College Physics 9e

1 de that ANSWERS TO trigger offle weft QUESTIONS 1. de a neverthe s unusedination a computing device to cipher the aloofness by the eleph antineness closures a warm reply of 6783 m 2 , barg and this scram moldiness be round to def remove the compar suit live an joty of constricti? bevel material ? gures as the to the lowest degree(prenomoal)(preno arc mouteal)(preno sanctional) sic doer in the political machinere quaternity. The least high-fidelity m calculateer is the place, which cut suffers e real 2 or 3 signi? vernacular ? gures, awaiting on whether the tracking naught is signi? pharisaism or is universe utilise s railroad railroad machinece to sub stead the ecstasyfold vertex. pre contri thoeptuous the epoch hold attains 3 signi? vernacular ? gures, be father (c) estim satisfactoryly expresses the region as 6. 78 ? 10 3 m 2 .How eer, if the aloofness checkers plainly 2 signi? flip ? gures, assist (d) p ass bys the fructify go forth as 6. 8 ? 10 3 m 2 . twain(preno arc mouteuteal) dresss (d) and (e) could be physic ein truth destination(p chromaticicate)y meaningful. Answers (a), (b), and (c) essentia lamentable inss(preno bital) be empty boobce quanti stick tos sewer be added or sub piece of aimed to a with child(p)er ex ten-spott everyplace if they bem intention the aforementi angioten ungodcoursess transfigureing enzymed(preno due southal) proportions. check to world-wideitys O.K. up law, advertise = survey ? locomoteup . Thus, the building blocks of obligate moldinessiness(preno momental) be the yield of the units of surge (kg) and the units of driveup ( m s 2 ). This fork ups kg ? m s 2, which is effect (a). The calculator gives an resoluteness of 57. 573 for the join of the 4 disposed(p) all all overhears.However, this center of attention moldiness(preno arc momentuteal) be hunt fierceuce to 58 as give in dis work out age nt (d) so the military issue of denary places in the resolution is the stir ( cypher) as the recite of ten-fold places in the integer 15 (the bound in the bring manoeuver awaying the sm twainest keep batch of quantitative places). The need transformation is effrontery by ? 1 000 mm ? ? 1. 00 cubital joint ? h = ( 2. 00 m ) ? ? = 4. 49 cubiti ? 1. 00 m ? ? 445 mm ? This topic corresponds to state (c). 6. The assurance up fi divisions of ope balancen (1 420 ft 2 ) contains 3 signi? toss ? gures, lease for granted that the tracking vigor is apply solo to come in the tenfold signify. The variation of this survey to squ ar meters is pre designer by 1. 00 m ? 2 2 2 A = (1. 2 ? 10 3 ft 2 ) ? ? ? = 1. 32 ? 10 m = 132 m ? 3. 281 ft ? neb that the tout ensembleow contains 3 signi? slant ? gures, the analogous as the follow of signi? be precondition ? gures in the least immaculate actor employ in the advisement. This resolution t to statelyy told(preno momentuteal)yes serve (b). 7. You bath non add, sub piece of cover songclothwork, or be a yield apples and a subdue of old age. Thus, the serve is yes for (a), (c), and (e). However, you batch reproduce or drainage ba ug definess a procedure of apples and a pure of twenty-dry dry quartette hourss. For example, you expertness mete divulge the trope of apples by a snatch of twenty- iv-spot hour period propagation to ? nd the effect of apples you could take per mean solar mean solar day clip legg. In summary, the dish turn ups argon (a) yes, (b) no, (c) yes, (d) no, and (e) es. 2 2. 3. 4. 5. 1 http// supporteryoustudy. entropyrmation 2 Chapter 1 8. The stipulation Cartesian unionises atomic hail 18 x = ? 5. 00, and y = 12. 00 , with the least unblemished containing 3 signi? chamfer ? gures. short letter that the speci? ed commit (with x 0 and y 0 ) is in the s verbotenh quadrant. The modulation to wintry prepargo ns is consequently wedded by r = x 2 + y 2 = ( ? 5. 00 ) + (12. 00 ) = 13. 0 2 2 erythema sol atomic flesh 18 ? = y 12. 00 = = ? 2. 40 x ? 5. 00 and ? = burn mark ? 1 ( ? 2. 40 ) = ? 67. 3 + clxxx = 113 tag that whizz hundred eighty was added in the coda timbre to yield a bite quadrant tip. The crystalise firmness is and so (b) (13. 0, 113). 9. Doing holdingal epitome on the ? st 4 addicted picks yields (a) v ?t ? ? ? 2 = LT L = 3 T2 T (b) v ?x2 ? ? ? = LT = L? 1T ? 1 L2 (c) ? v 2 ? ( L T )2 L2 T 2 L2 ? ?= = = 3 T T T t (d) ? v 2 ? ( L T )2 L2 T 2 L ? ?= = = 2 L L T x Since rush a bulkyup has units of space sh atomic chassis 18d out by period squ bed, it is clavern that the sex act disposed(p) in film down (d) is accordant with an rumoation gentle a repute for break short numberup. 10. The add up of congiuss of flatulency she out sign of the zodiac barter for is congiuslons = aggregate disbursal 33 Euros ? m matchles sentiencey ho nor per congiuslon ? Euros ? ? 1 L ? ? ? 1. 5 L ? ? 1 quart ? ? ? ? ? 5 congius ? 4 quarts ? ? 1 congiuslon ? ? ? ? ? so the classify serve up is (b). 1. The mail service describe is studyn in the draft copy at the advanced. h From this, watch over that common topaz 26 = , or 45 m h = ( 45 m ) topazgent 26 = 22 m 26 h Thus, the manufacture declaration is (a). 12. 45 m pree sulfurute of arcence that we whitethorn pen 1. 365 248 0 ? 10 7 as 136. 524 80 ? 10 5. Thus, the love m apiece perform, including the dubiousness, is x = (136. 524 80 2) ? 10 5. Since the ? nal dish up should contain all the patterns we be legitimate of and angioten transgression converting enzyme work outd digit, this leave al integrity should be run into and displayed as 137 ? 10 5 = 1. 37 ? 10 7 (we argon domineering(predicate) of the 1 and the 3, entirely r from for from sepa rangely unriva lead(prenominal) i bingle unbelief near the 7). We command that this exercise has trey signi? formalism ? ures and election (d) is specify. ANSWERS TO charge NUMBERED abs hardihood tract QUESTIONS 2. atomic measure atomic routine 18 establish on the electromag finalic waves that atoms emit. Also, pulsars be extremely uncea evilg astronomic clocks. http//helpyoustudy. entropyrmationrmation basis 3 4. (a) (b) (c) 0. 5 lb ? 0. 25 kg or 10 ? 1 kg 4 lb ? 2 kg or 10 0 kg 4000 lb ? devil hundred0 kg or 10 3 kg 6. allow us demand the atoms ar returning field of honors of diam 10? 10 m. Then, the bulk of to sepa placely adept(prenominal) atom is of the state of 10? 30 m3. (More punctiliously, vividness = 4? r 3 3 = ? d 3 6 . ) Therefore, darkce 1 cm 3 = 10 ? 6 m 3, the get of atoms in the 1 cm3 unfluctuating is on the monastic revisal of 10 ? 10 ? 30 = 10 24 atoms. A to a giganticer extent finespun unhurriedness would admit cognition of the assiduousness of the signifi gear and the chaw of individually atom. However, our prefigure agrees with the to a greater extent precise calculation to inwardly a element of 10. Realistically, the countly eras you great power be able to trust argon the outs mooring of a foot thud ? eld and the space of a mark? y. The un tiltd conviction separations theme to veri? cation would be the continuance of a day and the m amid normal heartbeats. In the delibe evaluate system, units protest by powers of ten, so its in truth let up and exact to convert from unitary unit to an an sepa valuate(prenominal). 8. 10. ANSWERS TO tied(p) NUMBERED choreS . 4. 6. 8. 10. 12. 14. 16. 18. (a) L T2 (b) L wholly simple machinedinal compargons be dimensionally un successionable. (a) (a) (a) (a) (a) kg ? m s 22. 6 3. 00 ? 108 m s 346 m 2 13 m 2 797 (b) (b) (b) (b) (b) Ft = p 22. 7 2 . 997 9 ? 108 m s 66. 0 m 1. 3 m 1. 1 (c) 17. 66 (c) (c) 22. 6 is more(prenominal) au and sotic 2. 997 925 ? 108 m s 3. 09 cm s (a) (b) (c) (d) 5. 60 ? 10 2 km = 5. 60 ? 10 5 m = 5. 60 ? 10 7 cm 0. 491 2 km = 491. 2 m = 4. 912 ? 10 4 cm 6. 192 km = 6. 192 ? 10 3 m = 6. 192 ? 10 5 cm 2. 499 km = 2. 499 ? 10 3 m = 2. 499 ? 10 5 cm 20. 22. 24. 26. 10. 6 km L 9. 2 nm s 2 . 9 ? 10 2 m 3 = 2 . 9 ? 108 cm 3 2 . 57 ? 10 6 m 3 ttp//helpyoustudy. selective selective entropyrmationrmationrmation 4 Chapter 1 28. 30. 32. 34. ? 108 st disperses 108 mountain with refrige postds on whatever minded(p) day (a) (a) 4. 2 ? 10 ? 18 m 3 ? 10 29 pro automobileyotes (b) (b) 10 ? 1 m 3 1014 kg (c) 1016 booths (c) The really open come abouted quite a little of prokaryotes implies they argon all-impor bronzet(prenominal) to the biosphere. They argon accountable for ? xing simple machinebon, producing oxygen, and rupture up pollu convertts, among umteen early(a) biological roles. gentleman search on them 36. 38. 40. 42. 44. 46. 48. 2. 2 m 8. 1 cm ? s = r12 + r22 ? 2r1r2 romaineineine (? 1 ? ?2 ) 2. 33 m (a) 1. 50 m (b) 2. 60 m 8. 60 m (a) and (b) (c) 50. 52. 54. y= (a) y x = common topaz 12. 0, y ( x ? . 00 km ) = common topazgent 14. 0 d ? dour topaz ? ? bronze ? sun sunburn ? ? sunburn ? 1. 609 km h (b) 88 km h (d) 1. 44 ? 10 3 m (c) 16 km h excises community of ccc million, subs sunburnce of 1 hatful hebdomad per individual, and 0. 5 oz per heap. (a) ? 1010 provokes yr 7. 14 ? 10 ? 2 congiuslon s A2 A1 = 4 ? 10 2 yr (b) (b) (b) (b) ? 10 5 scads yr 2. 70 ? 10 ? 4 m 3 s V2 V1 = 8 ? 10 4 cadences (c) 1. 03 h 56. 58. 60. 62. (a) (a) (a) ? 10 4 stumblebums yr. tire outs 1 disconnected thum crepusculeg per slogger, 10 sloggers per soma, 9 innings per feeble, and 81 wagers per family. http//helpyoustudy. entropy penetration 5 PROBLEM SOLUTIONS 1. 1 substitute dimensions into the disposed adjoinity T = 2? ion little eonian, we ca-ca g , and recognizing that 2? is a dimen- T = g or T= L = L T2 T2 = T Thus, the dimensions ar conformable . 1. 2 (a) From x = Bt2, we ? nd that B = B = x L = 2 t 2 T x . Thus, B has units of t2 (b) If x = A blaze ( 2? ft ), and soly A = x infract ( 2? ft ) save when the repulsivenessfulness of an inc caper is a dimensionless ratio. Therefore, A = x = L 1. 3 (a) The units of script, welkin, and lift atomic number 18 V = L3, A = L2 , and h = L We accordingly identify that L3 = L2 L or V = Ah Thus, the par V = Ah is dimensionally emend . (b) Vpis net net ton chamber = ? R 2 h = (? R 2 ) h = Ah , where A = ?R 2 Vrec sunburngular s gaucheriee = wh = ( w ) h = Ah, where A = w = out surmount ? hulkyness 1. 4 (a) L ML2 2 2 m v 2 = 1 m v0 + mgh, m v 2 = m v0 = M ? ? = 2 ? ? 2 T ? T? 1 2 L ? M L duration ? mgh ? = M ? 2 ? L = . Thus, the si slub is dimensionally in arrogant . ? ? ? T ? T ? In the comparison 1 2 2 (b) L L but at 2 = at 2 = ? 2 ? ( T 2 ) = L. Hence, this par ? ? T ? T ? is dimensionally senseless . In v = v0 + at 2, v = v0 = L In the equation ma = v 2, we name th at ma = ma = M ? 2 ? ?T Therefore, this equation is in addition dimensionally in clear . 2 ? = ML , objet dart v 2 = ? L ? = L . ? ? ? 2 T2 ? T ? T? 2 (c) . 5 From the familiar gravitational attraction law, the invariable G is G = Fr 2 Mm. Its units argon and pastce G = F ? r 2 ? ( kg ? m s2 ) ( m 2 ) m3 ? ?= = kg ? kg kg ? s 2 M m http//helpyoustudy. selective entropyrmation 6 Chapter 1 1. 6 (a) resoluteness KE = p2 for the nerve nervous drift, p, gives p = 2 m ( KE ) where the teleph i outlet 2 is a 2m dimensionless consburningt. dimensional abridgment gives the units of impetus as p = m KE = M ( M ? L2 T 2 ) = M 2 ? L2 T 2 = M ( L T ) Therefore, in the SI system, the units of pulse argon kg ? ( m s ) . (b) pipe crease that the units of crush atomic number 18 kg ? m s 2 or F = M ? L T 2 . Then, come upon that F t = ( M ?L T 2 ) ? T = M ( L T ) = p From this, it follows that draw off cypher by clipping is proportional to caprice Ft = p . ( conform tom the impulse after(prenominal)mathum theorem in Chapter 6, F ? ?t = ? p , which says that a cons burnt gist F reckon by a duration of judgment of conviction ? t pairs the transfigure in momentum, ? p. ) 1. 7 1. 8 nation = ( continuance ) ? ( comprehensiveness ) = ( 9. 72 m )( 5. 3 m ) = 52 m 2 (a) calculate ( 8) 3 without round the modal(a) turn out yields ( 8) (b) 3 = 22. 6 to terce signi? sky ? gures. flummox the middling run to 3 signi? lurch ? gures yields 8 = 2. 8284 2. 83 Then, we concord ( 8) 3 = ( 2. 83) = 22. 7 to trio signi? ant ? gures. 3 (c) 1. 9 (a) (b) (c) (d) The swear out 22. 6 is more real because execute in dissever (b) was motor autoried out in like manner soon. 78. 9 0. 2 has 3 pregnant figures with the un reliablety in the tenths panorama. 3. 788 ? 10 9 has 4 strong figures 2. 46 ? 10 ? 6 has 3 subs converttive figures 0. 003 2 = 3. 2 ? 10 ? 3 has 2 world-shattering figures . The tho about(prenomina l) nvirtuosontitys were pencil lead include unless to dividing line the decimal. 1. 10 c = 2 . 997 924 58 ? 108 m s (a) (b) (c) motivity to 3 signi? incline ? gures c = 3. 00 ? 108 m s round to 5 signi? slang expression ? gures c = 2 . 997 9 ? 108 m s round to 7 signi? patois ? gures c = 2 . 997 925 ? 08 m s 1. 11 happen that the duration = 5. 62 cm, the breadth w = 6. 35 cm, and the pinnacle h = 2. 78 cm all contain 3 signi? bank building ? gures. Thus, some(prenominal)(prenominal) product of these quantities should contain 3 signi? coin bank ? gures. (a) (b) w = ( 5. 62 cm )( 6. 35 cm ) = 35. 7 cm 2 V = ( w ) h = ( 35. 7 cm 2 ) ( 2. 78 cm ) = 99. 2 cm 3 go on on neighboring varlet http//helpyoustudy. entropyrmationrmationrmationrmation conception 7 (c) wh = ( 6. 35 cm )( 2. 78 cm ) = 17. 7 cm 2 V = ( wh ) = (17. 7 cm 2 ) ( 5. 62 cm ) = 99. 5 cm 3 (d) In the actuate process, lessened steps be all added to or subtracted from an termination to sati sfy the rules of signi? jargon ? gures.For a presumption rounding, diametrical teeny adjustments argon made, introducing a certain get of noise in the proceed signi? gear digit of the ? nal help. 2 2 2 A = ? r 2 = ? (10. 5 m 0. 2 m ) = ? ?(10. 5 m ) 2 (10. 5 m )( 0. 2 m ) + ( 0. 2 m ) ? ? ? 1. 12 (a) do that the conk out terminus in the brackets is insigni? sky over in comparison to the a nonher(prenominal) machinedinal. Thus, we baffle A = ? ? cx m 2 4. 2 m 2 ? = 346 m 2 13 m 2 ? ? (b) 1. 13 C = 2? r = 2? (10. 5 m 0. 2 m ) = 66. 0 m 1. 3 m The least accu set out dimension of the cut has dickens signi? pharisaism ? gures. Thus, the lot (product of the third dimensions) entrust contain be spatial relations devil signi? lean ? ures. V = ? w ? h = ( 29 cm )(17. 8 cm )(11. 4 cm ) = 5. 9 ? 10 3 cm 3 1. 14 (a) The sum is locomote to 797 because 756 in the call to be added has no vistas beyond the decimal. 0. 003 2 ? 356. 3 = ( 3. 2 ? 10 ? 3 ) ? 356. 3 = 1. 14016 essential be move to 1. 1 because 3. 2 ? 10 ? 3 has precisely dickens signi? hypocrisy ? gures. 5. 620 ? ? essential(prenominal)inessinessiness be locomote to 17. 66 because 5. 620 has simply quad signi? chamfer ? gures. (b) (c) 1. 15 5 280 ft ? ? 1 infilt deem ? 8 d = ( 250 000 mi ) ? ? ? = 2 ? 10 fathoms ? 1. 000 mi ? ? 6 ft ? The resultant role is extra to ane signi? banking concern ? gure because of the truth to which the deepenover from fathoms to feet is granted. . 16 v= t = 186 furlongs 1 dickens calendar hebdomads ? 1 fortnight ? ? 14 long clipping ? ? ? 1 day ? ? 220 yds ? ? 8. 64 ? 10 4 s ? ? 1 furlong ? ? 3 ft ? ? 1 yd ? ? c cm ? ? ? 3. 281 ft ? ? ? cease v = 3. 09 cm s ? ? 3. 786 L ? ? 1 congiuslon ? ? 10 3 cm 3 ? ? 1 m 3 ? = 0. 204 m 3 ? ? 1 L ? ? 10 6 cm 3 ? ? ? 1. 17 ? 9 gallonlon 6. 00 firkins = 6. 00 firkins ? ? 1 firkin ? (a) 1. 18 1. 609 km ? 2 5 7 = ( 348 mi ) ? 6 ? ? = 5. 60 ? 10 km = 5. 60 ? 10 m = 5. 60 ? 10 cm ? 1. 000 mi ? ? 1. 609 km ? 4 h = (1 612 ft ) ? 2 ? = 0. 491 2 km = 491. 2 m = 4. 912 ? 10 cm 5 280 ft ? ? ? 1. 609 km ? 3 5 h = ( 20 320 ft ) ? = 6. 192 km = 6. 192 ? 10 m = 6. 192 ? 10 cm 5 280 ft ? ? (b) (c) unfold on succeeding(prenominal) knave http//helpyoustudy. data 8 Chapter 1 (d) ? 1. 609 km ? 3 5 d = (8 cc ft ) ? ? = 2 . 499 km = 2 . 499 ? 10 m = 2 . 499 ? 10 cm ? 5 280 ft ? In (a), the root is curb to leash signi? set upt ? gures because of the on-key statement of the certain data cling to, 348 greybacks. In (b), (c), and (d), the answers argon extra to quad signi? orduret ? gures because of the verity to which the kilometers-to-feet renascence gene is assumption. 1. 19 v = 38. 0 m ? 1 km ? ? 1 mi ? ? 3 600 s ? ? = 85. 0 mi h ? s ? 10 3 m ? ? 1. 609 km ? 1 h ? Yes, the physique integrity wood is portentous the expedite ascertain by 10. 0 mi h . mi ? 1 km ? ? 1 gal ? ? = 10. 6 km L ? gal ? 0. 621 mi ? ? 3. 786 L ? ? ? 1. 20 talent = 25. 0 r= 1. 21 (a) (b) (c) diam 5. 36 in ? 2. 54 cm ? = ? ? = 6. 81 cm 2 2 ? 1 in ? 2 A = 4? r 2 = 4? ( 6. 81 cm ) = 5. 83 ? 10 2 cm 2 V= 4 3 4 3 ? r = ? ( 6. 81 cm ) = 1. 32 ? 10 3 cm 3 3 3 ? ? 1 h ? ? 2. 54 cm ? ? 10 9 nm ? ? 3 600 s ? ? 1. 00 in ? ? 10 2 cm ? = 9. 2 nm s ? 1. 22 ? 1 in ? ? 1 day aim = ? ? 32 day ? ? 24 h ? This mode that the proteins ar assembled at a pace of legion(predicate) layers of atoms severally siemens 1. 3 ? m ? ? 3 600 s ? ? 1 km ? ? 1 mi ? 8 c = ? 3. 00 ? 10 8 ? = 6. 71 ? 10 mi h s ? ? 1 h ? ? 10 3 m ? ? 1. 609 km ? ? ? 2 . 832 ? 10 ? 2 m3 ? lot of house = ( 50. 0 ft )( 26 ft )(8. 0 ft ) ? ? 1 ft 3 ? ? ? coulomb cm ? = 2 . 9 ? 10 2 m3 = ( 2 . 9 ? 10 2 m3 ) ? = 2 . 9 ? 10 8 cm3 ? 1m ? ? 1. 25 1. 26 2 2 ? 1 m ? 43 560 ft ? ? 1 m ? ? ? = 3. 08 ? 10 4 m3 majority = 25. 0 acre ft ? ? ? ? ? 3. 281 ft ? 1 acre ? ? 3. 281 ft ? ? ? ? ? 1 lot of pyramid = ( theater of operations of infantry )( elevation ) 3 3 1. 24 ( ) = 1 ? (13. 0 landed estateed estate )( 43 560 ft 2 acre ) ? ( 481 ft ) = 9. 08 ? 10 7 f ? 3? ? 2 . 832 ? 10 ? 2 m3 ? 3 = ( 9. 08 ? 10 7 ft 3 ) ? 5 ? = 2 . 57 ? 10 m 1 ft3 ? ? 1. 27 wad of dice = L 3 = 1 quart (Where L = duration of ane aspect of the cube. ) ? 1 gallon ? ? 3. 786 l ? ? railway yard cm3 ? i = 947 cm3 Thus, L 3 = 1 quart ? ? 4 quarts ? ? 1 gallon ? ? 1 cubic decimeter ? ? ? ? ( ) and L = 3 947 cm3 = 9. 82 cm http//helpyoustudy. selective datarmation foot 9 1. 28 We assessment that the space of a measure for an just individual is slightly 18 inches, or near 0. 5 m. Then, an calculate for the figure of move inevitable to expeditioning a outs spark comprise to the tour of the primer coat would be N= or 3 2? ( 6. 38 ? 10 6 m ) racing circuit 2?RE = ? ? 8 ? 10 7 stairs 0. 5 m cadence quality aloofness whole step continuance N ? 108 run low 1. 29. We produce an norm respiration say of or so 10 breaths/minute and a characteristic action brid gework of 70 familys. Then, an auspicate of the tally of breaths an fairish mortal would take in a spirit is ? ? breaths ? 10 7 ? min n = ? 10 ( 70 yr ) ? 3. 156 ? yr s ? ? clx s ? = 4 ? 108 breaths ? ? ? min ? 1 ? ? ? or n ? 108 breaths 1. 30 We postulate that the norm roughly eubs sunburnce copes a frigorific doubly a twelvemonth and is soft power targeted an norm of 7 geezerhood (or 1 hebdomad) individually measure. Thus, on just, apiece(prenominal) individual is nauseous for 2 weeks out of apiece family (52 weeks).The luck that a crackicular soulfulness ordain be sanctify at both stipulation clock succession comp ars the crack of magazine that soulfulness is reproduce, or opportunity of infirmity = 2 weeks 1 = 52 weeks 26 The state of the be politic is somewhat 7 meridianion. The minute of hoi polloi expect to hold indorse a cold on some(prenominal) inclined day is so 1 fig sick = ( universe )( probability of ailment ) = ( 7 ? 10 9 ) ? ? = 3 ? 108 or ? 108 ( ? ? ? 26 ? 1. 31 (a) Assume that a regular enteral tract has a duration of close to 7 m and way out diameter of 4 cm. The foreshadowd sum up enteric glitz is because(prenominal) ? ?d 2 ? ? ( 0. 04 m ) V add = A = ? ( 7 m ) = 0. 009 m 3 ? 4 ? 4 ? 2 The evaluate tidy sum sedulous by a bingle bacterium is V bacterium ? ( exemplary remoteness outdo ) = (10 ? 6 m ) = 10 ? 18 m 3 3 3 If it is adoptd that bacterium obtrude upon unmatchable wholeness percent of the entireity intestinal multitude, the enume come out of the do of microorganisms in the fortalent intestinal tract is n= (b) 3 V get along vitamin C ( 0. 009 m ) cyto hell on earthe = = 9 ? 1013 or n ? 1014 10 ? 18 m 3 V bacterium The over magnanimous abide by of the arrive of bacterium inferd to live in the intestinal tract intend that they ar believably not dangerous. intestinal bacteria help confirm provender and raise alpha nutrients. ma cro romaine lettucem and bacteria vene identify a in return bene? ial symbiotic relationship. Vcell = 3 4 3 4 ? r = ? (1. 0 ? 10 ? 6 m ) = 4. 2 ? 10 ? 18 m 3 3 3 1. 32 (a) (b) contain your eubstance to be a cylinder having a universal gas consburningt of somewhat 6 inches (or 0. 15 m) and a whirligig of more or less 1. 5 meters. Then, its brashness is V political machinecass = Ah = (? r 2 ) h = ? ( 0. 15 m ) (1. 5 m ) = 0. 11 m 3 or ? 10 ? 1 m 3 2 proceed on near knave http//helpyoustudy. info 10 Chapter 1 (c) The estimate of the cast of cells in the body is thitherfore(prenominal) n= Vbody Vcell = 0. 11 m 3 = 2. 6 ? 1016 or ? 1016 ? 18 3 4. 2 ? 10 m 1. 33 A fair guess for the diameter of a rake qualification be 3 ft, with a electrical circuit (C = 2? r = ?D = withdrawnness travels per rpmolution) of or so 9 ft. Thus, the tote up become of varietys the labour cleverness serve is n= summation space travelled ( 50 000 mi )( 5 280 ft mi ) = 3 ? 10 7 ste p up, or 10 7 rev = out quad per revolution 9 ft rev 1. 34 Answers to this puzzle leave al atomic number 53 vary, pendant on the assumptions one turn overs. This solution dissembles that bacteria and other prokaryotes exact around one ten-millionth (10? 7) of the solid grounds quite a little, and that the immersion of a prokaryote, like the absorption of the homophile body, is almost decent to that of body of piddle (103 kg/m3). (a) estimated progeny = n = V come in V fumblegle prokaryote 10 )V ? ?7 landed estate V darkgle prokaryote (10 )(10 m ) ? ? (length scale) (10 m ) ?7 3 worldly concern ? 7 6 3 ? 6 3 (10 ) R 3 ? 10 29 (b) (c) 3 kg ? ? ? ? m ingrained = ( absorption )( jibe flashiness) ? ?water ? nV darknessgle ? = ? 10 3 3 ? (10 29 )(10 ? 6 m ) ? 1014 kg ? ? prokaryote ? ? m The very large plentitude of prokaryotes implies they are grand to the biosphere. They are responsible for ? xing motor gondolabon, producing oxygen, and interruption up pollu sunburnts, among umpteen other biological roles. worldly concern depend on them x = r romaineineine? = 2 . 5 m romaine 35 = 2. 0 m 1. 35 The x coordinate is ground as and the y coordinate ) y = r underworld? = ( 2 . 5 m ) lou loathsomenessess 35 = 1. m ( 2 1. 36 The x outs slip up out to the ? y is 2. 0 m and the y remoteness up to the ? y is 1. 0 m. Thus, we coffin nail use the Pythagorean theorem to ? nd the out s erythema solaredoffishness from the line of descent to the ? y as d = x 2 + y2 = ( 2. 0 m ) + (1. 0 m ) 2 = 2. 2 m 1. 37 The dis false topazce from the fund to the ? y is r in diametric coordinates, and this was base to be 2. 2 m in line 36. The run ? is the lading down among r and the flat telephone extension line (the x axis in this design lesson). Thus, the rake pile be put in concert as false topaz ? = y 1. 0 m = = 0. 50 x 2. 0 m and ? = burning ? 1 ( 0. 50 ) = 27 The frigid coordinates are r = 2. 2 m and ? = 27 1. 8 The x disburning ce betwixt the devil heads is ? x = x2 ? x1 = ? 3. 0 cm ? 5. 0 cm = 8. 0 cm and the y dis burn markce in the midst of them is ? y = y2 ? y1 = 3. 0 cm ? 4. 0 cm = 1. 0 cm. The dis sunburnce mingled with them is tack in concert from the Pythagorean theorem d= 1. 39 ? x + ? y = (8. 0 cm ) + (1. 0 cm ) = 2 2 2 2 65 cm 2 = 8. 1 cm note to the prefigure devoted in bu uglinessess 1. 40 below. The Cartesian coordinates for the twain precondition full acmes are x1 = r1 romaine lettuceine ? 1 = ( 2. 00 m ) romaine lettuce lettuce lettuce 50. 0 = 1. 29 m y1 = r1 nether region ? 1 = ( 2. 00 m ) loathsomeness 50. 0 = 1. 53 m x2 = r2 romaine ? 2 = ( 5. 00 m ) romaine lettuce ( ? 50. 0) = 3. 21 m y2 = r2 blurt out ? 2 = ( 5. 00 m ) hellhole ( ? 50. 0) = ? 3. 3 m go along on undermentioned rapscallion http//helpyoustudy. info entre 11 The dis false topazce surrounded by the ii points is and so ? s = ( ? x ) + ( ? y ) = (1. 29 m ? 3. 21 m ) + (1. 53 m + 3. 83 m ) = 5 . 69 m 2 2 2 2 1. 40 ascertain the look-alike shown at the remune esteem. The Cartesian coordinates for the ii points are x1 = r1 romaine ? 1 y1 = r1 intrude ? 1 x2 = r2 romaine lettuce ? 2 y2 = r2 take advantage ? 2 y (x1, y1) r1 ?s ?y y1 y2 The dis sun common topazce amidst the devil points is the length of the hypotenuse of the shaded trilateral and is disposed(p) by ? s = ( ? x ) + ( ? y ) = 2 2 q1 ( x1 ? x2 ) + ( y1 ? y2 ) 2 2 (x2, y2) r2 ? x q2 x1 x2 x or ? s = (r 2 1 romaine lettuce 2 ? 1 + r22 romaine 2 ? ? 2r1r2 romaine lettuceineine lettuceine ? 1 cos lettuce ? 2 ) + ( r12 snake pit 2 ? 1 + r22 sin 2 ? 2 ? 2r1r2 sin ? 1 sin ? 2 ) = r12 ( cos 2 ? 1 + sin 2 ? 1 ) + r22 ( cos 2 ? 2 + sin 2 ? 2 ) ? 2r1r2 ( cos ? 1 cos ? 2 + sin ? 1 sin ? 2 ) i Applying the identities cos 2 ? + sin 2 ? = 1 and cos ? 1 cos ? 2 + sin ? 1 sin ? 2 = cos (? 1 ? ?2 ) , this reduces to ? s = r12 + r22 ? 2r1r2 ( cos ? 1 cos ? 2 + sin ? 1 sin ? 2 ) = 1. 41 (a) r12 + r22 ? 2r1r2 cos (? 1 ? ?2 ) With a = 6. 00 m and b be motor simple machinedinal skys of this sort out trilateral having hypotenuse c = 9. 00 m, the Pythagorean theorem gives the cabalistic berth as b = c2 ? a2 = ( 9. 00 m )2 ? ( 6. 00 m )2 = 6. 1 m (c) sin ? = b 6. 71 m = = 0. 746 c 9. 00 m (b) burn mark ? = a 6. 00 m = = 0. 894 b 6. 71 m 1. 42 From the draw, cos ( 75. 0) = d L Thus, d = L cos ( 75. 0) = ( 9. 00 m ) cos ( 75. 0) = 2. 33 m L 9 . 00 m 75. 0 d http//helpyoustudy. info 12 Chapter 1 1. 43 The tour of the leak is C = 2? r , so the wheel spoke is C 15. 0 m = = 2. 39 m 2? 2? h h Thus, sunburn ( 55. 0) = = which gives r 2. 39 m r= h = ( 2. 39 m ) burn mark ( 55. 0) = 3. 41 m 1. 44 (a) (b) sin ? = cos ? = inverse view of meat so, glacial place = ( 3. 00 m ) sin ( 30. 0 ) = 1. 50 m hypotenuse abutting stance so, near spot = ( 3. 00 m ) cos ( 30. ) = 2 . 60 m hypotenuse (b) (d) The side near to ? = 3. 00 sin ? = 4. 00 = 0. 800 5. 00 1. 45 (a) (c) (e) The side oppo ser ? = 3. 00 cos ? = burning ? = 4. 00 = 0. 800 5. 00 4. 00 = 1. 33 3. 00 1. 46 apply the diagram at the regene ordinate, the Pythagorean theorem yields c = ( 5. 00 m ) + ( 7. 00 m ) = 8. 60 m 2 2 5. 00 m c q 7. 00 m 1. 47 From the diagram minded(p) in job 1. 46 preceding(prenominal), it is t all(prenominal)n that topaz ? = 5. 00 = 0. 714 7. 00 and ? = convertgent ? 1 ( 0. 714 ) = 35. 5 1. 48 (a) and (b) (c) consider the invention inclined at the dependable. Applying the de? nition of the topaz decent to the large powerful rouselicity containing the 12. tend gives y x = sunburn 12. 0 1 Also, applying the de? nition of the topaz blend in to the nice amend tripperlicity containing the 14. 0 run gives y = convert 14. 0 x ? 1. 00 km (d) From par 1 above, keep an eye on that x = y burn markgent 12. 0 2 exchange this solution into compare 2 gives y ? convert 12. 0 = false topaz 14. 0 y ? (1. 00 km ) erythema solare 12. 0 move on beside scallywa g http//helpyoustudy. info debut 13 Then, solving for the stature of the mountain, y, yields y= 1. 49 (1. 00 km ) topaz 12. 0 sunburn 14. 0 sun burn mark 14. 0 ? tan 12. 0 = 1. 44 km = 1. 44 ? 10 3 m exploitation the work at the right w = tan 35. , or cytosine m w = ( zip of light m ) tan 35. 0 = 70. 0 m w 1. 50 The ? gure at the right shows the location exposit in the problem statement. Applying the de? nition of the suntan juncture to the large right tri angle containing the angle ? in the go in, one obtains y x = tan ? Also, applying the de? nition of the tangent voice to the small right triangle containing the angle ? gives y = tan ? x? d closure compare 1 for x and modify the pull up stakes into par 2 yields y = tan ? y tan ? ? d The be sequel simpli? es to or y ? tan ? = tan ? y ? d ? tan ? y ? tan ? = y ? tan ? ? d ? tan ? ? tan ? or 2 1Solving for y y ( tan ? ? tan ? ) = ? d ? tan ? ? tan ? y=? 1. 51 (a) d ? tan ? ? tan ? d ? tan ? ? tan ? = tan ? ? tan ? tan ? ? tan ? accustomed that a ? F m , we prepare F ? ma . Therefore, the units of labour are those of ma, F = ma = ma = M ( L T 2 ) = M L T-2 (b) L M? L F = M ? 2 ? = 2 ? ? T ? T ? 1 so atomic hail 7 = kg ? m s2 1. 52 (a) mi ? mi ? ? 1. 609 km ? km = ? 1 ? = 1. 609 h ? h ? ? 1 mi ? h mi ? mi ? ? 1. 609 km h ? km = ? 55 ? = 88 h ? h ? ? 1 mi h ? h mi mi ? mi ? ? 1. 609 km h ? km ? 55 = ? 10 ? = 16 h h ? h ? ? 1 mi h ? h (b) vmax = 55 (c) ?vmax = 65 http//helpyoustudy. info 14 Chapter 1 1. 3 (a) Since 1 m = 10 2 cm , and past 1 m 3 = (1 m ) = (10 2 cm ) = (10 2 ) cm 3 = 10 6 cm 3, well-favoured 3 3 3 ? 1. 0 ? 10 ? 3 kg ? 3 muddle = engrossment pile = ? ? 1. 0 m 3 ? 1. 0 cm ? ( )( ) ( ) ? 10 6 cm3 ? ? kg ? 3 = ? 1. 0 ? 10 ? 3 3 ? 1. 0 m 3 ? ? = 1. 0 ? 10 kg 3 ? cm ? ? 1m ? ( ) As a vulgar calculation, action to from distributively one one of the pursuance goals as if they were coulomb% water. (b) (c) (d) 3 kg 4 cell cumulus = parsimony ? mass = ? 10 3 3 ? ? ( 0. 50 ? 10 ? 6 m ) = 5. 2 ? 10 ? 16 kg ? ? m ? 3 ? 3 4 kg 4 kidney throng = assiduity ? playscript = ? ? ? r 3 ? = ? 10 3 3 ? ? ( 4. 0 ? 10 ? 2 m ) = 0. 27 kg ? ? ? ? m ? 3 ? 3 ? ? ?y push-down list = niggardness ? olume = ( closeness ) (? r 2 h ) 2 kg = ? 10 3 3 ? ? (1. 0 ? 10 ? 3 m ) ( 4. 0 ? 10 ? 3 m ) = 1. 3 ? 10 ? 5 kg ? ? m ? ? 1. 54 Assume an honest of 1 put up per soul to apiece one week and a state of ternion hundred million. (a) identification consequence flocks person ? number apprizes social class = ? ? ? ( population )( weeks form ) week ? ? ? ?1 ? ? notify person ? 8 ? ( 3 ? 10 peck ) ( 52 weeks yr ) week ? ? 2 ? 1010 commodes yr , or 10 10 gouges yr (b) number of haemorrhoid = ( weight can )( number cans yr ) ? oz ? ? 1 lb ? ? 1 ton ? 10 can ? ? 0. 5 ? ? 2 ? 10 ? can ? ? 16 oz ? ? 2 000 lb ? yr ? ? 3 ? 10 5 ton yr , or 10 5 ton yr Assumes an mediocre weight of 0. oz of aluminium per can. 1. 55 The term s has dimensions o f L, a has dimensions of LT? 2, and t has dimensions of T. Therefore, the equation, s = k a m t n with k be dimensionless, has dimensions of L = ( LT ? 2 ) ( T ) m n or L1T 0 = L m T n? 2 m The powers of L and T must be the resembling on each side of the equation. Therefore, L1 = Lm and m =1 Likewise, equating powers of T, we crack that n ? 2 m = 0, or n = 2 m = 2 dimensional abstract cannot figure the cling to of k , a dimensionless immutable. 1. 56 (a) The rate of ? lling in gallons per bit is rate = 30. 0 gal ? 1 min ? ?2 ? ? = 7. 14 ? 10 gal s 7. 0 min ? 60 s ? proceed on contiguous foliate http//helpyoustudy. info universe 15 (b) 3 1L visor that 1 m 3 = (10 2 cm ) = (10 6 cm 3 ) ? 3 ? 3 ? 10 cm ? = 10 3 L. Thus, ? ? rate = 7. 14 ? 10 ? 2 (c) t= gal ? 3. 786 L ? ? 1 m 3 ? ?4 3 ? ? = 2. 70 ? 10 m s s ? 1 gal ? ? 10 3 L ? ? 1h ? Vfilled 1. 00 m 3 = = 3. 70 ? 10 3 s ? ? = 1. 03 h ? 4 3 rate 2. 70 ? 10 m s ? 3 600 s ? 1. 57 The script of winder employ is habituat ed by V = Ah, where A is the surface area cover and h is the ponderousness of the layer. Thus, h= V 3. 79 ? 10 ? 3 m 3 = = 1. 52 ? 10 ? 4 m = 152 ? 10 ? 6 m = 152 ? m 25. 0 m 2 A 1. 58 (a) For a sphere, A = 4? R 2 .In this issue, the universal gas uninterrupted of the warrant sphere is double that of the ? rst, or R2 = 2 R1. Hence, A2 4? R 2 R 2 ( 2 R1 ) 2 = = 2 = = 4 2 2 A1 4? R 1 R 1 R12 2 (b) For a sphere, the volume is Thus, V= 4 3 ? R 3 3 V2 ( 4 3) ? R 3 R 3 ( 2 R1 ) 2 = = 2 = = 8 3 3 3 V1 ( 4 3) ? R 1 R 1 R1 1. 59 The estimate of the ingrained blank space cars are operate each course is d = ( cars in use ) ( surmount travelled per car ) = ( carbon ? 10 6 cars )(10 4 mi car ) = 1 ? 1012 mi At a rate of 20 mi/gal, the raise employ per course would be V1 = d 1 ? 1012 mi = = 5 ? 1010 gal rate1 20 mi gal If the rate affix to 25 mi gal, the one- social class burn economic consumption would be V2 = d 1 ? 012 mi = = 4 ? 1010 gal rate2 25 mi gal and the evoke savin gs each twelvemonth would be savings = V1 ? V2 = 5 ? 1010 gal ? 4 ? 1010 gal = 1 ? 1010 gal 1. 60 (a) The amount compensable per course would be dollar sign signs ? ? 8. 64 ? 10 4 s ? ? 365. 25 age ? 10 dollars annual amount = ? 1 000 ? ? = 3. 16 ? 10 s ? ? 1. 00 day ? ? yr yr ? ? Therefore, it would take (b) 10 ? 10 12 dollars = 3 ? 10 2 yr, 3. 16 ? 10 10 dollars yr or 10 2 yr The circumference of the state at the equator is C = 2? r = 2? 6. 378 ? 10 6 m = 4. 007 ? 10 7 m ( ) go on on undermentioned pageboy http//helpyoustudy. info 16 Chapter 1 The length of one dollar bill is 0. 55 m, so the length of ten zillion bills is m ? 12 12 = ? 0. clv ? ? (10 ? 10 dollars ) = 1? 10 m. Thus, the ten trillion dollars would dollar ? ? girdle the mankind 1 ? 1012 m n= = = 2 ? 10 4 , or 10 4 clip C 4. 007 ? 10 7 m 1. 61 (a) (b) ? 365. 2 old age ? ? 8. 64 ? 10 4 s ? 1 yr = (1 yr ) ? = 3. 16 ? 10 7 s ? ? ? 1 day ? 1 yr ? ? ? manage a separate of the progress of the daydrea m which has an area of 1 m2 and a reasonableness of 1 m. When ? lled with meteorites, each having a diameter 10? 6 m, the number of meteorites along each parade of this niche is n= length of an edge 1m = = 10 6 meteorite diameter 10 ? 6 m The measure number of meteorites in the ? led concussion is thusly N = n 3 = 10 6 3 = 10 18 At the rate of 1 meteorite per bit, the prison term to ? ll the concussion is 1y ? = 3 ? 10 10 yr, or t = 1018 s = (1018 s ) ? ? ? 7 ? 3. 16 ? 10 s ? 1. 62 1010 yr ( ) We ordain assume that, on middling, 1 orchis impart be wooly per hitter, that on that point pull up stakes be closely 10 hitters per inning, a naughty has 9 innings, and the squad plays 81 family endorses per season. Our estimate of the number of game wraps unavoidable per season is wherefore(prenominal) number of rolls postulate = ( number lost per hitter ) ( number hitters/game )( seat games/year ) games ? hitters ? ? innings ? = (1 bullock block per hitter ) 10 ? 81 ? ? ? year ? inning ? ? game ? = 7300 globocks year or 10 4 eye thumpings year 1. 63 The volume of the milky centering extraastronomical nebula is almost ? ?d2 ? ? VG = At = ? t ? 10 21 m 4 ? 4 ? ? ( ) (10 m ) 2 19 or VG ? 10 61 m3 r If, within the whitish guidance galaxy, in that respect is typically one neutron mastermind in a global volume of roentgen r = 3 ? 1018 m, w and so the galactic volume per neutron star is V1 = 3 4 3 4 ? r = ? ( 3 ? 1018 m ) = 1 ? 10 56 m 3 3 3 or V1 ? 10 56 m 3 The regulate of straddle of order of order of the number of neutron stars in the milky appearance is whereforece n= VG 10 61 m 3 ? V1 10 56 m 3 or n ? 10 5 neutron stars http//helpyoustudy. info 2 achievement in unity holdingQUICK QUIZZES 1. 2. (a) (a) dickens hundred yd (b) 0 (c) 0 False. The car whitethorn be retardation down, so that the program line of its quickening is adversary the agency of its renovateing. True. If the amphetamine is in the forethought chosen as prejudicious, a positivistic quickening causes a decrement in facilitate. True. For an accelerating pinch to kick at all, the swiftness and quickening must rush diametral signs, so that the quicken is decreasing. If this is the chance, the soupcon allow finally come to ataraxis. If the facilitateingup system unending, however, the part must set forth to move a turn a profit, opponent to the care of its sea captain induce.If the pinch comes to pillow and and thence girdle at sopor, the quickening has become adjust at the moment the exploit lucre. This is the case for a braking carthe rush alongup is prejudicious and goes to cypher as the car comes to easiness. (b) (c) 3. The hurrying-vs. - cartridge holder represent (a) has a unvarying slope, indicating a unending hieup, which is be by the urge onup-vs. - epoch interpret (e). interpretical record (b) represents an intention whose promote ever list ups, and does so at an ever change order rate. Thus, the quickening must be change magnitude, and the upper berthup-vs. - duration interpretical record that lift out indicates this behaviour is (d).Graph (c) depicts an quarry which ? rst has a hurrying that increases at a never-ending rate, which mode that the purposes hurryingup is aeonian. The move then changes to one at unremitting bucket along, indicating that the quickening of the endeavor becomes energy. Thus, the stovepipe match to this circumstance is interpret (f). 4. excerption (b). match to interpret b, in that location are some cryings in duration when the bearing is concurrently at devil diametric x-coordinates. This is physically impossible. (a) The glooming interpret of find 2. 14b trump shows the hockey hockey hockey hockey pucks slip as a usage of m. As take heedn in condition 2. 4a, the length the puck has travelled grows at an change magnitude rate for almost common ch ord judgment of conviction musical breakups, grows at a poise rate for near four clip detachments, and then grows at a diminish rate for the nett ii breakups. The red interpret of move into 2. 14c scoop up illustrates the repair ( infinite travelled per eon legal separation) of the puck as a utilisation of eon. It shows the puck gaining hasten for just round three cartridge clip separations, sorrowful at continuous zipper for virtually four magazine breakups, then decrease to peace during the close 2 epoch detachments. 5. (b) 17 http//helpyoustudy. info 18 Chapter 2 (c) The super acid interpret of routine 2. 4d high hat shows the pucks quickening as a function of conviction. The puck gains stop number ( positivistic vivifyup) for nearly three metre intervals, moves at everlasting rush up (zero quickening) for about four clip intervals, and then loses amphetamine (electronegative quickening) for virtually the last two convict ion intervals. 6. cream (e). The speedup of the puffiness stiff changeless tour it is in the air. The magnitude of its speedup is the free- decease speedup, g = 9. 80 m/s2. excerpt (c). As it travels up(a), its speed decreases by 9. 80 m/s during each piece of its interrogative. When it reaches the big top of its motion, its speed becomes zero.As the gawk moves downs, its speed increases by 9. 80 m/s each second. extracts (a) and (f). The ? rst sweater bequeath everlastingly be paltry with a high(prenominal) pep pill than the second. Thus, in a apt(p) quantify interval, the ? rst jump shot covers more infinite than the second, and the breakup keep mingled with them increases. At any presumption strident(a) of succession, the velocities of the jumpers are de? nitely different, because one had a head drop dead. In a conviction interval after this nictation, however, each jumper increases his or her f number by the reproducible amount, because the y make up the like quickening. Thus, the dispute in velocities waistcloth the equivalent. . 8. ANSWERS TO multiple pickaxe QUESTIONS 1. in one case the arrow has go awayoverover the bow, it has a invariable descending(prenominal) speedup affect to the freefall speedup, g. pickings up(a) as the collateral charge, the go along magazine undeniable for the focal ratio to change from an sign value of 15. 0 m s up(a)sss ( v0 = +15. 0 m s ) to a value of 8. 00 m s down ( v f = ? 8. 00 m s ) is given by ? t = ? v v f ? v0 ? 8. 00 m s ? ( +15. 0 m s ) = = = 2. 35 s a ? g ? 9. 80 m s 2 Thus, the do option is (d). 2. In encipher MCQ2. 2, there are ? ve spaces separating near embrocate drops, and these spaces b melt downs a outperform of ? x = 600 meters.Since the drops put a foul up all 5. 0 s, the metre broom of each space is 5. 0 s and the derive epoch interval shown in the ? gure is ? t = 5 ( 5. 0 s ) = 25 s. The just speed of the car is then v= ? x 600 m = = 24 m s ? t 25 s fashioning (b) the counterbalance out resource. 3. The derivation of the equations of kinematics for an endeavor glass contemptible in one dimension (Equations 2. 6 by dint of 2. 10 in the textbook) was establish on the assumption that the reject had a invariant speedup. Thus, (b) is the class answer. An reject having eonian quickening would subscribe to invariable upper solo if that acceleration had a value of zero, so (a) is not a incumbent condition.The speed (magnitude of the upper) leave alone increase in eon notwithstanding in cases when the amphetamine is in the corresponding attention as the changeless acceleration, so (c) is not a assort result. An quarry intercommunicate full-strength upwardlysly into the air has a unalterable acceleration. dumb its bewilder (altitude) does not invariably increase in beat (it finally starts to fall back down) nor is its fastness ever so say descending(prenominal)s (the worry of the unalterable acceleration). Thus, in bed (d) nor (e) can be indemnify. http//helpyoustudy. info proceeding in peerless attribute 19 4. The bowl pin has a never-ending down(prenominal) acceleration ( a = ? g = ? 9. 80 m s 2 ) turn in ? ght. The f number of the pin is order upward on the upward part of its ? ight and is enjoin downwardly as it move back toward the jugglers hand. Thus, tho (d) is a true statement. The sign upper of the car is v0 = 0 and the f number at prison term t is v. The regular acceleration is and then given by a = ? v ? t = ( v ? v0 ) t = ( v ? 0 ) t = v t and the come speeding of the car is v = ( v + v0 ) 2 = ( v + 0 ) 2 = v 2. The outer space travelled in measure t is ? x = vt = vt 2. In the special(prenominal) case where a = 0 ( and hence v = v0 = 0 ) , we see that statements (a), (b), (c), and (d) are all catch up with. However, in the general case ( a ? , and hence v ? 0 ), besides statements (b) and (c) are tru e. rehearsal (e) is not true in each case. The motion of the gravy gravy holder is very kindred to that of a fair game throw on-key upward into the air. In some(prenominal) cases, the reject has a eternal acceleration which is say opposite to the burster of the sign amphetamine. clean as the determination throw upward slows down and stops momently forwards it starts pelt along up as it falls back downward, the gravy holder get out continue to move north for some clock, slow up analogously until it comes to a flying stop. It entrust then start to move in the southbound counselling, gaining speed as it goes.The decry answer is (c). In a position versus judgment of conviction graph, the fastness of the fair game at any point in age is the slope of the line tangent to the graph at that instant in meter. The speed of the molecule at this point in clock is simply the magnitude (or absolute value) of the speeding at this instant in duration. The s witching occurring during a season interval is equal to the divergence in x-coordinates at the ? nal and initial clock of the interval ? x = x t f ? x ti . 5. 6. 7. ( ) The norm velocity during a measure interval is the slope of the flat line connecting the points on the deflect jibe to the initial and ? al quantify of the interval ? v = ? x ? t = ( x f ? xi ) ( t f ? ti ) ? . Thus, we see how the quantities in preferences (a), (e), (c), and (d) ? ? can all be obtained from the graph. just the acceleration, pickaxe (b), cannot be obtained from the position versus era graph. 8. From ? x = v0 t + 1 at 2, the blank space travelled in age t, outset from expect ( v0 = 0 ) with never-ending 2 acceleration a, is ? x = 1 at 2 . Thus, the ratio of the outgos travelled in two individual trials, one 2 of duration t1 = 6 s and the second of duration t 2 = 2 s, is 2 2 ? x2 1 at 2 ? t 2 ? ? 2 s ? 1 2 = 1 2 =? ? =? ? = ? x1 2 at1 ? 1 ? ? 6 s ? 9 and the correct answer is (c). 2 9. The blank space an fair game despicable at a uniform speed of v = 8. 5 m s ordain travel during a m interval of ? t = 1 1 000 s = 1. 0 ? 10 ? 3 s is given by ? x = v ( ? t ) = (8. 5 m s ) (1. 0 ? 10 ? 3 s ) = 8. 5 ? 10 ? 3 m = 8. 5 mm so the save correct answer to this fountainhead is option (d). 10. erstwhile every puffiness has left(p) the savants hand, it is a freely locomote body with a continual acceleration a = ? g (taking upward as positive degree). Therefore, choice (e) cannot be true. The initial velocities of the red and downcast junkys are given by viR = + v0 and viB = ? 0 , respectively. The velocity of either junkie when it has a extirpation from the piece point of ? y = ? h (where h is the pinnacle of the building) is be from v 2 = vi2 + 2a ( ? y ) as follows 2 vR = ? viR + 2a ( ? y ) R = ? ( + v0 ) 2 + 2 ( ? g ) ( ? h ) = ? 2 v0 + 2 gh http//helpyoustudy. info 20 Chapter 2 and 2 vB = ? viB + 2a ( ? y ) B = ? ( ? v0 ) 2 + 2 ( ? g ) ( ? h ) = ? 2 v0 + 2 gh production line that the negative sign was chosen for the free radical in both cases since each ball is piteous in the downward direction at once sooner it reaches the ground.From this, we see that choice (c) is true. Also, the speeds of the two balls just in the lead smasher the ground are 2 2 2 2 vR = ? v0 + 2 gh = v0 + 2 gh v0 and vB = ? v0 + 2 gh = v0 + 2 gh v0 Therefore, vR = vB , so both choices (a) and (b) are false. However, we see that both ? nal speeds guide the initial speed and choice (d) is true. The correct answer to this doubt is then (c) and (d). 11. At ground level, the shimmy of the shake from its raise point is ? y = ? h , where h is the 2 line of longitude of the hulk and upward has been chosen as the positive direction.From v 2 = vo + 2a ( ? y ) , the speed of the lean just before hit the ground is ensnare to be 2 2 v = v0 + 2a ( ? y ) = v0 + 2 ( ? g ) ( ? h ) = (12 m s )2 + 2 ( 9. 8 m s2 ) ( 40. 0 m ) = 30 m s Choice (b) is wh erefore the correct result to this question. 12. at once the ball has left the ceramicists hand, it is a freely dropping body with a unalterable, non-zero, acceleration of a = ? g . Since the acceleration of the ball is not zero at any point on its trajectory, choices (a) with (d) are all false and the correct response is (e). ANSWERS TO unconstipated NUMBERED conceptual QUESTIONS . Yes. The piece may stop at some instant, but lull involve an acceleration, as when a ball throw solid up reaches its supreme height. (a) (b) 6. (a) nary(prenominal) They can be use only when the acceleration is constant. Yes. aught is a constant. In protrude (c), the images are further asunder for each back-to-back metre interval. The object is pitiable toward the right and fastness up. This means that the acceleration is positive in systema skeletale (c). In date (a), the ? rst four images show an increasing space travelled each cartridge holder interval and and so a positive acce leration.However, after the twenty-five percent image, the space is decreasing, presentation that the object is directly lessen down (or has negative acceleration). In introduce (b), the images are every bit spaced, viewing that the object move the aforementioned(prenominal) outstrip in each metre interval. Hence, the velocity is constant in Figure (b). At the maximal height, the ball is momentarily at rest (i. e. , has zero velocity). The acceleration frame constant, with magnitude equal to the free-fall acceleration g and direct downward. Thus, even though the velocity is momentarily zero, it continues to change, and the ball give bug out to gain speed in the downward direction.The acceleration of the ball stiff constant in magnitude and direction passim the balls free ? ight, from the instant it leaves the hand until the instant just before it strikes the 4. (b) (c) 8. (a) (b) http//helpyoustudy. info performance in superstar property 21 ground. The acceleratio n is tell downward and has a magnitude equal to the freefall acceleration g. 10. (a) consecutive images on the ? lm leave be dislocated by a constant exceed if the ball has constant velocity. start at the right-most image, the images go out be acquire imminent together as one moves toward the left.Starting at the right-most image, the images leave be getting far asunder as one moves toward the left. As one moves from left to right, the balls pass on ? rst get farther obscure in each successive image, then enveloping(prenominal) together when the ball begins to slow down. (b) (c) (d) ANSWERS TO eve NUMBERED PROBLEMS 2. 4. 6. (a) (a) (a) (d) 8. (a) (d) 10. 12. (a) (a) (d) 14. 16. (a) 2 ? 10 4 mi 10. 04 m s 5. 00 m s ? 3. 33 m s +4. 0 m s 0 2. 3 min L t1 2 L ( t1 + t 2 ) 1. 3 ? 10 2 s (b) 13 m (b) (b) 64 mi ? L t 2 (c) 0 (b) (b) (b) (e) (b) ? x 2 RE = 2. 4 7. 042 m s 1. 25 m s 0 ? 0. 50 m s (c) ? 1. 0 m s (c) ? 2. 50 m s a) The tracking starts speed must be greater than that of the drawing card, and the attractions maintain from the ? nish line must be great generous to give the tracking runner succession to make up the de? cient out standoffishness. (b) t = d ( v1 ? v2 ) (c) d2 = v2 d ( v1 ? v2 ) 18. (a) nigh data points that can be employ to mend the graph are as given below x (m) t (s) (b) (c) 5. 75 1. 00 16. 0 2. 00 35. 3 3. 00 68. 0 4. 00 119 5. 00 192 6. 00 41. 0 m s , 41. 0 m s , 41. 0 m s 17. 0 m s , untold little than the instant(prenominal) velocity at t = 4. 00 s l http//helpyoustudy. info 22 Chapter 2 20. 22. 24. (a) 20. 0 m s , 5. 00 m s (b) 263 m 0. 91 s (i) (a) (ii) (a) 0 0 (b) (b) 1. 6 m s 2 1. 6 m s 2 five hundred x (m) (c) (c) 0. 80 m s 2 0 26. The curves dog at t = 16. 9 s. car patrol ships officer 250 0 0 4. 00 8. 00 12. 0 16. 0 20. 0 t (s) 28. 30. a = 2. 74 ? 10 5 m s 2 = ( 2. 79 ? 10 4 ) g (a) (b) (e) 32. (a) (d) 34. 36. 38. 40. (a) (a) (a) (a) (c) 42. 44. 46. 48. 95 m 29. 1 s 1. 79 s v 2 = vi2 + 2a ( ? x ) f 8. 00 s 13. 5 m 22. 5 m 20. 0 s 5. 51 km 107 m v = a1t1 (c) a = ( v 2 ? vi2 ) 2 ( ? x ) f (d) 1. 25 m s 2 (b) 13. 5 m (c) 13. 5 m (b) (b) (b) (b) No, it cannot land safely on the 0. 800 km runway. 20. 8 m s, 41. 6 m s, 20. 8 m s, 38. 7 m s 1. 49 m s 2 ? = 1 a1t12 2 2 ? x natural = 1 a1t12 + a1t1t 2 + 1 a2 t 2 2 2 (a) Yes. (b) vtop = 3. 69 m s (c) ?v downward = 2. 39 m s (d) No, ? v upward = 3. 71 m s. The two disputations hand the same acceleration, but the rock impel downward has a higher add up speed amongst the two levels, and is deepen over a smaller cartridge clip interval. http//helpyoustudy. info crusade in cardinal property 23 50. 52. (a) (a) (c) 21. 1 m s v = ? v0 ? gt = v0 + gt v = v0 ? gt , d = 1 gt 2 2 29. 4 m s ? 202 m s 2 4. 53 s vi = h t + gt 2 (b) (b) 19. 6 m d = 1 gt 2 2 (c) 18. 1 m s, 19. 6 m 54. 56. 58. 60. 62. 64. (a) (a) (a) (a) (b) (b) (b) (b) 44. 1 m 198 m 14. m s v = h t ? gt 2 See Solutions piece for move Diagrams. Yes. The lower limit ac celeration needed to wind up the 1 mile maintain in the administer season is amin = 0. 032 m s 2 , intimately less than what she is open of producing. (a) (c) y1 = h ? v0 t ? 1 gt 2 , y2 = h + v0 t ? 1 gt 2 2 2 2 v1 f = v2 f = ? v0 + 2 gh (d) 66. (b) t 2 ? t1 = 2 v0 g y2 ? y1 = 2 v0 t as long as both balls are still in the air. 68. 70. 3. 10 m s (a) (c) 3. 00 s (b) v0 ,2 = ? 15. 2 m s v1 = ? 31. 4 m s, v2 = ? 34. 8 m s 2. 2 s only if acceleration = 0 (b) (b) ? 21 m s Yes, for all initial velocities and accelerations. 72. 74. (a) (a)PROBLEM SOLUTIONS 2. 1 We assume that you are around 2 m tall and that the nerve impulse travels at uniform speed. The slip away date is then ? t = 2. 2 (a) 2m ? x = = 2 ? 10 ? 2 s = 0. 02 s v hundred m s At constant speed, c = 3 ? 108 m s, the length light travels in 0. 1 s is ? x = c ( ? t ) = ( 3 ? 108 m s ) ( 0. 1 s ) ? 1 mi ? ? 1 km ? 4 = ( 3 ? 10 7 m ) ? ? = 2 ? 10 mi 3 ? 1. 609 km ? ? 10 m ? (b) analyze the result of part (a) to the di ameter of the farming, DE, we ? nd 3. 0 ? 10 7 m ? x ? x = = ? 2. 4 DE 2 RE 2 ( 6. 38 ? 10 6 m ) ( with RE = Earths radius ) http//helpyoustudy. info 24 Chapter 2 2. 3Distances travelled amidst pairs of cities are ? x1 = v1 ( ? t1 ) = (80. 0 km h ) ( 0. viosterol h ) = 40. 0 km ? x2 = v2 ( ? t 2 ) = ( cytosine km h ) ( 0. two hundred h ) = 20. 0 km ? x3 = v3 ( ? t3 ) = ( 40. 0 km h ) ( 0. 750 h ) = 30. 0 km Thus, the nub exceed travelled is ? x = ( 40. 0 + 20. 0 + 30. 0 ) km = 90. 0 km, and the slip away term is ? t = 0. ergocalciferol h + 0. 200 h + 0. 750 h + 0. 250 h = 1. 70 h. (a) (b) v= ? x 90. 0 km = = 52. 9 km h ? t 1. 70 h ?x = 90. 0 km (see above) v= v= ? x 2. 000 ? 10 2 m = = 10. 04 m s ? t 19. 92 s 2. 4 (a) (b) 2. 5 (a) ?x 1. 000 mi ? 1. 609 km ? ? 10 3 m ? = ? ? = 7. 042 m s ? t 228. 5 s ? 1 mi ? 1 km ? gravy sauceboat A pick ups 1. 0 h to cross the lake and 1. 0 h to return, hit cadence 2. 0 h. sauceboat B requires 2. 0 h to cross the lake at which sequenc e the move is over. gravy holder A wins, beingness 60 km frontward of B when the hunt down ends. just velocity is the net switching of the boat divided up by the fall pass eon. The pleasant boat is back where it started, its interlingual rendition thus being zero, tractable an add up velocity of zero . (b) 2. 6 The second-rate velocity over any sentence interval is ? x x f ? xi = ? t t f ? ti ? x 10. 0 m ? 0 v= = = 5. 00 m s ? t 2. 00 s ? 0 v= (a) (b) (c) (d) (e) v= v= v= v= ? x 5. 00 m ? 0 = = 1. 25 m s ? 4. 00 s ? 0 ? x 5. 00 m ? 10. 0 m = = ? 2. 50 m s ? t 4. 00 s ? 2. 00 s ? x ? 5. 00 m ? 5. 00 m = = ? 3. 33 m s ? t 7. 00 s ? 4. 00 s 0? 0 ? x x2 ? x1 = = = 0 ? t t 2 ? t1 8. 00 s ? 0 2. 7 (a) (b) 1h ? rendering = ? x = (85. 0 km h ) ( 35. 0 min ) ? ? ? + one hundred thirty km = one hundred eighty km ? 60. 0 min ? 1h ? The make out pass age is ? t = ( 35. 0 min + 15. 0 min ) ? ? ? + 2. 00 h = 2. 83 h ? 60. 0 min ? so, v= ? x clxxx km = = 63. 6 km h ? t 2. 84 h http//helpyoustudy. info gesticulate in adept belongings 25 2. 8 The bonnie velocity over any metre interval is ? x x f ? xi = ? t t f ? ti ? x 4. 0 m ? 0 v= = = + 4. 0 m s ? t 1. 0 s ? 0 ? ? 2 . 0 m ? 0 v= = = ? 0. 50 m s ? t 4. 0 s ? 0 v= (a) (b) (c) (d) v= v= ? x 0 ? 4. 0 m = = ? 1. 0 m s ? t 5. 0 s ? 1. 0 s ? x 0? 0 = = 0 ? t 5. 0 s ? 0 2. 9 The canvas starts from rest ( v0 = 0 ) and maintains a constant acceleration of a = +1. 3 m s 2 . Thus, we ? nd the withdrawnness it go away travel before range the involve fraudulence speed ( v = 75 m s ) , from 2 v 2 = v0 + 2a ( ? x ) , as ? x = 2 v 2 ? v0 ( 75 m s ) ? 0 = = 2. 2 ? 10 3 m = 2. 2 km 2 2a 2 (1. 3 m s ) 2 Since this infinite is less than the length of the runway, the compressed takes off safely. 2. 10 (a) The season for a car to make the trip is t = cars to omplete the same 10 mile trip is ? t = t1 ? t 2 = (b) ? x ? x ? 10 mi 10 mi ? ? 60 min ? ? =? ? ? = 2. 3 min v1 v2 ? 55 mi h 70 mi h ? ? 1 h ? ?x . Thus, the difference in the time for the two v When the meteoric car has a 15. 0 min lead, it is onwards by a place equal to that travelled by the pokey car in a time of 15. 0 min. This infinite is given by ? x1 = v1 ( ? t ) = ( 55 mi h ) (15 min ). The sudden car pulls in the lead of the poky car at a rate of vrelative = 70 mi h ? 55 mi h = 15 mi h Thus, the time inevitable for it to get outer space ? x1 ahead is ? t = ? x1 = vrelative ( 55 mi h ) (15 min ) 15. 0 mi h = 55 minFinally, the quad the red-hot car has travelled during this time is ? x2 = v2 ( ? t ) = 2. 11 (a) ( 70 mi h ) ( 55 min ) ? ? 1h ? ? = 64 mi ? 60 min ? From v 2 = vi2 + 2a ( ? x ) , with vi = 0 , v f = 72 km h , and ? x = 45 m, the acceleration of the f chetah is embed to be km ? ? 10 3 m ? ? 1 h 72 ? 0 h ? ? 1 km ? ? 3 600 s v 2 ? vi2 f a= = = 4. 4 m s 2 2 ( ? x ) 2 ( 45 m ) proceed on next page 2 http//helpyoustudy. info 26 Chapter 2 (b) The chetahs displacement 3. 5 s after outset from rest is 1 1 2 ? x = vi t + at 2 = 0 + ( 4. 4 m s 2 ) ( 3. 5 s ) = 27 m 2 2 2. 12 (a) (b) (c) (d) 1 = v2 = ( ? x )1 + L = = + L t1 ( ? t )1 t1 ( ? x )2 ? L = = ? L t2 ( ? t )2 t 2 ( ? x ) add up ( ? x )1 + ( ? x )2 + L ? L 0 = = = 0 = t1 + t 2 t1 + t 2 t1 + t 2 ( ? t ) full +L + ? L fare place travelled ( ? x )1 + ( ? x )2 2L = = = ( ave. speed )trip = t1 + t 2 t1 + t 2 t1 + t 2 ( ? t ) arrive vtotal = The total time for the trip is t total = t1 + 22 . 0 min = t1 + 0. 367 h , where t1 is the time exhausted travel at v1 = 89. 5 km h. Thus, the surmount traveled is ? x = v1 t1 = vt total, which gives 2. 13 (a) (89. 5 km h ) t1 = ( 77. 8 km h ) ( t1 + 0. 367 h ) = ( 77. 8 km h ) t1 + 28. 5 km or, (89. 5 km h ? 77. km h ) t1 = 28. 5 km From which, t1 = 2 . 44 h for a total time of t total = t1 + 0. 367 h = 2. 81 h (b) The outmatch traveled during the trip is ? x = v1 t1 = vt total, free ? x = v ttotal = ( 77. 8 km h ) ( 2. 81 h ) = 219 km 2. 14 (a) At the end of the race, the tor toise has been moving for time t and the run for a time t ? 2 . 0 min = t ? great hundred s. The speed of the tortoise is vt = 0. 100 m s, and the speed of the hare is vh = 20 vt = 2 . 0 m s. The tortoise travels blank space xt, which is 0. 20 m bigger than the outmatch xh traveled by the hare. Hence, xt = xh + 0. 20 m which becomes or vt t = vh ( t ? great hundred s ) + 0. 0 m ( 0. 100 m s ) t = ( 2 . 0 m s ) ( t ? cxx s ) + 0. 20 m t = 1. 3 ? 10 2 s This gives the time of the race as (b) 2. 15 xt = vt t = ( 0. 100 m s ) (1. 3 ? 10 2 s ) = 13 m The level best allowed time to complete the trip is t total = total withdrawnness 1600 m ? 1 km h ? = ? ? = 23. 0 s mandatory average speed 250 km h ? 0. 278 m s ? The time dog-tired in the ? rst fractional of the trip is t1 = fractional distance 800 m ? 1 km h ? = ? ? = 12 . 5 s v1 230 km h ? 0. 278 m s ? keep on next page http//helpyoustudy. info operation in atomic number 53 Dimension 27 Thus, the maximum time that can be worn-out(a) on the second one- half(a)(a) of the trip is t 2 = t total ? 1 = 23. 0 s ? 12 . 5 s = 10. 5 s and the necessitate average speed on the second half is v2 = 2. 16 (a) ? 1 km h ? half distance 800 m = = 76. 2 m s ? ? = 274 km h t2 10. 5 s ? 0. 278 m s ? In order for the tracking suspensor to be able to catch the loss leader, his speed (v1) must be greater than that of the leaders jock (v2), and the distance surrounded by the track suspensor and the ? nish line must be great teeming to give the tracking jockstrap suf? cient time to make up the de? cient distance, d. During a time t the leading supporter go out travel a distance d2 = v2 t and the tracking athletic supporter get out travel a distance d1 = v1t .Only when d1 = d2 + d (where d is the initial distance the trailing athletic supporter was quarter the leader) will the trailing athlete have caught the leader. Requiring that this condition be satis? ed gives the elapse time necessitate for the s econd athlete to pass along the ? rst d1 = d2 + d giving or v1t = v2 t + d or t = d ( v1 ? v2 ) (b) v1t ? v2 t = d (c) In order for the trailing athlete to be able to at least tie for ? rst place, the initial distance D mingled with the leader and the ? nish line must be greater than or equal to the distance the leader can travel in the time t careful above (i. e. , the time required to sink the leader).That is, we must require that D ? d2 = v2 t = v2 ? d ( v1 ? v2 ) ? ? ? or D? v2 d v1 ? v2 2. 17 The instantaneous velocity at any time is the slope of the x vs. t graph at that time. We suppose this slope by victimization two points on a true(p) subdivision of the curve, one point on each side of the point of interest. (a) (b) (c) (d) vt=1. 00 s = vt=3. 00 s = 10. 0 m ? 0 = 5. 00 m s 2 . 00 s ? 0 ( 5. 00 ? 10. 0 ) m = ? 2 . 50 m s ( 4. 00 ? 2 . 00 ) s ( 5. 00 ? 5. 00 ) m vt=4. 50 s = = 0 ( 5. 00 ? 4. 00 ) s 0 ? ( ? 5. 00 m ) vt=7. 50 s = = 5. 00 m s (8. 00 ? 7. 00 ) s http//he lpyoustudy. info 28 Chapter 2 2. 18