hw2_sol

hw2_sol - 2. 4- Blood pressure is usually given as a ratio...

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Unformatted text preview: 2. 4- Blood pressure is usually given as a ratio of the maximum pressure (systolic pressure) to the minimum pressure (diastolic pressure). As shown in Video V2.2 such pressures are commonly measured with a mercury mano- meter. A typical value for this ratio for a human would be 120/70, where the pressures are in mm Hg. (:1) What would these pressures be in pnscals? (b) If your ear tire was inflated to 120 mm Hg. would it be sufficient for normal driving? 76: M (a) For Mom/fit 73: {/33X/0%)(01/10m)=/é.94ég FEW 70 mum ,' 73: .033X/03fl;)(0l 070m): ij/afifi (In) For /20 M4" Hg _' F: flélp Kip-5%.)fl4fDX/DJ 15/191?) I‘m N/mz- —‘ 2. :72 705:; 5/553 a. fgp/ka/ 4-)}: Pressure 1.; fray-7055') /20mm1% 13 2751‘ sufifiéic'wt 76/" mar/m/ JHL/m'j. 2:8 Sometimes when riding an elevator or driving up or down a hilly. road a person s ears “pop” as the pressure difference between the msrde and out51de of the ear is equalized. Determine the pressure difference (in psi) associated with this phenomenon if it occurs during a 150 ft elevation change. 2.14 (See Fluids in the News article titled “Giraffe’s blood pres- sure,“ Section 2.3.1.) (3) Determine the change in hydrostatic pres- sure in a giraffe’s head as it lowers its head from eating leaves 6 m above the ground to getting a drink of water at ground level as shown in Fig. P114. Assume the specific gravity of blood is SO = 1. (b) Compare the pressure change calculated in part (a) to the normal 120 mm of mercury pressure in a human’s heart. m: Ear bydmsi—wha prawn: Change] _ .. 42M _ 1‘1" = 81912.8. Alp—3’19. _ Mainly”)- 53.8w 3 (-19) 7; Cam/art. «2.31 PreJJHR In 1114111111 heart Con Vert— presxurc Ir} P4145 (4) in Mum H5 ,' 151v - = it! 52.3 $1 _ 313%”; (33 m, 4.1+? 4L = (0.442%)(103M)= WZ mm H5 #3, M ' Thus! ’flle. prenun (slung: Hi Tlu giraffe} he“! In I: (1‘92 MM” “2 Camper-ell lat-T" 1219M“. “:5 +7”. Klimt“: heart 2.21) On a given day, a barometer at the base of the Washington Monument reads 29.97 in. of mercury. What would the barometer reading be when you carry it up to the observation deck 500 ft above the base of the monument? Le?! ( )b and ( )od Correspond 2‘0 #18 base and observation deck) respectively. 771W, with H =hei9/n‘ 01F the-pizflvmem‘, lb p05 ‘flad = 62.3.14 = fax/0' H3 (5004“): 38,5 fig Btfi f = (5;! h , w/aere 32.7 = 847% and It =baramefer read/177, T/m J 29.97 _ - _, lg Xi’éhod -38.5fl2 {fig/52H) — 0.01/55 HRH—’15.?) : —O.5W) in. 2.2.7 A mercury manometer is connected to a large reservoir of water as shown in Fig. P227. Determine the ratio. Paw/hm, of the distances 11,, and hm indicated in the figure. “FIGURE P2.27 rh— =2(l3.56)-'I = 25,: m __ . .2 . 2? A closed cylindrical tank filled with water has a hem- : ispherical dome and is connected to an inverted piping system 5 as shown in Fig. P217. The liquid in the top part of the piping I system has a specific gravity of 0.8, and the remaining parts of ' the system are filled with water. If the pressure gage reading at ; A is 60 kPa, determine: (a) the pressure in pipe B, and (b) the ; pressure head, in millimeters of mercury. at the top of the dome ‘ (point C). §§§ itine- {d) a + (Sé)fa;za)[3rm) + £23 (2m) =‘PB 75 3 6,043 + (asMinnffiJfiMM(Moxloafigfim) = /03 AB. a”) 5/:= 75A— 63,20 (5m) _- LD-kPa‘(9.80X103£3>[3/m) 3 = 529.4 xm 1’— 4?? 1. 3 N (g 45c h 30.4. xn we ("I = 3 N 133x ’0 7.1—3 = 0.230m1 (“34””) = Z30m~m 2. 5 3 A 6-in.—~diameter piston is located within a cylinder which is connected to a é—inndiameter inclined—tube manometer as shown in Fig. P2.-53.Theflui_d in the cylinder and the ma- nometer is oil (specific weight =“ 59 lb/ft3 ). Whgn a weight W is placed on the top of the cylinder the fluid level in the ma— I nometer tube rises from point (1) to (2). How heavy is the weight? Assume that the change in position of the piston is negligible. I FIGURE P'2.53 Wr'f‘h phi-on fi/one /ei‘ plasma. an 14(2 075/315-3301 = 1P? J and mqflomeifr eguaém become: 7570 ._ x” 7% Sin 30°=o (’2 {rt/’7}, cue/75¢ added frag/re 7% Menage, 75 a) when 4%,: f + H area. 91: F AF am»! manomeifir £514.59»: [grammes 73°! ‘ Jon MN“ 72%) 513' 35% (2’ SuHmzi 5311) 75mm £3.62) 19, (Jim): 76'”; ‘ 32-; ‘HJs/no‘=o 73 a 75- 0? M : 32L] $131 30° A? 50 77rd- ..W—. = I G: 1 ‘I- 75153) and W = 2.90/1: an: .tm-‘r. A 2.55 2.55 A vertical rectangular gate is 8 ft wide and 10 ft long and weighs 6000 lb. The gate slides in vertical slottin the side of a reservoir contain— ing water. The coeficient of friction between the slots and the gate is 0.03. Determine the mini- mum vertical force required to lift the gate when the water level is 4 ft above the top edge of the gate. FR: X 4RD A = (Laviflsafltifituoh) = M1 400 lb Z thfiapfl‘l'ni :0 F Thus’ F at; 62 l1, N = = 00 R ’ L (0.03~)=F+ ZFveri-fcgl = 0 FR EN Thus, boob Lb F: :.- (cone [in + (0.03)(34flaolb) F; ~ maximum ifichom! «farce Fm farce +0 “H git-be : 7200 “a 2.58 Free surlace 2.58 A SII‘LICUJI'B is 3113(3th [O the ocean floor as shown in p:- -__.::.-:_-:..::::-:-:-{age-Egg?{gaze{;:;:£-:a:::_::,;55_;_»_;§_._:_A_::. P258. A film—diameter hatch is located in an inclined wall and hinged on one edge. Determine the minimum air pressure. p,, within the container to open the hatch. Neglect the weight of the hatch and friction in the hinge. -:-:- - ---- l FIGURE P2.58 Fkré’gcfl allure ft: [0,", + :éfzm).rl}:3fi' «n+3 = /0.5rm F1; ’79‘ 77111.4J 5'6: (la/1103321; )(lo.€m)(_g){z,m)l RA = 353x105” 73 Ami: FRI in: 7:3: 7‘ 5c Lei/Jere ya: 4:3. + 1,”, : 2!,“ .50 mi 1e: (’Ey/Mfl 1‘ 11,»; = «2/.012/m (21m MM») 1 5!" eguilibr/I'Im/ K 27%;:0 a7: ‘fha-é A; (em/2m — 20 m) = farm/Mr (7m) ff: mm {133x105M){/.012m) 7 (14,071“) =' /0‘7’é/?a 2- 6' A homogeneous. 4-ft-wide. 8-ft-long rectangular gate weighing 800 lb is held in place by a horizontal flexible cable as shown in Fig. P2. 61 Water acts against the gate which is hinged at point A, Friction in the hinge is negligible. Determine the tension in the cable. Eé : XAcA Caller: Ac: 5M 50” 771M, pl; _—_- flz'qgs ){éfil'jflsméo'2fl/i‘x 475i!) ‘0 Aer: ya = 57575 + 3H -‘ 6.15 £11 (3&JK41‘ZXW) T /Sf£)6;héa‘) = a) [4H)(coséo°)+ 52 {2.4) 7—_ 6W I“ (X H)(5/In £0”) ...
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hw2_sol - 2. 4- Blood pressure is usually given as a ratio...

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