Phys 110. Answers to even numbered problems on Midterm Map


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1 Phys Answers o een numbered problems on Miderm Mp. REASONING The word per indices rio, so.35 mm per dy mens.35 mm/d, which is o be epressed s re in f/cenury. These unis differ from he gien unis in boh lengh nd ime dimensions, so boh mus be conered. For lengh, m = 3 mm, nd f =.348 m. For ime, yer = dys, nd cenury = yers. Muliplying he resuling growh re by one cenury gies n esime of he ol lengh of hir longlied dul could grow oer his lifeime. Muliply he gien growh re by he lengh nd ime conersion fcors, mking sure unis cncel properly: mm m Growh re.35 d 3 mm f.348 m d y y cenury 4 f/cenury 4. REASONING. To coner he speed from miles per hour (mi/h) o kilomeers per hour (km/h), we need o coner miles o kilomeers. This conersion is chieed by using he relion.69 km = mi (see he pge fcing he inside of he fron coer of he e). b. To coner he speed from miles per hour (mi/h) o meers per second (m/s), we mus coner miles o meers nd hours o seconds. This is ccomplished by using he conersions mi = 69 m nd h = 36 s.. Muliplying he speed of 34. mi/h by fcor of uniy, (.69 km)/( mi) =, we find he speed of he bicycliss is mi Speed = h mi h.69 km mi km 54.7 h b. Muliplying he speed of 34. mi/h by wo fcors of uniy, (69 m)/( mi) = nd ( h)/(36 s) =, he speed of he bicycliss is mi Speed = h mi h 69m mi h 36s m 5. s. REASONING To coner from gllons o cubic meers, use he equilence U.S. gl = m 3. To find he hickness of he pined lyer, we use he fc h he pin s olume is he sme, wheher in he cn or pined on he wll. The lyer of pin on he wll cn be hough of s ery hin bo wih olume gien by he produc of he surfce re (he bo op ) nd he hickness of he lyer. Therefore, is hickness is he
2 rio of he olume o he pined surfce re: Thickness = Volume/Are. Th is, he lrger he re i s spred oer, he hinner he lyer of pin.. The conersion is.67 U.S. gllons m U.S. gllons m b. The hickness is he olume found in () diided by he re, 3 3 Volume.5 m 4 Thickness.9 m Are 3 m 4. REASONING The Pyhgoren heorem (Equion.7) cn be used o find he mgniude of he resuln ecor, nd rigonomery cn be employed o deermine is direcion.. Arrnging he ecors in ilohed fshion, we cn see h he ecor A gies he resuln weserly direcion nd ecor B gies he resuln souherly direcion. Therefore, he resuln A + B poins souh of wes. b. Arrnging he ecors in ilohed fshion, we cn see h he ecor A gies he resuln weserly direcion nd ecor B gies he resuln norherly direcion. Therefore, he resuln A + ( B) poins norh of wes. Using he Pyhgoren heorem nd rigonomery, we obin he following resuls:. Mgniude of A B 63 unis 63 unis 89 unis n 63 unis 63 unis 45 souh of wes b. Mgniude of A B 63 unis 63 unis 89 unis 63 unis n 45 norh of wes 63 unis
3 6. REASONING The riple jump consiss of double jump in one direcion, followed by perpendiculr single jump, which we cn represen wih displcemen ecors J nd K (see he drwing). These wo perpendiculr ecors form righ ringle wih heir resuln D = J + K, which is he displcemen of he colored checker. In order o find he mgniude D of he displcemen, we firs need o find he mgniudes J nd K of he double jump nd he single jump. As he hree sides of righ ringle, J, K, nd D (he hypoenuse) re reled o one noher by he Pyhgoren heorem (Equion.7) The double jump moes he colored checker srighline disnce equl o he lengh of four squre s digonls d, nd he single jump moes lengh equl o wo squre s digonls. Therefore, J 4 d nd K d () Le he lengh of squre s side be s. Any wo djcen sides of squre form righ ringle wih he squre s digonl (see he drwing). The Pyhgoren heorem gies he digonl lengh d in erms of he side lengh s: d s s s s () Firs, we pply he Pyhgoren heorem o he righ ringle formed by he hree displcemen ecors, using Equions () for J nd K: D J K 4d d 6d 4d d d (3) Subsiuing Equion () ino Equion (3) gies D d s s 4 4. cm 4 5 cm D K J d s s 36. REASONING The ringle in he drwing is righ ringle. We know one of is ngles is 3., nd he lengh of he hypoenuse is 8.6 m. Therefore, he sine nd cosine funcions cn be used o find he mgniudes of A nd A y. The direcions of hese ecors cn be found by emining he digrm. A y # 3. N W S 8.6 m E A # 9 A Ne
4 . The mgniude A of he displcemen ecor A is reled o he lengh of he hypoenuse nd he 3. ngle by he sine funcion (Equion.). The drwing shows h he direcion of A is due es. A A sin 3. m sin 3. m, due es b. In similr mnner, he mgniude A y of A y cn be found by using he cosine funcion (Equion.). Is direcion is due souh. A A cos 3. m cos 3. m, due souh y 5. REASONING According o he componen Norh mehod for ecor ddiion, he componen of he resuln ecor is he sum of he componen of A B nd he componen of B. Similrly, he y componen of he resuln ecor is he sum of he R y componen of A nd he y componen of B. The A mgniude R of he resuln cn be obined from he nd y componens of he resuln by using 3. he Pyhgoren heorem. The direcionl ngle cn be obined using rigonomery. We find he following resuls: R R y 44 km cos km 36 km 44 km sin 3. km km y R R R A A y B y B 36 km km 7 km R y km n n 74 R 36 km
5 Chper 4. REASONING The erge ccelerion is defined by Equion.4 he chnge in elociy diided by he elpsed ime. The chnge in elociy is equl o he finl elociy minus he iniil elociy. Therefore, he chnge in elociy, nd hence he ccelerion, is posiie if he finl elociy is greer hn he iniil elociy. The ccelerion is negie if he finl elociy is less hn he iniil elociy. () The finl elociy is greer hn he iniil elociy, so he ccelerion will be posiie. (b) The finl elociy is less hn he iniil elociy, so he ccelerion will be negie. (c) The finl elociy is greer hn he iniil elociy ( 3. m/s is greer hn 6. m/s), so he ccelerion will be posiie. (d) The finl elociy is less hn he iniil elociy, so he ccelerion will be negie. Equion.4 gies he erge ccelerion s s Therefore, he erge ccelerions for he four cses re: () = (5. m/s. m/s)/(. s) = (b) = (. m/s 5. m/s)/(. s) = (c) = [ 3. m/s ( 6. m/s)]/(. s) = +.5 m/s.5 m/s +.5 m/s (d) = ( 4. m/s 4. m/s)/(. s) = 4. m/s 6. REASONING Alhough he plne follows cured, wodimensionl ph hrough spce, his cuses no difficuly here becuse he iniil nd finl elociies for his period re in opposie direcions. Thus, he problem is effeciely problem in one dimension only. Equion.4 reles he chnge in he plne s elociy o is erge ccelerion nd he elpsed ime =.6 yers. I will be conenien o coner he elpsed ime o seconds before clculing he erge ccelerion.. The ne chnge in he plne s elociy is he finl minus he iniil elociy:
6 8.5 km/s.9 km/s 39.4 km/s 39.4 km s m km m/s b. Alhough he plne s elociy chnges by lrge moun, he chnge occurs oer long ime inerl, so he erge ccelerion is likely o be smll. Epressed in seconds, he inerl is.6 yr 365 d yr 4 h d 6 min h 6 s min s Then he erge ccelerion is m/s s m/s 8. REASONING We cn use he definiion of erge ccelerion / (Equion.4) o find he spriner s finl elociy he end of he ccelerion phse, becuse her iniil elociy ( m/s, since she srs from res), her erge ccelerion, nd he ime inerl re known.. Since he spriner hs consn ccelerion, i is lso equl o her erge ccelerion, so.3 m/s Her elociy he end of he.s period is m/s.3 m/s. s.8 m/s b. Since her ccelerion is zero during he reminder of he rce, her elociy remins consn.8 m/s. 6. REASONING The erge ccelerion is defined by Equion.4 s he chnge in elociy diided by he elpsed ime. We cn find he elpsed ime from his relion becuse he ccelerion nd he chnge in elociy re gien.. The ime h i kes for he VW Beele o chnge is elociy by n moun = is (nd noing h.447 m/s = mi/h)
7 .447 m / s 6. mi / h m / s mi / h.35 m / s.4 s b. From Equion.4, he ccelerion (in m/s ) of he drgser is.447 m / s 6. mi / h m / s mi / h.6 s s 44.7 m / s 8. REASONING AND. From Equion.4, he definiion of erge ccelerion, he mgniude of he erge ccelerion of he skier is 8. m/s m/s 5. s.6 m/s b. Wih represening he displcemen reled long he slope, Equion.7 gies: ( ) (8. m/s m/s)(5. s) =. m 38. REASONING AND. The elociy he end of he firs (7. s) period is = + = (. m/s )(7. s) A he end of he second period he elociy is = + = + (.58 m/s )(6. s) And he elociy he end of he hird (8. s) period is 3 = = + (.49 m/s )(8. s) = +5.6 m/s b. The displcemen for he firs ime period is found from = + = ( m/s)(7. s) + (. m/s )(7. s) = +49. m
8 Similrly, = m nd 3 = m, so he ol displcemen of he bo is = = +33 m 44. REASONING Becuse here is no effec due o ir resisnce, he rock is in free fll from is lunch unil i his he ground, so h he ccelerion of he rock is lwys 9.8 m/s, ssuming upwrd o be he posiie direcion. In (), we will consider he inerl beginning lunch nd ending. s ler. In (b), we will consider he inerl beginning lunch nd ending 5. s ler. Since he displcemen isn required, Equion.4 suffices o sole boh prs of he problem. The sone slows down s i rises, so we epec he speed in () o be lrger hn 5 m/s. The speed in (b) could be smller hn 5 m/s (he rock does no rech is mimum heigh) or lrger hn 5 m/s (he rock reches is mimum heigh nd flls bck down below is heigh he.s poin).. For he inerl from lunch o =. s, he finl elociy is = 5 m/s, he ccelerion is = 9.8 m/s, nd he iniil elociy is o be found. Soling Equion.4 for gies 5 m/s ( 9.8 m/s )(. s) 35 m/s Therefore, lunch, Speed 35 m/s b. Now we consider he inerl from lunch o = 5. s. The iniil elociy is h found in pr (), = 35 m/s. The finl elociy is 35 m/s ( 9.8 m/s )(5. s) = 4 m/s (.4) Insnneous speed is he mgniude of he insnneous elociy, so we drop he minus sign nd find h Speed 4 m/s 48. REASONING The minimum ime h plyer mus wi before ouching he bskebll is he ime required for he bll o rech is mimum heigh. The iniil nd finl elociies re
9 known, s well s he ccelerion due o griy, so Equion.4 ( ) cn be used o find he ime. Soling Equion.4 for he ime yields m / s 4.6 m / s 9.8 m / s.47 s 56. REASONING The bll is iniilly in free fll, hen collides wih he pemen nd rebounds, which pus i ino free fll gin, unil cugh by he boy. We don he enough informion o nlyze is collision wih he pemen, bu we re only sked o clcule he ime i spends in he ir, undergoing freefll moion. The moion cn be conenienly diided ino hree inerls: from relese (h = 9.5 m) o impc, from impc o he second highes poin (h = 5.7 m), nd from he second highes poin o h 3 =. m boe he pemen. For ech of he inerls, he ccelerion is h due o griy. For he firs nd ls inerl, he bll s iniil elociy is zero, so he ime o fll gien disnce cn be found from Equion.8 y. The second inerl begins he pemen nd ends h, so he iniil elociy isn zero. Howeer, he symmery of freefll moion is such h i kes he bll s much ime o rise from he ground o mimum heigh h s i would ke for bll dropped from h o fll o he pemen, so we cn gin use Equion.8 o find he durion of he second inerl. Tking upwrd s he posiie direcion, we he = 9.8 m/s for he ccelerion in ech of he hree inerls. Furhermore, he iniil elociy for ech of he inerls is = m/s. Remember, we re using symmery o re he second inerl s if he bll were dropped from res heigh of 5.7 m nd fell o he pemen. Using Equion.8 y, wih = m/s, we cn sole for he ime o find h y Applying his resul o ech inerl gies he ol ime s ol 9.5 m 5.7 m 5.7 m. m 9.8 m/s 9.8 m/s 9.8 m/s s nd rd inerl inerl 3 inerl 3.43 s
10 Posiion (m) Noe h he displcemen y for ech inerl is negie, becuse upwrd hs been designed s he posiie direcion. 66. REASONING On posiionersusime grph, he elociy is he slope. Since he objec s elociy is consn nd i moes in he + direcion, he grph will be srigh line wih posiie slope, beginning = 6 m when = s. A = 8 s, is posiion should be = 6 m + 48 m = +3 m. Once he grph is consruced, he objec s elociy is found by clculing he slope of he grph:. The posiionersusime grph for he moion is s follows: The Time (s) objec s displcemen is +48 m, nd he elpsed ime is 8 s, so is elociy is 48 m 8 s.7 m/s 68. REASONING The erge elociy for ech segmen is he slope of he line for h segmen.
11 Tking he direcion of moion s posiie, we he from he grph for segmens A, B, nd C,. km 4. km A. km/h.5 h. h B C. km. km. km/h.5 h.5 h 4. km. km 4 km/h 3. h.5 h
A 1.3 m 2.5 m 2.8 m. x = m m = 8400 m. y = 4900 m 3200 m = 1700 m
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