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PHYS 122 Midterm 1 Solutions

# PHYS 122 Midterm 1 Solutions - Why/22 6’0qu5 Pilgrim"...

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Unformatted text preview: Why /22 - 6’0qu5 Pilgrim" 698\$ Wdtwm / 10 October 2007 C2 MULTIPLE CHOICE I. For the pressure—depth relationship for a ﬂuid, what is assumed? a) The pressure increases with depth. 13) A pressure difference on an object depends on the point at which it’s measured. Q The ﬂuid density is constant. (1) The relationship applies only to liquids. 2. Rod 1 has length L and is heated by and amount AT. If rod ‘2 is twice as long as rod 1, made of the same material and heated by the same amount, its change in length will he 6) 2AL 1)) AL C) AL/Z d) There is not enough information to tell. 5. The same amount of heat, Q, is added to two objects of the same mass. If object I experienced a greater temperature change than object 2, then d) There is not enough information to tell. @ SHORT ANSWER Why is it easier to ﬂoat in the Red Sea (very salty) than in the Pacific Ocean (not so salty), which is easier to ﬂoat in than Lake Michigan (not at all salty)? Explain completely. TIM. §MW§S 44A; MsL—hj‘, 50 m (2:4 Jean. 2 to JM\$+ M92 mat LII-ta. Wat/Juﬂm 1n +vu L:a.s+- ' /MC&.'T1/u_ WWW +15,“ M044; aw» std/=1.— MSS 0.40. WA ‘ou's‘xaacét’wl’uw wLM eq/me m mass 04 44M. abaéat m tbac'o—t in {emu—Qua. Mwl‘yg HM. watt/t 30 W M90, Jugs \MdLS-t‘obvc obts'pZaad Asp—GOa-Jt ow O‘czgtot‘, and So yam wLLl L’choM M May {M sum web‘- 1 M. Ouellette, 10 October 9007 Page 1 of 3 ‘H’ “‘6”: 33m). = A19 “6M3 PROBLEMS @ 1. Airplanes ﬂu (in part) due to Bernoulli’s Principle. a) Find the pressure difference on an airplane wing where air ﬂows over the upper surface with a 6/ speed of 115 m/ s and along the bottom surface with a speed of 105 m/ s. ‘5“‘5M’5 Pm-‘a-sw.“ 332/7 = ?2+'39V21+?24_ k?“ . . u;=los'“ls ’ M mm “WWW A/vm/zm/ ‘/ A?‘ P113. = iyvﬁ- Jing‘ 1/ ‘= 429052 " “’12) = 'EC4-193E4l532 — (40531] = 447.0 Pa.— 2 ./ b) H the me“ Of the WinB is :5? 1112, what is the net upward force exerted on the wing? 4 / P=E % F: 92: ,_i£ ~+ A 32 m1 Ea- = 44— N 2. 4.9910414 2 a) Estimate the number of air molecules in a cubic room 10 m on a side. (Hint: Make assumptions 4 about the pressure and temperature in the room.) / V" MOM/~33 = 403m pV=Nk3T .l/ F=\a.+w\= Lorlo‘Pa. ‘/ Mg \LaT T=ZO°C= 293K ‘/ : (4.04'109PaX/103M33 N= ? (4.331043%)(L93 k) N -= 1.49 102% pave-h}ch '/ b) What is the average speed of an air molecule in this room? The mass of a nitrogen molecule is ‘r 468x10951<3 / . . Jimms = '2? “ST 2/ Wm; = 7 .. ' ’15 3L T m ' 4.68 [0 kg» 0:“? —v—a—4 = IBLA-ag'uo'23JAX19BK) 4.038'10'wk9 D'Mf- 508 "‘"Is . 1/ M Ouellette, 10 October 2007 Page 2 of 5 PROBLEMS (cont) 5. Students on a spring break trip bring 5.5 kg of ice at 00°C in a Sty rotoam cooler which has walls 42 cm thick. The thermal conductivity of Styrofoam is 0.050 W/(m's'K). The surface area of the cooler is 1.5 m9 and it rests in the shade where the air temperature is 91°C. a) Does heat ﬂow into or out of the cooler? 7 / 'H'CAA' {laws “kl-o 4484. cools/v- firm «We WM a5»; oudﬁ’faLa. b) What is the rate of this heat ﬂow? 5 “4,;55‘4? P: 1%? = kZAT = (aoaoX/ISX’Ll—oo) 1": 0.006 2" A = 0.04Zm e9. - k = 0.030 W/(wu) m: ’ 27-5 W; A = A-QMZ Ta." 2‘06 99 = Q At c) How much heat is required to melt all of the ice? 4 - SJ / —,—. 9 :4- : 56k- 333 to Q) , (p ML—z ( 3X u = 3.3905 J/ co = 4.85‘lObJ + "'6’ 3 cl) How many hours does it take for the ice to melt? g / ’t =? b A6? A0 4.83m J ——~ = 226W — = wa— = “’4 Lifﬂ 9 A" new 225w 2. At = g.\3‘\o4s /Z. = 21.6 \mrs M Ouellet te, 10 October 2007 Page 5 of 5 Useful Information g = 9.8 m/SZ pwarm = 1.00x103 kg/m3 cice = 2100 J/(kg-K) kB = 1.3 8x10'Z3 J/K pseawate, = 1.03x103 kg/m3 cm, = 4186 J/(kg-K) NA = 6.022xlo23 mol'l pica = 0.92x103 kg/m3 Lﬁeemate, = 3.33xlo5 J/kg R = 8.31 J/(mol-K) pair = 1.29 kg/m3 Lboﬂ, water = 22.6x105 J/kg pa =1.01x105 Pa Formulae m E 2 4 3 Fg = mg p = 7 P = 7 Acircle = m Vsphere = gm Fluids F . p=—Ay p=pa+pgh Fb=mfg=pngf wa=w-Fb 1 2 1 2 plAlv1 = p2A2v2 = constant p1 + Epv1 + pgyl = p2 +§pv2 + pgy2 = constant Thermodynamics 9 TF =—5—TC+32 TK =TC+273.15 m pV=NkBT=nRT N=nNA n:— M 1 3 3 K =—mv2 =—k T =—NkT < > ﬁns 2 B 2 B AL = aLiAT AA = 2004,.AT AVsolid = 3aV,AT AVHquid/gas = ﬁKAT Q = chT Q = :mL A_Q_ ._. ﬂy; _A_Q = UAeT“ At 0' At M nllpllpﬁn 10 nr‘fnl-wr 7007 Damn 1 NF] ...
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PHYS 122 Midterm 1 Solutions - Why/22 6’0qu5 Pilgrim"...

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