , 50 C : H 2 O l , 0 C H 2 O l , 50 Cz0.0754 kJ50H = C p dT = mol C1 mol 1000 g18.02 g 1 kg =
Penn State - CH E - 210
8.63 (Contd)S balance on mixing point:0.12 27273 + 0.45 172730 = 2.000 105 X 2 X 2 100% = 40.5% Sb gbg b gbg(b) Draw system boundary for every balance to enclose freezer and mixing point (Inputs:fresh feed and recycle streams; output; slurry leavin
Penn State - CH E - 210
8.64 (contd)References: B(l, 10oC), I(l, 10oC)substancenin mol / hH in kJ / molbB (l)B (v)I (l)I (v)gb8575-15925-zzbgnout mol / h0-0-g-8575-15925bH out kJ / molg-42.21-41.01d H i = dH i + dC i = 42.21 kJ / mol.d H i = dH i
Penn State - CH E - 210
8.65 (contd)(b) T = 75 C pC6 H 6 = 648 mm Hg , pC7 H 8 = 244 mm Hg (from Table 6.1-1)bgb g bgb gRaoult's law ptank = xC6 H 6 pC6 H 6 + xC7 H 8 pC7 H 8 = 0.439 648 + 0.561 244bg= 284 + 137 mm Hg = 421 mmHg Ptank = 0.554 atmyC 6 H 6 =bg284 mm Hg
Penn State - CH E - 210
8.66 a. Basis: 1 mol feed/sn V mo l vapor/s @ T, P1 mo l/s @ TFo Cy mol A/mol(1-y) mol B/mo lxF mol A/mol(1-xF) mol B/mo ln L mol vapor/s @ T, Pvapor and liquid streamsin equilibriumx mol A/mol(1-x) mol B/mo lbg b g bgb T g y = x p bT gRaoul
Penn State - CH E - 210
8.66 (contd)c.C*C*1C*C*C*C*22025330PROGRAM FOR PROBLEM 8.66IMPLICIT REAL (N)READ (5, 1) A1, B1, C1, A2, B2, C2ANTOINE EQUATION COEFFICIENTS FOR A AND BFORMAT (8F10.4)READ (5, 1) TRA, TRBARBITRARY REFERENCE TEMPERATURES (DEG.C.) FOR A A
Penn State - CH E - 210
8.66(contd)DELH = NL *(XL*HAL + (1.0 XL)*HBL) + NV*(XV*HAV + (1.0 XV)*HBV) F1WRITE (6, 4) T, NL, NV, DELH4FORMAT (1Hb, 5X' Tb=', F6.1, 3X' NLb=', F7.4, 3X' NVb=', F7.4, 3X'DELHb=',* E11.4)WRITE (6, 5) PAV, PBV, XL, HAL, HBL, XV, HAV, HBV5FORMAT (
Penn State - CH E - 210
8.67 Basis:2500 kmol product 1 kmol condensateh.25 kmol product= 10,000 kmol h fed to condenser.m1 , (kg/h) at T11090 kmol/h C 3 H 8 (v )7520 kmol/h i -C 4H10 (v )1390 kmol/h n -C 4H10 (v )saturated vapor at Tf, P1090 kmol/h C 3 H 8 ( l )7520
Penn State - CH E - 210
8.67 (contd)(b) Cooling water: Tout = 40 C , T2 = 34 C , T1 = 25 CP = xi pi* ( 40C ) = 0.109 (11,877 ) + 0.752 ( 3961) + 0.139 ( 2831) = 4667 mm Hgif (T f ) = 1 P bgiT +Eyi= 0 T f = 45.7Cp (T f )*ibgbgRefs: C 3 H 8 l , C 4 H 10 l @ 40C, H 2
Penn State - CH E - 210
8.68 (contd)HCHO balance on absorber n6a , CH 3 OH balance on absorber n6bWt. fractions of product solution x1 , x 2 , x 3HCHO balance on distillation column n7CH 3 OH balance on distillation column n8CH 3 OH balance on recycle mixing point n1Energy
Penn State - CH E - 210
8.68 (contd)HCHO balance on distillation column (include the condenser + reflux stream within thesystem for this and the next balance):19.89 = 0.262n7 n7 = 75.9 mol productCH 3 OH balance on distillation column:bg8.34 = 0.006 75.9 + n8 n8 = 7.88 mol
Penn State - CH E - 210
8.68 (contd)bgGas cooler: Same refs. as above for product gas, H 2 O l , 30 C for cooling watersubstanceninH innoutH outHCHOCH 3 OH19.98.3430.30.835.035.6003.513.603.474.0919.98.3430.30.835.035.61.782.382.192.242.162.54m
Penn State - CH E - 210
Vroom = 141 ft 3 . DA = dry air.8.70mDA =ha =140 ft 3lb - mol o R 29 lb m DA 1 atm.= 101 lb m DAlb - mol 550 o R0.7302 ft 3 atm0.205 lb m H 2 O= 0.0203 lb m H 2 O / lb m DA101 lb m DA.From the psychrometric chart, Tdb = 90 o F, ha = 0.0903h
Penn State - CH E - 210
400 kg 2.44 kg water= 10.0 kg water evaporates / min97.56 kg air10 kg H 2 O min(b) ha == 0.025 kg H 2 O kg dry air , Tdb = 50 C400 kg dry air min8.73 (a)minFig. 8.4-1bgH = 116 11 = 115 kJ kg dry air , Twb = 33 C, hr = 32%, Tdew point = 28.5 C
Penn State - CH E - 210
8.74 (contd)The outside air is first cooled to a temperature at which the required amount of water iscondensed, and the cold air is then reheated to 55F. Since ha remains constant in thesecond step, the condition of the air following the cooling step m
Penn State - CH E - 210
8.74 (contd)Energy balance on cooler-reheaterReferences: Same as Fig. 8.4-2 [including H2O(l, 32oF)]minH in10.93 45.565.57 26.2SubstanceFresh air feedRecirculated air feedDelivered airH out21.4oCondensed water (49 F)76.50.165 17.0Q = H =
Penn State - CH E - 210
8.75 (contd)Moisture content of exiting chips:0.065 kg water 100% = 9.8% &lt; 15% meets design specification0.600 kg dry chips + 0.065 kg waterbg(c) References: Dry air, H 2 O l , dry chips @ 0C.substanceminAirH 2O ldry chipsHin15.02 100.20.400
Penn State - CH E - 210
8.78 a.UV= 4 C Wbh gTdb = 45 CTdew pointa in= 0.0050 kg H 2 O kg D.A.Twb = 20.4 C, V = 0.908 m 3 kg D.A.Fig. 8.4-1bgTwb = Tas = 20.4 C, saturated haout= 0.0151 kg H 2 O kg D.A.b. Basis: 1 kg entering sugar (S) solutionm1 (kg D.A.)0.0050 k
Penn State - CH E - 210
8.79 (contd)( 20 - 6.4) Btu / lb m dry airc. QBA = H = H B H A QDC = H = H D H C 12.2 ft 3 / lb m dry air(23 - 20) Btu / lb m dry air312.2 ft / lb m dry air= 1.1 Btu / ft 3= 0.25 Btu / ft 3d.70%52%35%CDAB58.52070758.80 Basis: 1 kg D.
Penn State - CH E - 210
8.80 (contd)Inlet air:Tdb = 37 Chr = 50%Moles dry air: ma =Rh = 0.0198 kg H O kg DASH = b88.5 - 0.5g kJ kg DA = 88.0 kJ kg DATFig. 8.4-12a111250 kg 1 kg DA= 1226 kg DA hh1.0198 kgOutlet air: Tdb = 15 C, sat' dRh = 0.0106 kg H O kg DASH =
Penn State - CH E - 210
8.83 Basis: 100 mol solution 20 mol NaOH, 80 mol H2O80 mol H 2 O= 4.00 mol H 2 O mol NaOH20 mol NaOH r=bgRefs: NaOH(s), H 2 O l @25 CsubstanceninNaOH sbgH Obl gNaOHbr = 4.00g2bH innoutH out20.0 0.0n in mol80.0 0.0H in kJ mol20.0 34.4
Penn State - CH E - 210
di8.85 2 mol H 2SO 4 = 0.30 2.00 + nH 2 O nH 2 O = 4.67 mol H 2 O r =a.For this closed constant pressure system,bgQ = H = nH 2SO 4 H s 25 C, r = 2.33 =b. msolution =2 mol H 2SO 4mol H 2SO 498.08 g H 2SO 4molbgb280.6 + 150gg88.6 kJ +Basis:
Penn State - CH E - 210
8.86 (contd)bgbgH HCl, n = 5.75 = H s 25 C, n = 5.75 += 64.87 kJ mol +ejH HCl, 20o C =z2025z401nHClmC p dT251120 g 0.66 calg C8 molsb40 25g C4.184 JkJcal103 J.0.02913 01341 10 5 T + 0.9715 10 8 T 2 4.335 10 12 T 3 dT= -0.15 kJ
Penn State - CH E - 210
8.87 (contd)Air @ 200C: Table B.8 H = 515 kJ mol.Air (dry) @ 50C: Table B.8 H = 0.73 kJ molbgH 2 O v , 50 C : Table B.5 H =b2592 104.8g kJ1 kg18.0 g3kg10 g 1 molsubstancebgH Obv gninHinnoutHoutNaOH aq0.1542.470.152.85 n in mol min
Penn State - CH E - 210
8.89 P2 O5 + 3H 2 O 2H 3 PO 4Bmol H 3 PO 4a.wt% P2 O 5 =bg 100% ,n 14196.mtb98.00g 100%2nwt% H 3 PO 4 =Bg H 3 PO 4 molmcAg totalwhere n = mol P2 O5 and mt = total mass .wt% H 3 PO 4 =bg wt% P O2 98.0014196.25= 1381 wt% P2 O 5.
Penn State - CH E - 210
8.89 (contd)R|H||Table B.6 S|H||TFGGHH2 O ( l ) (149oblb mmgF = 65.4 o C) = (274 kJ / kg) 0.4303 = 118 Btu / lb mLMNlb mBtu) (118 1141)minlb mQ = mH = (46.3IJJ = 1141 Btu / lbkJkg KBtuooH2 O ( v ) (186 F = 85.6 C) = (265
Penn State - CH E - 210
8.90 Basis: 200 kg/h feed solution. A = NaC 2 H 3 O 2.n 1 (kmol H2 O(v )/h)50C, 16.9% of H 2Oin feed200 kg/h @ 60C.n 0 (kmol/h)0.20 A0.80 H 2OProduct slurry @ 50C.n 2 (kmol A-3H 2 O(v )/h).n 3 (kmol solution/h)0.154 A0.896 H 2 OQ (kJ/hr)
Penn State - CH E - 210
8.90 (contd)bgProduct solution: nH = n A H s 25 C + mb0.154g1.095 kmol A=z5025z5025C p dT (hydrate at 25 C , heat to 50 C )bg1.13 kmol A 3H 2 O sh= 36700 kJ h=bzLMNgH 2 O v , 50 C : nH = n H v +=OPQ5025b neglect E gC p dT (va
Penn State - CH E - 210
8.92 a.mA (g A) @ TA0 (oC)nA (mol A)nS (mol solution) @ Tmax (oC)mB (g B) @ TB0 (oC)nB (mol B)Refs: A(l), B(l) @ 25 Csubstance nin H in noutH outAnA H An in molBnB H BH in J / molSnAH S (J mol A)mA (g A)m, nB = BM A (g A / mol A)MBM
Penn State - CH E - 210
8.93 Refs: Sulfuric acid and water @ 25 CsubstanceninHinH2SO4H2OH 2 SO 4 aqb.1rM A C pA T0 25M w C pw T0 25bgbbggnout1Houtn in molH in J/molH m r + M A + rM w C ps Ts 25bg bgbg(J/mol H2SO4)bg bgb gbg= H br g + b98 + 18r gC bT
Penn State - CH E - 210
8.94 a.Ideal gas equation of state n A 0 = P0V g / RT0Total moles of B: n B 0 ( mol B) =(1)bgdVl ( L) SG B 1 kg / L 10 3 g / kgi(2)M B (g / mol B)Total moles of A: n Ao = n Av + n AlHenrys Law: rFG mol A(l) IJ = k pH mol B Ks(3)Abgn Al
Penn State - CH E - 210
8.94 (contd)b.Vt20.0MA47.0CvA0.831MB26.0CvB3.85SGB1.76Vl3.03.03.03.03.03.03.03.0T0300300300300330330330330P01.05.010.020.01.05.010.020.0Vg17.017.017.017.017.017.017.017.0nB0203.1203.1203.1203.1203.12
Penn State - CH E - 210
8.94 (contd)30029135.0150.018.0Program OutputT (assumed)(K)321.10296.54296.5715.00.02911.54E30.07542.6E64.2E0374P(atm)8.0197.4157.416Nav(mols)4.5794.5714.571Nal(mols)1.7031.7111.711T(calc.)(K)296.542296.568296.568P
Penn State - CH E - 210
8.96 a. 2.30 lb m 15.0 wt% H 2 SO 4@ 77 F H1 = 10 Btu / lb moU||| m ( lbV|||Wadiabatic mixing3m2 (lb m ) 80.0 wt% H 2 SO 4@ 60o F H 2 = 120 Btu / lb m60.0 wt% H 2 SO 4 @ T o F, H 3U |m|RVSmass balance: 2.30b0.150g + m b0.800g = m (0.60
Penn State - CH E - 210
8.98T = 140 FFig. 8.5-2Vapor: 80% NH 3 , 20% H 2 OCLiquid: 14% NH 3 , 86% H 2 OABBasis: 250 g system mass mv (g vapor), mL (g liquid).14.60.80 x NH3Mass Balance: mv + mL = 250.NH3 Balance: 0.80m g + 014mL = ( 0.60)(250) mv = 175 g, mL = 75g
Liberty - BUS - 621
Contract Laws4. Contract Law What contract law issues could your business face? What types ofcontracts should your business have with its suppliers, customers, and evenemployees (employment law later will fill this out in greater detail) and owners(bu
Liberty - BUS - 550
CRISIS COMMUNICATION1CRISIS COMMUNICATIONBUSI_550_D02Professor NeedhamApril 9, 2011CRISIS COMMUNICATION1AbstractCrisis is defined as an unstable or crucial time or state of affairs in which a decisivechange is impending. United States economic c
Liberty - BUS - 550
Running Head: CRISIS COMMUNICATIONCRISIS COMMUNICATIONBUSI_550_D02Professor NeedhamApril 19, 2011CRISIS COMMUNICATION1AbstractCrisis is defined as an unstable or crucial time or state of affairs in which a decisivechange is impending. United Stat
Liberty - BUS - 550
Running Head: CRISIS COMMUNICATIONCRISIS COMMUNICATIONBUSI_550_D02Professor NeedhamMay 11, 2011CRISIS COMMUNICATION1AbstractCrisis is defined as an unstable or crucial time or state of affairs in which a decisivechange is impending. United States