# set3-answers - 7 2.64 A thin 4-ft-wide right-angle gate...

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Unformatted text preview: 7 2.64 A thin 4-ft-wide, right-angle gate with negligible mass is free to pivot about a frictionless hinge at point 0, as shown in Fig. P2.64. The horizontal portion of the gate covers a 1—ft— diameter drain pipe which contains air at atmospheric pressure. Determine the minimum water depth, 12, at which the gate will pivot to allow water to flow into the pipe. Width = 4 ft\ Right-angle gate Hinge W FSGURE £92.64 , For agar/i brmm : 204000 5;» ﬁle 44%: «9n 7"}?! her/ganba/ pore/an 41L Wjake {Uh/6’! bit/QI’IKEJ by preﬁgre Q”, 1301‘)“ “4&5 CXCEP'é £0? 711: ﬁre». of We pipe) »- ; ~ F Mm/m m )m ”Thus From [:7 (I) huh I. , 74m: 1 3H 0 outwe- MW/M) h /5’£’/=Z .....______..__. g 27g CQ’CM’H-e icgrces (MC HWH’“ H———____ (9/! Bulge VQr'hch “Pd/:53 WGCﬁH’ 0‘! Wﬁ+€f i’l GAA «Love Lu'ojﬁ W“: rm V 3 3m % (YT—(3m!) 1H 1'12, F: 929;:1».’z3.9m2-1H = 832”? Horf2¢n+al {WC-C : ﬁnite or: {fa/“hos! froju'hm (97C curuea/ Surface. Fl»! : Kﬂco‘ L'c' A :7 (fl-(f R3 ‘W (Co/‘5 (14‘; ( C(n‘lrofd 3 ﬁt) it»; = 3370 25! 2.89 When a hydro-meter (see Fig. P289 and Viden V2.6) liavv ing a stem diameter of 0,30 in. is placed in water; the stem pro~ ﬂuid trudes 3.15 in. above the water surface. If the water is replaced Hydrometer—r surface with a liquid having a speciﬁc giavity of 1 10 how much of the stem would piotiude above the liquid smface? The hydiometei weighs 0.042 11) [FIGURE 92,39 When +6: éhgdmmeéey IS Hoahhg 17‘s Wezghi ‘ZLJ’ 13 1941414164 by 4712 [914094114 érce F73 Fay €3mhbnum , ~ 2— 5m, 3 7741(5 74;, m4” ( F =7“) _ , {3V )Jv‘= ‘10. (I) _ ,_ ,, ”10 when +4; 1‘s ﬁve Jul/verge»! volume Wm: “hie new lL aid 3 (56) (him) @u a) (ambmmj 555 0) 4174(2) M754 7:) (maxim 25 (925,0)7L (55)”?! o>'VL2.' ﬁnd 1 Jv‘, . g 5217 (Con/f) 1-4? (ﬁfanéf) F;vnw,£ig.(7) M :2 .2? 0.0411, = (a-73X/o: If” I 374—27; ‘62,"Il-ﬁ3’ I 5., M ﬁrm #1 0) 3 a .5:= 473x10 (fact '2;- QIZXM “1C1; Ila 3 +3 Vz=—(a73-417)x19“3ﬁ = 051mg: 7; obi-ma 777:3 duffel/mac 7’hc chant)! m hum A1 (‘17-)(0. 30m) Aﬂ= (0.1.! x» “##ang j) A]: PH? in [391751 T”: new lcﬁmdl 'hm sfem WWI/d ProiLrude 3/5“). + 1.44 In 2 3.494171 «km 7719 Jaw/ac ce Thus, ——._.———~_-_———_—— 1-‘10 *2.91 An inverted hollow cylinder is puShed into the water as is shown in Fig. P291. Determine the distance, (3, that the water rises in the cylinder as a function of the depth, d, of the lower edge of the cylinder. Plot the results for 0 s: a! s H, when H is equal to 1 m. Assume the temperature of the air within the cylinw der remains constant. IFIGURE P2.91 F?” canszéani- {cm/Janeéure Camp/(\$515}: wﬁm}, +3»: 9/1,}de . , 2." I e v: 1:2 e i z » (4/7!” '17“ I3 7'7): on? Volume, and 4' 4,;[74 reﬁne «A: ﬂue. ”72.7947 4nd 7434/ 5711.795, Vt’S/Jec/vile/y . I!“ rec/law: 7714}— (Sec ﬁg'we) 72.: 79% ' ﬁ:a’/cl-1)+ﬁl_m ﬂ: ”3—92” +£=%~D2(H~£) 771145) 1/3”,” 11350) , TI" 1. 6'17", [gﬁizﬂ)= (Y{d‘l) +4iyLm) Iii-«D (”w-’0.) (2) and udrn . . _#M Til-”:M’Apé} <5”- 73’0/7’1—gjam Hzlm £72.52) ﬁrm/\$1.133.) 117: (62‘ (d+ll.3)),€ ‘1‘ Oi (1m) :30 5o ’f'hwi' (5151.329 ﬁve guadmé/cl igrmu/a) ﬂ (UL-r113!) 1." 1 dz+l8.lplal+l113 _ __ _ __2 Shake £07 4:0) i=0; 177d ”ejAIHt/e .519." Should be. Used drzaL _ (Emmi) — WEI—2+1“: d-HZé? " ____________.,__.Z_______,_____.___ Tab/ale}! déﬂéﬂ. W/F’h‘ ’ﬂre @rfggmﬂcj/gj plol‘ org Jhaum an dive ﬂo/Iow/hj Paqe. (Can 2:) 1—?3 Depth. d (m) Water rise, I, (m) 0.000 0.000 0.100 0.007 0.200 0.016 0.300 0.024 0.400 0.033 0.500 0.041 0.600 0.049 0.700 0.057 0.800 0.065 0.900 0.073 Water rise} (m) 0.200 0.400 0.600 0.800 Depth, d (m) 1.000 1~74 237 ’ The U—tube of Fig. P14?" is partially ﬁlled with water and rotates around the axis a— , a. Determine the angular velocity that will cause the water to start to vaporize at the bottom of the tube (point A). 3155”“ I}, at rofaéihj Hal‘s! var/e: m ac‘corahnre unﬁt \$50 The .2 sun-£1014, FIGURE 92m = ﬂw‘r“ JP 1 W177: 7712 word/Mk 575%”) ahownj =0 q'é 1": 641.11. 0144/ 2:” [207.150 7714‘}: I :L 2 '5‘. z Consﬁmé: “Fww ("7%) + X git) 5' .. 6—2)“!- X - 3‘2 1- Consfﬂnzé (53.9.33) «p: %2/P1-%)g“ b’fi‘u-I) A)? pmhi‘ A) f‘=0 and 2-30, and a)‘ ' 7534.- 407; + X (1) 17C F4 = vapor pressure 2 0. 251. psda, W’ 34 .- (0.254 Ps:.'- Impsg )(Iwgi): -2030 ii (me) “ﬁle" from E3111» zwloa ...
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