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# Note for PHYS 1321 with Professor Pan at UH

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This 6 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Houston taught by a professor in Fall. Since its upload, it has received 54 views.

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Date Created: 02/06/15

HOMEWDRK Chapter 1 Solution Mob 7 Hydrauhc ehgmeers m the United States d1teh use as a umt blvmume alwaner the acretbut detmed as the VD umE ulwamrlhalwm uverl acre b1 ahd td a depth d1 1 01A severethuhderstbrm dumbed 20 m at ram m 30 mm an a lawn at area 26 kmz Whatvmume b1 waber m acre4eet ten ah the tbwn7 Snllltinn 1 acre a 43560 M a 7 3560 n3 Smce z m went the Vu ume dtwaterthatleu dunnglhe stbrm rs V26 km216 0426 km23281ftkm216 ft 4 66x107 ft Thusr 466x107 n3 fawn acre a 43560 x10 ft acre ft Mob 5 Harvard Endge whmh cbhhects MIT wrth rts lralemmes acrbss the charwes Rwer has a Englh b13644 Smuuls p us bhe ear The umt b1 bhe Smuul rs based ah the Englh b1 ohver Reed Smuul In ass d1 1962 whb was named br dragged Englh by Englh acrbss the bndge SD that bther wedge members dtthe lambda chr A pha tratermty Lm d mark an wwlh pawl 15mm Englhs E ng the brrdge The marks have been rebamued brahhuauy by tratermty wedges smcethe mma measurement usuaHy dunngumes d1tratnc EDNgESUDn SD thatthe puhce cahhbt easrw mtertere Presumabw the puhce were ungmaHy upset because the Swan rs not an 5 base umt butthese days they seem td have accepted the umu rrgureg shbws three barauex baths measured m Smuuls is WHMES W and Zemas 1 What rs the Englh at 500 Smuuls m laWHhEs ahd b emas rrgure 174 Snllltinn rrdm rrg14we seelhallll 5 rs Eqmva emm 253 w and 2127 32 130 5 rs Equwa enllu 216 7 so 156 z The mlurmalmn audws us m dpuen s m w drz a In arms at w we have 50055005 258w 608w 2125 Mlnumlsullrwehave 50055005 433z 1805 Mob 13 Three dwgna ducks A a and 5 run at drwerem rapes and dd nut have srmuuanedus readmgs dnerp ngure shuws srmukanedus readmgs pp pairs UflhEE DEkS39ar39Dur DEEaSmnS LAupe EarhesluccasanurExamp E sreads 250 s andcreads 920 s quotth Events are sum s apart pp duck A haw lar apart are they up a duck a and p duck 7 c When E DEkA readsADD swhaldues duck sread id When duck creads 150 swhaldues unk a read Assume pegauue readmgs 1dr prezerd umes rrgure 1 s Snllltinn Theume up any dmese ducks rs a shawghbhnelunmun ulthal pp andmerwrm S DpES 1 and yrmlercepts n rrdm the dam m the gure we deduce These are used m ubtammg the dudwmg resuus 13 We find tgrt tgrtA4955 1b We obtain r575 495i4is 1c Cluck B reads rs 133401400e 15525 s 198 swhen cluckA reads xi 400 s 101me re 15 127r515947we get rss 7245 s Prob 18 Because Earth39s rotation is gradually slowing die length of each day increases The day atthe end 0f10 century is 10 ms longer um die day at me start of die century In 20 centuries whatis die total omie daily increases in time Solution The Iasr day ufthe 20 centuries is Iungerman hefirst day by 20 century 0001 seeumry 002 s The average day during the 20 centuries is 10 0022 001 s lungerthan me first day Since the increase uccurs unifurrmy the cumulative effect Tis T average increase in length of a dayuumber of days 001 s M 2000 y day y 7305 s urruughlytwu huurs Prob 25 During heavy rain a section of a mountainside measuring 25 km horizontally 080 km up along the slope and 20 m deep slips into a valley in a mud slide Assume that the mud ends up uniformly distributed over a surface area of the valley measuring 040 km x 040 km and that mud has a density of 1900 kgm3 What is the mass of the mud sitting above a 40 m2 area of the valley floor Solution The volume of the section is 2500 m800 m20 m 40 x 106 m3 Letting quotdquot stand for the thickness of the mud after it has uniformly distributed in the valley then its volume there would be 400 m400 md Requiring these two volumes to be equal we can solve for d Thus d 25 m The volume of a small part of the mud over a patch of area of 40 m2 is 40d 100 m3 Since each cubic meter corresponds to a mass of 1900 kg stated in the problem then the mass ofthat small part ofthe mud is 19gtlt105 kg Prob 31 A vertical container with base area measuring 140 cm by 170 cm is being filled with identical pieces of candy each with a volume of 500 mm3 and a mass of 00200 g Assume that the volume of the empty spaces between the candies is negligible If the height of the candies in the container increases at the rate of 0250 cms at what rate kilograms per minute does the mass of the candies in the container increase Solution The mass density of the candy is p M 400 gtlt10394 gmm3 400 gtlt10394 kgcm3 V 500 mm3 If we neglect the volume of the empty spaces between the candies then the total mass of the candies in the container when filled to height h is M pAh where A 140 cm170 cm 238 cm2 is the base area of the container that remains unchanged Thus the rate of mass change is given by duijh pAg 400 gtlt1074 kgcm3238 cm20250 cms 00238 kgs 143 kgmin Prob 32 In the United States a doll house has the scale 0fl12 ofa real house that is each length of the doll house is Iquot n that of the real house and a miniature house a doll house to frtwrthrh a doll house has the scale ofl 144 ofa real house Suppose a real house Flg 177 has a from length of 20 m a depth of 12 m a herght of6 o m and a standard sloped roofver1lcal Lrlangular faces on the ends ofhelght 3 o rh 1h cublc meters what are the Volumes othe correspondmg a doll house and b rhrruature house7 Flgure 177 S ulutinn Thetotal volume vs the real house rsthatoratrrahgular prlsm or herghth 30m and base areaA 20 x 12 240 mu m addrtroh to a rectangular box herght h 50 m and same haset Thererore V hAh A 6 L39A1800m3 3 Each dlmensmn rsreduceu bv aractoror112ahd we hhu Va lsoo ru 10 ml by In thrscase each drmehsroh relatrvetothe real house ls reduced bv aractor or 1144 Thererore 3 V 1soo m3i s 60 104 ur 144 Prob 53 Ah astronomleal uhlt AU ls equalto the average dlstanee from Earth to the Sun about 92 9 x lo6 ml A parsee pe ls the drstahee at whlch a length ofl AU would subtehd an angle oferactly 1 second ofarc Flg 178 A llghtryear 1y ls the drstahee that lxght pavelmg through a vacuum thh a speed of186 000 mxs would cover m 1 0 year Express the EarthrSun dutance m a parsecs and b lxghtryears n mg rm mm I wnmd 1 1 pm pc ngure 1 s Snllltinn The ppjemye mp1s prub em 1s m Lmverllhe sarmmp mstapcem parsecs and thtryears 1p re ale parse1pm Au we hate thalth 51s measured m ramaps 111s equa1 m the arc Englh s amped by the radms 17 Fur a yery1arge radms awe and smaH yame at a the arc may be apprpxirpaoed as the straw hnersegmenl p1 Englh 1 Au Thus 91 arcsec1arcsec w 17 7 60 arcsec 60 arcmm TherelurE ppe parsems rad 5 1AU 5 1 206 10 AU 9 R a 485x111quot X Next we re ale AU m hgmryear 11y Smce a year 15 abuut 316 x 1117 s we have 11y 186000m1s 31s x1075 9 x10 ml 1a151pce1pe 2 06 x105 AU1nErnnglhe rexauppspp gwes R1AU1AU L s 206x10 AU 49x10 pe memepmat 1AU 92 939 10 m1 and My 5 9 x10 m1 lhetwu Expressmnslugelher mm a 57x10 51y16x10 51y 1AU929x10 m1929x10 ml 59x10 ml surresuns can belunher pmpmed as give 1pc 3 21y

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