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Quiz3_solutions

# Quiz3_solutions - PHY 317K I Quickies Consider a 1 inch ice...

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Unformatted text preview: PHY 317K I) Quickies Consider a 1 inch ice cube ﬂoating in a pitcher of water. Take the density of ice to be pice = 917 kg/m2 and that of water to be p = 1000 kg/m2; the ice cube ﬂoats with its top face level to the water. 1.) How high It abOVe the surface of the water is the top face of the ﬂoating ice cube? @411 = 0.08 in - _ £19: _ _ am a «Emma, l p “I T—ﬂﬂ/ 00” “ii/MP B.)h=0.12in C.)h=0.181n (See. PraMem Set 9, Potlem (6*?) D.)h=0.251n A hose of 2 cm inside diameter feeds water at a speed of 0.5 In / s to a. drip irrigations system composed of 25 drip tubes, each with an inside diameter of 2 mm. Assume the water ﬂows with the same speed in all drip tubes. 2.) What is the speed 1; of the water ﬂowing in the drip tubes? A.) v = 0.4 m/s Alin 2 3‘5 AMI} B.) 1) = 1.0 m/s \fa: P‘x‘vl :I-‘(angS "VS as @ 'U = 2 m/s 951% 3g}? (_, 091 M)1 D.) e = 4.0 m/s A simple pendulum censisting of a. small mass suspended from a nearly massless string of length l oscillates with period T = 2.01 s. A second pendulum consists of a uniform rod of the same length l which pivots from one end. Recall that the moment of intertia of a. rod rotating about one end is I = mlE/B. —~ { ﬂ 1 i =1 MT R 5 1 i: (W 3 3.) What is the period of oscillation Trad of the rod pendulum? de=1.64s .. mp: .. W :ﬂddi M969 3% B.) Trad : 2.01 s C.) de = 2.46s D.) Cannot be determined from information giVEn (Problem continues on next: page.) PHY 317K Quiz 3 3 (Problem continues from previous page.) In the next two questions, consider a loudspeaker being driven at a ﬁxed frequency f = 500 Hz. The maximum displacement of the loudspeaker’s diaphragm is d z 1 pun = 10—6 m. 4.) What is the maximum acceleration amax of the diaphragm in units of g? A.) aim/g : 0.1 KG): Xm (ﬁat/i w) B.) emu/g = 0.5 @amax/g = 1.0 08) 1% 9 bxmwaws 0’” “ill I D.) amax/g : 2'0 ﬁlmy: Kuwa‘ :0 xmveyn)[( m‘XSOQ JED}: \$86 {Vi/5“ 5.) What is the wavelength A of sound waves generated by the loudspeaker? A.)/\=34em A=69cm C->A=1arcm w L: WM” :: 0.63M deﬁed) D.) A:4.31m We 500 Hz xww PHY 317K Quiz 3 4 II) A Door Consider a door of mass .M = 40 kg, width W = 0.9 In and height H = 2.5 m supported by two hinges as indicated in the figure below: Top View The hinges are mounted a distance D : 0.20 In from the top and bottom of the door. You may assume that the door has a uniform mass density everywhere. ' 6.) What is the net vertical component of force on the door provided by the two hinges? A.) 84N FM”: Ewe ’mj : I B.) 168 N ﬂ ﬂ w» I C.)196N Elmer ’ m3 ’” (be kjimg "Tn/5;) D.) 392 N 7.) What is the horizontal component of force on the door provided by the top hinge? Cal cu [mic Tarp/e. abaui N directed away from the center of the door bmnm "a a B.) 196 N directed toward from the center of the door C.) 196 N directed away from the center of the door {Ch _ : {Cm- m in? Q 1 D.) Cannot be determined from information provided , , l H (Pl‘obiem continues on next page.) x it} : ‘3 ﬁts») :1: M3 i “3" Q“ m” we!) (HEW); _____(;.s.- a (MOM) i PHY 31 7K Quiz 3 5 (Problem continues from previous page.) 8.) What is the horizontal component of force on the door provided by the bottom hinge? 84 N directed toward the center of the door B.) 196 N directed toward from the center of the door C.) 196 N directed away from the center of the door D.) Cannot be determined from information provided ,r— Fret/x: Ewing”? :0 are: — e :- W, oppose dire-(«Hm PHY 317K Quiz 3 6 III) The Earth—Moon System As is well known, the moon orbits the earth in an approximately Circular orbit of radius REM = 3.82 x 108 m in a period T m 28 days. The mass of the mOOn m is about 1/80 that of the earth, while its radius is about 3/11 that of the earth. The velocity for objects to escape the gravitational attraction of the earth is vase = 11.2 km/s. 9.) What is the speed of the moon 11mm,“ in its orbit around the earth? “moon = 1-0 km/S \l .1: _ 9:“- [email protected] X10ng :1 man T EN Jays 3% LNG) 3600 ; vmoon = 2.4 km/s C.) amen = 3.5 km/s I ) D.) 11mm“: 11 km/s 10.) At What fraction fCM Of the earth-moon distance REM along the line between the centers of the earth and the moon is the center-of-mass of the earth-moon system? V‘m'd m {M 6 77 mm fCM 2 0.012 9 \L 13-) fCM :012 yam: (“669) t mmGWl ,X mm) ram :61“) rm 0.) fCM = 0-52 WWW, "‘6 W .- D.) foM = 0.90 i ’0'. a! as Fm» _...s 11.) At what fraction fG of the distance REM along the line between the centers of the earth and the moon does the gravitational attraction of the earth on a spaceship (or any other object) equal that of the moon? A.) fg = 0.012 imdml ii: EM :3 :ﬁé/l/lﬂﬂl B.)fG;0.12 0 X X (M 'X) c.) fa :0.52 Meﬁgﬂ TX)"; mm X1 - fG = 0.90 1 r \4 me {>0 4 fan X (Problem continues 011 next page.) #1 x0 33%,“ X: ‘ MW # Kiwi)" # MC rom "’ PHY 317K Quiz 3 (Problem continues from previous page.) 12.) What is the escape velocity vesm for objects on the moon? A.) vesmzlﬂkm/S 1 E4: ’uesm = 2.4 km/s ‘ I 0 O C.) vesm = 3.5 km/s 1" - fyg D.) vesm = 11 km/s ‘ ﬁmW: Eﬂ R Vega: R Um I V‘ GMM‘] Y/th Re) if; 1/ M3” 1 Kmﬂ Rm Afr/712‘ PHY 317K Quiz 3 8 IV) Drinking Water Through a Straw Consider the physics of drinking water from a glass or bottle containing more than a liter of water using a straw. For this problem, we assume the straw is at least 20 cm long with a uniform cross sectional area of .5 cm2. Let the local atmospheric pressure be represented by p0 and describe the pressure 39 applied by you to drink from the straw in terms the “gauge pressure” pg = p — 390. Note that gauge pressure can either be positive or negative. 13.) What gauge pressure do you have to apply at the top of the straw to blow bubbles out the end of the straw when it is submerged 15 cm below the surface of the water? A.) pg : —1.03 x 103 Pa l): P4 bojd ., B.) pg = —0.98 X 103 Pa [39 «7.. 9. Pa :J’gol : Qaaokyﬂ {ﬁrm/gaxi gm) lemﬁpﬁ C.) pg = +1.03 X 103 Pa pg = +1.47 x 103 Pa 14.) What gauge pressure do you have to apply to draw the water just to the top of the straw when it is elevated 10 cm above the surface? A.) pg = —1.03 x 103 Pa pg : —0.98 X 103 Pa F3: Pﬂd 1-0000Kj/m3)(qI?W/5l9(-'0.l m) im; (1)109 = +1.03 X 103 Pa u D.) pg = +1.47 x103 Pa (Problem continues on next page.) PHY 317K Quiz 3 9 (Problem continues from previous page.) To actually drink water from a straw, it must have a positive velocity 1; when drawn to the top of the straw. In what follows, assume the water ﬂows in the straw with speed 1) t 32 0111/5. 15.) What gauge pressure do you have to apply to draw the water to the top of the straw with speed 32 cm/s when it is elevated 10 cm above the surface? A pg = —1.03 x 103 Pa 13.) pg 2 —0.98 X 103 Pa (1)199 = +1.03 x 103 Pa P5 +5;ng U93 1‘3 '1: PA lifV: +1”? l»; D. = 1.47 103 P a 3 )pg + x 3‘ r3 1» 9341, I ‘3st +P3£hrl43l Pg 3 \$ /,. lam l(j/rv\’)(.3§4r'1/s)A +, 1000 rmeMm/sl) C". l M) :ixx ['03 W 16.) How long does it taeltce to draw 1 liter of wate from the container? "WWW- A.)t=313i 7 7—: 71mg; Vol V B.)t=45s voliler 7 T RI AV @=625 ‘ moo 0mg '5 : .)t=118s «1r: @‘Soma) (gaCM/S> ...
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