BOBOdHHHHHHH&E -Z x HHHV,(ld'0 6  Bordersl@/\{OptTurn Aue B ShapezNy Can PaintzN- Pattern Gradien >2\1H ide Sel 4(n:DSETb ?7&K  |?pYt| x#)P!!  h ! ? @  ( M b    P h i j             k    - S                . J          $ / : ;  < o q r ,    K L M N O P Q R S T U V W X Y Z    o q       !    U s      A j   $ T     ; [ |     < c             >       J q   ) [    M    :    e      `    / r !< ! ! "+ ", "G "H "I " " #y # # # $ $< $= $h $ % %T %U %r % % &G &H &I &J &K R > ? @ D v w                           & ' ) * 0 1 2 @ A K L R S T U V W X [ \ ] ^ ` c d h i j k l m n q r s t v w                   / /  /                                    ) * + , / 0 1 2 7 9 : ; < > H J L M N Q R i /j k l w x                                                                                                 7 8 h i l m t u | }              !                          " # - . 0 9 ; < = > E F G H S T V a b c d   "                                                         % & * !1 J K O !U ~                                                                             " # ( , - . / 1 2 3 < > C D H I K L P Q S T U W X Y b c f g h m u v w         #               !    $  %  $ : &                                             !  $  %  '  (  )  +  ,  .  0  1  5  6  :  ;                                                    o  q  r ' s  u  v ( w  y ) z ' {  |   )    '      (    )  '    )    '      (    '      )    '      (    )  '    )      )    )    '      (    )  '    '      (    )  '    )    )    '      (    )  '    (        (    (    '      (    )  '    (    '      (    )  '    (    '        )  '    )  '    ,   0 !Z     o  &                                          d e i j                       !      " # $ 2 3 B C O Q S T n o q r t u                '  '  '   (  ) '   )  '      (   ) '  )  '   ( ( ') * - ). 2 )3 4 7 '8 : ; (< > )? '@ A 'B D E (F H )I 'J M 'N P Q (R ^ '_ ` c )d h )i j k )l 'm p { '| }  )  )  ( )       !     % & ) * . / 8 : ; < = G H U V Y Z ^ _ h j k l m w x                                           ) '   '    )  )  ( )   ) '   + ', 5 '6 8 '9 ): ; < )= '> G 'H J 'K T 'U W 'X )Z [ \ )] '^ g 'h j 'k t 'u w 'x ){ | } )~ '  '  ' )  )    !  !   )  )  )  7 )8 = )> M 'O Q 'R d )e q 's u 'w } '~  '   '  '  '   '                               ? @ X ^ a g                                " $ % ) * . / 2 3 5 6 7 9 : < > ? C D H I d e g h l m o p r s { |                    ) * , - . 0 1 3 5 6 : ; ? @ C D F G H J K M O P T U Y Z [ \ ^ _ ` h i j k l n o s t x y | }                                                                       & ' ( ) , 5 6 : ; < C D H I J K L M Q R V W X \ ] ^ b c d f g h t u y z {                      *                    # $ % & ' / 1 2 3 5 *: +; ,I -K ,M -N ,O -P ,[ -] ,_ -` ,a -b ,p -r ,t -u ,v -w , - , - , - ,                    # $ - . 0 1 9 : I L N O \ _ a b c i j v w y z {                             ; < ` a c d e g h j l m q r v w z { } ~                                                                                                      $  %  &  )  *  ,  -  /  0  1  2  3  4  7  8  ?  @  A  o  p  r  s  t  u  v  w                               ! ! ! ! !( !* !, !- !7 !K !M !O !P !] !^ ! !! ", !"F "I "N " " " " " " " " " #7 #8 #r #t # # # # # # # # # # # $ $! $" $% $' $* $+ $/ $I !$N $i $m $s $t $ $ $ $ $ $ $ $ $ $ $ $ $ !% %  % % % %% %& %1 %4 %6 %7 %8 %A %B %N %O %R %S %v %y %z %{ % % % % % % % % % % % % % % % % % % % % % % % % % .% & !& &@ !&E &H &J &K   6u oh @@ ?  ?o )1BVPPd   "PHYsics 1150 Chapter 11 Homework Ch 11: 1, 6, 7, 11,20, 22, 37 11.1 Water flows through a pipe at the rate of 0.35 m3/min. What is its speed where the pipe has a diameter of 1.5 cm? Where the diameter is 2.5 cm?  Dm/Dt = r A v = constant r = 1000 kg/m3 A1 = r12 = (0.75 cm)2 = 1.77 cm2 A2 = r22 = (1.25 cm)2 = 4.91 cm2 v = [Dm/Dt ] [1/r A] v1 = [Dm/Dt ] [1/r A1] = [0.05 m3/s] [1/(1000 kg/m3)( 1.77 cm2)] Be very careful of the units! A1 = 1.77 cm2 = 1.77 cm2 [1 m / 100 cm]2 = 0.000177 m2 = 1.77 x 104 m2 v1 = [Dm/Dt ] [1/r A1] = [0.05 m3/s] [1/(1000 kg/m3)( 1.77 x 104 m2)] v1 = 0.28 m/s = 28 cm/s A2 = 4.91 cm2 = 4.91 cm2 [1 m / 100 cm]2 = 0.000 491 m2 = 4.91 x 104 m2 v2 = [Dm/Dt ] [1/r A2] = [0.05 m3/s] [1/(1000 kg/m3)( 4.91 x 104 m2)] v2 = 0.102 m/s = 10.2 cm/s 11.6 What must be the gauge pressure at a fire hydrant if water from a fire hose is to reach a height of 12 m?  We can immediately apply Bernouliis Equation, 1/2 r v2 + r g h + p = constant or 1/2 r v12 + r g h1 + p1 = 1/2 r v22 + r g h2 + p2 In our case, the hydrant is position 1 and the top of the stream of water is position 2, v1 = 0, h1 = 0, p1 = p1g = ? v2 = 0, h2 = 12 m, p2 = p2g = 0 (the guage pressure at the top of the stream is zero). 1/2 r (0)2 + r g (0) + p1g = 1/2 r (0)2 + r g (12 m) + (0) p1g = r g (12 m) p1g = (1000 kg/m3) (9.8 m/s2) (12 m) p1g = 117,600 N/m3 = 117,600 Pa = 117.6 kPa p1g = 117.6 kPa p1g = 117.6 kPa [1 atm / 101.3 kPa] = 1.16 atm 11.7 Air (r = 1.3 kg/m3) blows past a 1.0 m x 2.5 m window with a speed of 20 m/s. What is the difference in pressure on the two sides of the window? What is the net force on the window?  We can (again) immediately apply Bernouliis Equation, 1/2 r v2 + r g h + p = constant or 1/2 r v12 + r g h1 + p1 = 1/2 r v22 + r g h2 + p2 In our case, position 1 is outside the window and position 2 is inside the window. The height is the same so h1 = h2 = h, 1/2 r v12 + r g h1 + p1 = 1/2 r v22 + r g h2 + p2 1/2 r (20 m/s)2 + r g h + p1 = 1/2 r (0)2 + r g h + p2 1/2 r (20 m/s)2 + [r g h] + p1 = 1/2 r (0)2 + [r g h] + p2 1/2 r (20 m/s)2 = p2 p1 p2 p1 = 1/2 r (20 m/s)2 = 1/2 (1.3 kg/m3) (20 m/s)2 = 260 N/m2 = 260 Pa Dp = 260 Pa F = Dp A = (260 N/m2)(1.0 m x 2.5 m) = 650 N F = 650 N 11.11 Below are data for the cross-sectional area of open pipes and the pressure head behind each. (The pressure head is the gauge pressure measured in meters of water). Find the volume flow rate through each pipe. cross-section area pressure head pipe (cm2) ( m ) a 1.0 2.5 b 1.2 2.5 c 1.5 2.5 d 1.8 3.0 e 3.0 4.0 f 3.0 5.0 g 5.0 6.0 This will be a good problem to solve once algebraicly and then use a spreadsheet to solve for each particular set of numbers. We begin, again, with Bernouliis Equation, 1/2 r v2 + r g h + p = constant or 1/2 r v12 + r g h1 + p1 = 1/2 r v22 + r g h2 + p2 We are not concerned with any difference in height, so we may set h1 = h2 = h. Then our equation becomes 1/2 r v12 + p1 = 1/2 r v22 + p2 We can think of the flow rate due to a pressure difference as the flow out of a large tank. Inside the tank the flow velocity is essentially zero. A diagram may help  1/2 r v12 + p1 = 1/2 r v22 + p2 1/2 r (0)2 + p1 = 1/2 r v22 + p2 p1 p2 = 1/2 r v22 1/2 r v22 = p1 p2 1/2 r v22 = D p However, this difference in pressure is often refered to as a pressure head and is then stated in terms of the height of a column of water that would provide this difference in pressure; that is D p = r g h 1/2 r v22 = D p 1/2 r v22 = r g h 1/2 v22 = g h v22 = 2 g h  (This is also known as Torecellis Theorem). Now that we know the velocity, the volume flow rate is just the velocity multiplied by the cross-sectional area, volume flow rate = A v  Now we are ready to create a spreadsheet to do the actual calculations. However, a word of caution is appropriate here. Be careful with the units! As an example, the formula for pipe a, which appears in cell E3, is =C3*SQRT(2*9.8*D3) 11.20 The inside diameter of the larger part of the pipe in Figure 11.22 is 2.50 cm. Water flows through the pipe at a rate of 1.80 l/s. What is the inside diameter of the constriction?  A better diagram might look more like this  We begin, again, with Bernouliis Equation, 1/2 r v2 + r g h + p = constant or 1/2 r v12 + r g h1 + p1 = 1/2 r v22 + r g h2 + p2 We are not concerned with any difference in height, so we may set h1 = h2 = h. Then our equation becomes 1/2 r v12 + p1 = 1/2 r v22 + p2 We know the volume flow rate of 0.180 l/s. In the larger part of the pipe, we know the cross-section area A1, A1 = r12 = (1.25 cm)2 = (0.0125 m)2 = 4.91 x 104 m2 volume flow rate = Av = A1 v1 v1 = [volume flow rate]/A1 = [0.180 x 103 m3/s]/[ 4.91 x 104 m2] v1 = 0.367 m/s where we have made use of 1 liter = 1 l = 103 m3 1/2 r v12 + p1 = 1/2 r v22 + p2 1/2 r (0.367 m/s)2 + p1 p2 = 1/2 r v22 1/2 r v22 = 1/2 r (0.367 m/s)2 + p1 p2 v22 = (0.367 m/s)2 + 2 (p1 p2)/r We know the pressures p1 and p2 in terms of the pressure head, the height of the column of water that is equivalent to these pressures. p = r g h p1 = r g h1 = (1000 kg/m3) (9.8 m/s2) ( 0.10 m) p2 = r g h2 = (1000 kg/m3) (9.8 m/s2) ( 0.05 m) p2 p1 = r g h2 r g h1 2 (p1 p2)/r = 2 (r g h2 r g h1)/r = g (h2 h1) = (9.8 m/s2)(0.05 m) = 0.49 m2/s2 v22 = (0.367 m/s)2 + 2 (p1 p2)/r v22 = (0.367 m/s)2 + 0.49 m2/s2 v22 = 0.13 m2/s2 + 0.49 m2/s2 v22 = 0.13 m2/s2 + 0.49 m2/s2 v22 = 0.62 m2/s2 v2 = 0.79 m/s This is the velocity through the constriction. What area (or diameter) will provide this velocity? volume flow rate = Av = A2 v2 A2 = [volume flow rate]/v2 = A2 = [0.180 x 103 m3/s]/[0.79 m/s] = A2 = 2.28 x 104 m2 = r2 r2 = 7.25 x 105 m2 r = 8.5 x 103 m d = 2r = 1.70 x 102 m d = 1.7 cm (finally!)11.22 The dimensions given in Figure 11.24 are all inside diameters. When the faucet valve is wide open, it does not constrict the smaller pipe. If water enters the larger pipe under a pressure of 3800 torr, what volume emerges from the faucet per second. Compare this exact result with the restult obtained by disregarding the speed of the water in the larger pipe. Assume the pressure in the open faucet is 760 torr (1 atm).  In the large pipe, at position 1, p1 = 3800 torr = 3800 torr [133 Pa / 1 torr] = 505,000 Pa = 505 kPa h1 = 0 A1 = r12 = (0.0025 m)2 = 1.96 x 105 m2 In the small pipe, at position 2, p2 = 760 torr = 760 torr [133 Pa / 1 torr] = 101,000 Pa = 101 kPa h2 = 10 m A2 = r22 = (0.001 m)2 = 3.14 x 106 m2 As always, it would seem, we shall apply Bernouliis Equation, 1/2 r v2 + r g h + p = constant or 1/2 r v12 + r g h1 + p1 = 1/2 r v22 + r g h2 + p2 VFR = volume flow rate = A1 v1 = A2 v2 v1 = VFR/A1; v12 = [VFR/A1]2 v2 = VFR/A2; v22 = [VFR/A2]2 Now we know everything in Bernouliis Equation, except the volume flow rate (VFR) so we can (finally) solve for that! 1/2 r v12 + r g h1 + p1 = 1/2 r v22 + r g h2 + p2 1/2 r [VFR/A1]2 + r g h1 + p1 = 1/2 r [VFR/A2]2 + r g h2 + p2 [VFR/A1]2 + 2 g h1 + 2 p1/r = [VFR/A2]2 + 2 g h2 + 2 p2/r [VFR/A1]2 [VFR/A2]2 = 2 g (h2 h1) + 2 (p2 p1)/r VFR2 [ (1/A1)2 (1/A2)2 ] = 2 g (h2 h1) + 2 (p2 p1)/r VFR2 [(A22 A12)/(A12 A22)] = 2 g (h2 h1) + 2 (p2 p1)/r VFR2 = [ 2 g (h2 h1) + 2 (p2 p1)/r] [(A12 A22)/(A22 A12)] Now (finally!) we can plug in numbers, VFR2 = [2 (9.8 m/s2) (10 m 0) + 2 ( 404,000N/m2)/(1000 kg/m3)] x x [(( 3.14 x 106 m2)2 (1.96 x 105 m2)2)/(( 3.14 x 106 m2)2 (1.96 x 105 m2))2)] (messy, eh? Sure, but fairly straightforward, nonetheless). VFR2 = [196 m2/s2 808 m2/s] [(3.79 x 1021 m8)/(9.86 x 1012 384 x 1012) m4] VFR2 = [196 m2/s2 808 m2/s] [(3.79 x 1021 m8)/( 374 x 1012 m4)] VFR2 = [ 612 m2/s2 ] [ 1.01 x 1011 m4 ] VFR2 =6.202 x 109 m6/s2 VFR = 7.88 x 105 m3/s VFR = 7.88 x 105 m3/s [ 1000 l/s ] =0.0788 l/s (liters per second) Regard the larger pipe as a large tank with v1 = 0. Then Bernouliis Equation is 1/2 r v12 + r g h1 + p1 = 1/2 r v22 + r g h2 + p2 0 + r g h1 + p1 = 1/2 r v22 + r g h2 + p2 1/2 r v22 + r g h2 + p2 = r g h1 + p1 1/2 r v22 = r g (h1 h2) + (p1 p2) v22 = 2 g (h1 h2) + 2 (p1 p2)/r v22 = 2(9.8 m/s2) (0 10 m) + 2 (505 kPa 101 kPa)/(1000 kg/m3) v22 = 196 m2/s2 +808 m2/s2 v22 = 612 m2/s2 v2 = 24.7 m/s This number seems large, but believable. Now, what is the volume flow rate? VFR = volume flow rate = A2 v2 = (3.14 x 106 m2)(24.7 m/s) = VFR = 7.77 x 105 m3/s [ 1000 l/m3 ] = 0.077 l/s (liters per second) And that seems like a reasonable number. The answers in the two cases are very similar so treating the larger pipe as a large tank is a very reasonable approximation. Be sure to check on units! 11.37 The heart of a person sitting quietly pumps blood at the rate of 4x103cm3/min through an aorta of cross section 0.8cm2. a) What is the average blood speed in the aorta? The blood then spreads out into a network of about 5 x 109 (!) tiny capillaries whose average radius is about 8 x 104 cm. b) What is the speed of the blood through these capillaries? VFR = volume flow rate = A v v = VFR/A = [4 x 103 cm3/min] / [0.8 cm2] = 5,000 cm/min = 50 m/min v = 50 m/min [min/60 s] = 0.83 m/s What is the total area of the capillaries? Aeach = r2 = (3.14)( 8 x 104 cm)2 = (3.14)( 8 x 106 m)2 = 2.01 x 1010 m2 While each capillary has a very small cross section, there are five billion of them! Atotal = n Aeach = (5 x 109)(2.01 x 1010 m2) = 1.00 m2 = 1.00 x 104 cm2 VFR = volume flow rate = A v v = VFR/A = [4 x 103 cm3/min] / [1.00 x 104 cm2] = 4 x 101 cm/min v = 0.4 cm/min [min/60 s] = 6.67 x 103 cm/s As you should expect, that velocity is very slow because the total cross section area is very large. ZNDSET.Hx 6*DSET .Ht~ A\ :@L$ /\\{$ /A$ 0\$ 0-0-$ 10F0n$ 12 21FGf  2*8  2>|K 2c~ p6* #xprWEwdxpr  WEWE" WEWE pSymbolWE qDWEp! Avant GardeWE q!m, Symbol .* D,! Avant Garde!)mWEWE qDWE q!t(D!)tWE pTimesWE q WE q!=,Times( WE q! WE q!3WE q! WE" *WE!)= 3 WEPWEWE q!m( ,mWEWEWE q!3 ( 73WEWE q!mWE q!iWE q!n (*minWE q! WE q!=WE q! WE q!3WE q! WE" VWE(? = 3 WEPWEWE q!m( XmWEWEWE q!3 ( c3WEWE q!mWE q!iWE q!n (VminWE q! WEq"p##"##WE(k WE" q'WEWE q!mWE q!iWE q!n( zminWEWE q!6WE q!0WE q! WE q!sWE q!eWE q!c(q60 secWE q! WE q!=WE q! WE q!0WE q!.WE q!0WE q!5WE q! WE" WE( = 0.05 WEPWEWE q!m( mWEWEWE q!3 ( 3WEWE q!sWE q!eWE q!c (secDSET.Hl///6*DSET.Hh///6*DSETT\dT` 36 7 Xv1ZNDSETTHP@L  36 7 Dv2ZNDSETT4<,8*  3  5 01.5 cmZNDSETT  ($*  3 5 5 2.5 cmZNDSET.H  6*DSET .H BB ?Z8GZ8 4 2-?ZZ8#GZZ85$$/ 0 22P/1w 26*DSETT! ! 00333H 5 7 5 7 5 4 ""p = po = atmospheric pg = 0 v = 0ZNDSETT  &33$ 5 7 5  pg = ? v = 0ZNDSETT   3 5 5 12 mZNDSET.H 6*DSETh.H0x8!:W1H3j~2Y|{2]29F*26*t@$H$H$H@HHH@uuu@DSETT K""3 3< 5 7 5  7 5 v1 = 20 m/s p1 = ?ZNDSETT l""33< 5 7 5  7  5 v2 = 0 p2 = atmosphericZNDSETT#  3  5 1.0 mZNDSETT* h 3  5 2.5 mZNDSET.H 6*DSET .H  9@M  2 8CZ 8QZ 8Z : 2 2SS`x  26*DSET:.H| 8  /  / ! 6*DSET.Hpx  Z 8 ZZ 86*DSET.Hpt  Z 8 ZZ 86*DSETThp`l d 3  5 dhZNDSETTT\LX ` 3  5 PAZNDSETT@H8D \ 3  5 <vZNDSETT,4$0% X 3  5 (v = 0ZNDSET.H!#mE >,6*> &#xprWEwdxprWEWE pPalatinoWE qvWE q WE q=WE q WE2"># ##WE, Palatino .*v = WE q2WE q WE qgWE q WE qh) 2 g hWEDSET.H63'($6* #xprWEwdxprWEWE pPalatinoWE qvWE qoWE qlWE quWE qmWE qeWE q WE qfWE qlWE qoWE qwWE q WE qrWE qaWE qtWE qeWE q WE q=WE q WE qAWE q WE q WE2"# ##WE, Palatino .*volume flow rate = A WE q2WE q WE qgWE q WE qh)2 g hWEDSET.H  @@@! 6*DSETb.H! F@%@ 8k@%@ 8i1@v@" 2a@@# t@@$ 2Fa% & .E.\ 8?E?\ 8` \ 8` \ 8.> \ 8>C \ 8%C% \ 9A]\ 9GU' 2$2#( 26*DSETT "! l T 3  5  d = 2.50 cmZNDSET.H#$!+  80{ : 86*DSETT$!$ P 3  5 d = ?ZNDSET:.H%! 8 8 8!h 8h 86*DSET:.H&! 8 8 8!h 8h 86*DSETT'!#L3 4 5 cmZNDSETT(!#H3 4 4 10 cmZNDSETh.H)* H% 8n$ 8k1x. 2Hl/ Xe0 26*DSET .H*q)GG 8 8k 8kk  8k!i 8i 8.F.\ 8?E?\ 8.>\ 8%D% \ 9(D \ 9F  + %3# - 26*DSET .H+BF* 1E1\ 8BEB\ 8#1(A\ 8,:#, 26*DSETT,+#D3 4 5 cmZNDSETT-*#@ 3 4 4 10 cmZNDSETT .) xl  3  5 | d = 2.50 cmZNDSET.H/t$)$  8e i 86*DSETT0)hp`l$  3  5 dd = ?ZNDSET.H1\PO2 6*DSET.H2XM1<Z 88Z 83Z 8''3Z 8M3 2O4 2&ZZ 8EZZ 8ZZ 8 ZZ 8 ZZ 8 $ 8!   ZZ 9>8 ZZ 8 75 26*DSETT 32 LTDPl  3  5 H d = 2.00 cmZNDSETT 42 8@0<G  3  5 4 d = 5.00 cmZNDSETT524<,8,  3  5 010.0 mZNDSET.H6,(P7 X)  8 ;6*DSET .H7$6̰̰̰ ̰ ̰ 0̰0̰@̰@̰P ̰6*DSET082  <H$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$9h6 2CTAB  COLMCHNK=a=b=c=d=e=f=gl\h&  P P ! P `P 6.0 m00 COLMCHNK =c-s area =cm*cm>>?>?>?fffffh>>> COLMCHNK I%q=c-s area=m*m? ' ?ѷX? ' ?j? ' ?IQ? ' ?a? ' ?IQ? ' ?IQ? ' ?nO? ' ?nO;? ' ?IQ ? ' ?IQ ? ' ?au? ' ?IQ ? ' ?jC? ' ?ѷXe,_`p  COLMCHNK = press head=m>@>@>@>>>> 9"xprWEwdxpr"WEWE" 9WEWEp"New Century SchlbkWE q"rWE q"aWE q"dWE q"iWE q"aWE q"nWE q"s,"New Century Schlbk" .99* radiansWEWE q" COLMCHNK tIyE@ vol flow rate @m^3/s1A G @ ?4m1A G @ ?3rS1A G @ ?'RT`1A G @ ?` 1A G @ ?xd1A G @ ?¡1A G @ ?V$2A G @ ?V$ 2A G @ ?¡H2A G @ ?xd?62A G @ ?` l2A G @ ?'RT`2A G @ ?3rSb2A G @ ?4m:f LL>PC  L >pL >p@ C W?>pՖ  \    "YY  "xprWEwdh DSETT.H9ner86*PFNTMCUTSDSUMHDNISTYL;@STYLr\< .       4  \                  ]              !  "  #  $  %  &  B '  !(  3")  #*  $+  %,  &-  '.  (/ )0 *1 +2 N,3 -43.53/63 073<18 29  3:  4;  ,5< 6=<7><8?<9@<:A<r;HASH( %( ) & ( $)*+& '8 ( $( ( () * * *'V V (  (  )  *  ( #&  %4%x3v+%(& ( -(6( 9 ( . (/ * 0 (5%v1&*')'v2& ,&u3(7 (8 (:A( A( A) A* "A ( !C& L:Qhjazo}Bod CHAR   "         @ @  /S [ /C @   @C @   HASH      //      CELLl3:HASH!!!! 6  GRPH,0nlkkj~X HASH ommm l l RULRD@@}l@.@.D@.}$@.@.@.@l@.@.D@.$@.@.@.pHASH   - @ @BBLKUP  !"#$%&'()*+,-./0123456789: $NAMEDefault Default SSHeaderBodyFooterFootnoteFootnote IndexdDFNTM HelveticaGenevaPalatinoSymbol! Avant GardeETBL@FNTM8CUTS@DSUMHHDNIhSTYLrETBLT