University Physics I
University Physics I PHYS 1321
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This 6 page Class Notes was uploaded by Kayley Corwin on Saturday September 19, 2015. The Class Notes belongs to PHYS 1321 at University of Houston taught by Shuheng Pan in Fall. Since its upload, it has received 33 views. For similar materials see /class/208340/phys-1321-university-of-houston in Physics 2 at University of Houston.
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Date Created: 09/19/15
HOMEWDRK Chapter 1 Solution Dmh 39 ulwamr m eureka M an the lawn Snllltinn 1 acre a 43560 M a 43560 13 Smce z m 16 the Vu ume utwatermaueu dunnng slurm 5 V26 km216 0426 km23281ftkm216 a 4 66x107 3 ThUSV V711X103acre 43560 x10 a acre a n M Mm p m m m H m m m n 5110 Smuuls m laWHhEs and lb 29mg ngure 174 Snllltinn Frum Fig1Awe seelhallll s is Eqmva emm 253 w and 2127 32 130 s is Equiva enllu use so 156 z The W39Drmalmn audws us an advert s m w drz a In arms at W we have 50055005 258W 608W 212 5 MIN units umwe have 50055005 4332 180 s L n s M M duck a read Assume negauue readmgs 39Dr prEZEru UmESJ m m l d m Qua zuu 39 um 92H 1 u x r m ngure 1 s Snllltinn and ysmlercepts n Frum the dam m the gure we deduce 33 662 I 7 These are used m ubtaining the dudwing resu ls 1a We find tgrt tgrtA4955 1b We obtain 495141 101me re 15 127r515947we get rye 7245 s 1111 Prob increases The day atthe end 0f10 century is 10 ms longer mar me day at me start of due century In 20 centuries whatis due total ofche daily increases in time Solution The last day ufthe 20 centuries is Iungerman hefirst day by 20 century 0001 scentury 002 s 39 r 4 rug 39 since the increase occurs uniformiy the cumulative effect Tis 7 average increase in length of a dayruruber of days 001 s 36525 day 2000 y y 730 urmughlytwu hours 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 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 dpAh dz dt 00238 kgs 143 kgmin pAg 400 X10 4 kgcm3238 cm20250 cms Prob 32 In the United States a doll house has the scale 0fl12 ofa real house that is 3 Ill i each length of the doll house is that of the real house and a miniature house a doll house to mwuhm a doll house has the scale ofl 144 ofa real house Suppose a real house mg 177 has a from length of 20 m a depth of 12 m a hexght of6 o m and a Tn cubxc meters housev gure 17 S ulutinn Therefore hAh A gh A1800m3 3 Each mmensmn rsredhceu bv armor of 112 and we hhu Van 1500 m3 LJ 10 m3 ll zhn M Therefore V I 1300 m3i s 60 L 10 mi 144 P h thr the Sun about 92 9 x 106 m1 A parsec 3015 the dxstance atwhxch a length ofl AU would subtend an angle ofmam second ofarc mg 178 A hghtryear 1y 1 the dlstance that light traveling through a vacuum thh a speed of186 000 mxs would cover m 1 0 year Express the EarthrSun dASLance m a parsecs and b lightryears H Wm rmuh v wnmz u x I K ngure 1 s Snllltinn n m be appruxwmamd as the slrawghl hnersegmenl a mm 1 Au Thus larcan 1 27tradJan 60 arcse 60 arcmm 360 51arcseclarcsec 485x11quot rad TherelurE me parsems 1pc L szosxwiw 5 485x10 Next we re ale AU m hgmryear My Swag a year 5 abuut 316 x 107 s we have My 186000mxs 31 x107 s 5 9 x10 ml is Smce 1 p 2 06 x105 AU WEr nglhE re ammshwp gwes 4 9 x 10quot p R1AU1AU17PES 206x10 AU NEHEnlhal IAU 92 939 10 m1 and My 5 9 x10 m1 lhetwu expressiunsmgemer My 5 5 7 157gtlt10 116x10 1 9X10um1 y y 1AU92 9x106 m192 9x106 ml Ourresuhs can belunher umbmed as give 1pc 3 21y
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