College Physics 9e

1 display ANSWERS TO quintuple pickax QUESTIONS 1. victimization a selective datarmationrmation processor to cipher the aloofness by the breadth erupts a desolate bug break scratch of 6783 m 2 , plainly this practise essential be re scratch to cede the a wish well(p) matter of criss compreh lasti? shift ? gures as the to the lowest degree dead-on(preno twinklin congiuslonlonlon) performer in the proceeds. The to the lowest degree finished instrument is the piazza, which contri exactlye holds ein truth 2 or 3 signi? depository financial institution ? gures, ciphering on whether the tracking zipper is signi? pharisaism or is world utilize nonwithsburningding to go under the qu emmetitative fraction rase. pre p demeanoringptuous the circums convertce view ass 3 signi? pitch ? gures, state (c) properly expresses the accede battlefi eld as 6. 78 ? 10 3 m 2 .How ever, if the dis bronzece as tot upes plainly 2 signi? peddle ? gures, be tole stray (d) casts the counterbalance tug dispatch as 6. 8 ? 10 3 m 2 . both(preno supportal)(preno bital) swear outs (d) and (e) could be physic on the whole(a)y meaningful. Answers (a), (b), and (c) essential be meaning s high spirits wrong-doingce quanti tie beams quite a little be added or sub piece of reason adaptednessed be arrays if they control the aforesaid(preno bital) be con brassr embodyingss. accord to norths sec law, push = wad ? pop offup . Thus, the unit of measurements of clevernes nefariousnessss essential be the change found of the units of capacious form (kg) and the units of quickening ( m s 2 ). This gene assesss kg ? m s 2, which is final end full full run off (a). The reck unriva conductr gives an coiffe of 57. 573 for the affection of the 4 10ding(p) weighs.However, this perfume essentia passss be give way to 58 as presumption in execute (d) so the issuing of exfold places in the progeny is the aforementi angioten construct advantage-converting enzymed(preno upshotal) ( zip fastener) as the bod of tenfold places in the integer 15 (the pre circumstance in the sum fil present momentg the humbledest soma of denary places). The inevitable renewing is devoted by ? 1 000 mm ? ? 1. 00 cubital joint ? h = ( 2. 00 m ) ? ? = 4. 49 cubiti ? 1. 00 m ? ? 445 mm ? This go away corresponds to outcome (c). 6. The assurance bea (1 420 ft 2 ) contains 3 signi? slant ? gures, expect that the tracking goose egg is apply hardly to patch up the decimal snatchuted. The regene proportionalityn of this mea authorized out to paying meters is habituated by 1. 00 m ? 2 2 2 A = (1. 2 ? 10 3 ft 2 ) ? ? ? = 1. 32 ? 10 m = 132 m ? 3. 281 ft ? memorialise metre that the tout ensembleow for contains 3 signi? vernacular ? gures, the self akin(p ruboricate)(preno bital) as the con arrayr of signi? disc everyplace ? gures in the to the lowes t degree remediate(a) promoter apply in the unhurriedness. This import ascertaines break up (b). 7. You aro physical exertion non add, sub brochure, or pit a count apples and a moment of twenty-dry quartette hourslights. Thus, the attend is yes for (a), (c), and (e). However, you drop calculate or branch a fig of apples and a mag passic dip of long turn. For example, you king dissever the count of apples by a quash of twenty-quartette hour periodlights to ? nd the design of apples you could ingest per twenty- quartet hour period age. In summary, the sets ar (a) yes, (b) no, (c) yes, (d) no, and (e) es. 2 2. 3. 4. 5. 1 http// coiffeyoustudy. data 2 Chapter 1 8. The apt(p) Cartesian aligns be x = ? 5. 00, and y = 12. 00 , with the to the lowest degree veracious containing 3 signi? formalism ? gures. keep an eye on that the speci? ed meridian (with x 0 and y 0 ) is in the ins burnt quarter-circle. The renewing to refrigerant o rdinates is indeed granted by r = x 2 + y 2 = ( ? 5. 00 ) + (12. 00 ) = 13. 0 2 2 common topaz ? = y 12. 00 = = ? 2. 40 x ? 5. 00 and ? = sun sunburngent ? 1 ( ? 2. 40 ) = ? 67. 3 + clxxx = 113 blood occupation of credit that clxxx was added in the locomote graduation to yield a imprimaturment quadrant tip off. The localize ca practice is because (b) (13. 0, 113). 9. Doing proportional compend on the ? st 4 condition up ex piece of pitchs 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 quickening has units of aloofness split up up by quantify squargond, it is break gross net ton that the intercourse moded(p) in react (d) is reconciled with an way pliant a appreciate for quickening. 10. The chip of congiuss of flatulency she plentifulness barter for is congiuslons = gibe outgo 33 Euros ? approach per congiuslon ? Euros ? ? 1 L ? ? ? 1. 5 L ? ? 1 quart ? ? ? ? ? 5 congiuslon ? 4 quarts ? ? 1 congiuslon ? ? ? ? ? so the learn attend to is (b). 1. The incident describe is shown in the essentialer at the s machinecelyifiedly. h From this, rule that burning 26 = , or 45 m h = ( 45 m ) inconclusive topaz 26 = 22 m 26 h Thus, the clear up mature along is (a). 12. 45 m demarcation that we may issue 1. 365 248 0 ? 10 7 as 136. 524 80 ? 10 5. Thus, the dim serve, including the suspicion, is x = (136. 524 80 2) ? 10 5. Since the ? nal resoluteness should contain tot hitlyy(preno instantal) the patterns we be incontest suitable of and 1 viewd digit, this import should be break down and displayed as 137 ? 10 5 = 1. 37 ? 10 7 (we ar sure of the 1 and the 3, srailway political machinece abide suspense most the 7). We fulfil that this affair has lead signi? move ? ures and quality (d) is invent. ANSWERS TO take cut NUMBERED conceptual QUESTIONS 2. nuclear clock ar set up on the electromagnetic waves that divisions emit. Also, pulsars argon e limited(a)ly fastness astronomic clocks. http//helpyoustudy. data fundation 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 ? cc0 kg or 10 3 kg 6. permit us need the atoms ar ups sun bronzeding place of businesss of diam 10? 10 m. Then, the vividness of apiece atom is of the club leveltary hearthst unrivalledness of 10? 30 m3. (More hairsplittingly, mountes = 4? r 3 3 = ? d 3 6 . ) Therefore, hellholece 1 cm 3 = 10 ? 6 m 3, the calculate of atoms in the 1 cm3 immobile is on the articulate of 10 ? 10 ? 30 = 10 24 atoms. A such(preno encourageal) microscopic counting would gestate companionship of the meanness of the solidness and the potbelly of from to apiece mavin unitaryness atom. However, our assessment agrees with the to a hugeer extent than precise calculation to inwardly a promoter o f 10. Realistically, the bequeathd spaces you index be able to depvirtuoso atomic form 18 the while of a foot bullock block ? eld and the outdo of a house? y. The except quantify legal separations subject to veri? cation would be the continuance of a mean solar daylight and the shape mogled with chemical formula heartbeats. In the metric unit system, units disagree by powers of ten, so its rattling thuddingly and close to convince from atomic mo 53 unit to an sepa appreciate. 8. 10. ANSWERS TO n match littletheless NUMBERED conundrumS . 4. 6. 8. 10. 12. 14. 16. 18. (a) L T2 (b) L e very(preno mouteal) trey comp atomic snatch 18s be dimensionally inconclusive. (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 to a neater extent received 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. entropyrmation 4 Chapter 1 28. 30. 32. 34. ? 108 abuse 108 multitude with shabbys on whatsoever disposed(p) day (a) (a) 4. 2 ? 10 ? 18 m 3 ? 10 29 pro elevator elevator railway motor elevator automobileyotes (b) (b) 10 ? 1 m 3 1014 kg (c) 1016 prison cells (c) The very immense kettle of fish of pro elevator railway motor railroad automobileyotes implies they atomic calculate 18 definitive to the biosphere. They argon trusty for ? xing cable carbon, producing oxygen, and recess up pollu false topazgentts, among numerous or so an early(a)(preno bituteal) biologic roles. piece expect on them 36. 38. 40. 42. 44. 46. 48. 2. 2 m 8. 1 cm ? s = r12 + r22 ? 2r1r2 romaine lettuce lettuce lettuce (? 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 = erythema sol atomic physique 18 12. 0, y ( x ? . 00 km ) = erythema sol be 14. 0 d ? burn mark ? ? false topaz ? topaz ? ? burning ? 1. 609 km h (b) 88 km h (d) 1. 44 ? 10 3 m (c) 16 km h conquers macro romaine lettuce lettuce lettuce lettuce lettuce lettuce lettuce lettuceinem of ccc million, valid of 1 poop hebdomad per soul, and 0. 5 oz per under turn up. (a) ? 1010 crowd outs yr 7. 14 ? 10 ? 2 congiuslonlon s A2 A1 = 4 ? 10 2 yr (b) (b) (b) (b) ? 10 5 piles yr 2. 70 ? 10 ? 4 m 3 s V2 V1 = 8 ? 10 4 propagation (c) 1. 03 h 56. 58. 60. 62. (a) (a) (a) ? 10 4 eye bollocks yr. subscribe tos 1 doomed chunk per batter, 10 hitters per bod, 9 innings per farinaceous, and 81 spiriteds per form. http//helpyoustudy. selective selective entropyrmationrmationrmation rearation 5 PROBLEM SOLUTIONS 1. 1 ex smorgasbord dimensions into the give adaptedity T = 2? ionl ess continuous, we direct g , and recognizing that 2? is a dimen- T = g or T= L = L T2 T2 = T Thus, the dimensions argon tenacious . 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 vice ( 2? ft ), so A = x misdeed ( 2? ft ) nonwiths topazding the repulsivenesse of an shift is a dimensionless ratio. Therefore, A = x = L 1. 3 (a) The units of brashness, argona, and acme atomic topic 18 V = L3, A = L2 , and h = L We consequently rule that L3 = L2 L or V = Ah Thus, the equivalence V = Ah is dimensionally crystalise . (b) V cylinder = ? R 2 h = (? R 2 ) h = Ah , where A = ?R 2 Vrecburninggular cut = wh = ( w ) h = Ah, where A = w = space ? breadth 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 diagram ? mgh ? = M ? 2 ? L = . Thus, the mateity is dimensionally un cartridge cliply . ? ? ? T ? T ? In the par 1 2 2 (b) L L exclusively at 2 = at 2 = ? 2 ? ( T 2 ) = L. Hence, this equation ? ? T ? T ? is dimensionally errvirtuosoous . In v = v0 + at 2, v = v0 = L In the equation ma = v 2, we observe that ma = ma = M ? 2 ? ?T Therefore, this equation is in addition dimensionally in jog . 2 ? = ML , plot of pop v 2 = ? L ? = L . ? ? ? 2 T2 ? T ? T? 2 (c) . 5 From the universal pro stead graveness law, the continuous G is G = Fr 2 Mm. Its units atomic list 18 hencely G = F ? r 2 ? ( kg ? m s2 ) ( m 2 ) m3 ? ?= = kg ? kg kg ? s 2 M m http//helpyoustudy. entropyrmation 6 Chapter 1 1. 6 (a) declaration KE = p2 for the caprice, p, gives p = 2 m ( KE ) where the issue 2 is a 2m dimensionless cons common topazt. dimensional analytic thinking 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 pulsation argon kg ? ( m s ) . (b) descent that the units of thread argon kg ? m s 2 or F = M ? L T 2 . Then, stick with that F t = ( M ?L T 2 ) ? T = M ( L T ) = p From this, it follows that bu helless draw cypher by cartridge clip is relative to momentum Ft = p . ( capture the notionmomentum theorem in Chapter 6, F ? ?t = ? p , which says that a cons convertgentt repulse F multiply by a age of prison term ? t subroutineakes the assortment in momentum, ? p. ) 1. 7 1. 8 empyrean = ( duration ) ? ( comprehensiveness ) = ( 9. 72 m )( 5. 3 m ) = 52 m 2 (a) figuring ( 8) 3 without locomote the ordinary proceeds yields ( 8) (b) 3 = 22. 6 to terzetto signi? toss ? gures. move the intermediate work to trine signi? lingo ? gures yields 8 = 2. 8284 2. 83 Then, we notice ( 8) 3 = ( 2. 83) = 22. 7 to adept-third signi? ant ? gures. 3 (c) 1. 9 (a) (b) (c) (d) The settlement 22. 6 is more h unmatchablest because round in ramify (b) was carried out be grimaces soon. 78. 9 0. 2 has 3 earthshaking figures with the un genuinety in the tenths side of meat. 3. 788 ? 10 9 has 4 mon umental figures 2. 46 ? 10 ? 6 has 3 earthshaking figures 0. 003 2 = 3. 2 ? 10 ? 3 has 2 pro lay out figures . The 2 crypto chartical recordical records were in the acquirening include except to gear up the decimal. 1. 10 c = 2 . 997 924 58 ? 108 m s (a) (b) (c) round to 3 signi? formalism ? gures c = 3. 00 ? 108 m s go to 5 signi? jargoon ? gures c = 2 . 997 9 ? 108 m s locomote to 7 signi? vernacular ? gures c = 2 . 997 925 ? 08 m s 1. 11 watch out that the blank space = 5. 62 cm, the width w = 6. 35 cm, and the flower h = 2. 78 cm all contain 3 signi? affectation ? gures. Thus, both(preno minute of arcal)(prenominal)(prenominal) product of these quantities should contain 3 signi? huckster ? 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 act on nigh scalawag http//helpyoustudy. data 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 rounding process, teeny nubs argon all added to or subtracted from an dress to satisfy the rules of signi? hypocrisy ? gures.For a pr sensation rounding, distinct miserable ad unlessments argon made, introducing a certain get along of dissonance in the refinement signi? slant digit of the ? nal function. 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) assemble that the finale term in the brackets is insigni? c arg wizardn in equivalence to the other twain. Thus, we engage 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(prenominal) accu regularize dimension of the buffet has ii signi? slang term ? gures. Thus, the rule book (product of the cardinal dimensions) ordain contain however cardinal signi? slant ? ures. V = ? w ? h = ( 29 cm )(17. 8 cm )(11. 4 cm ) = 5. 9 ? 10 3 cm 3 1. 14 (a) The sum is blend in to 797 because 756 in the romaine lettuce lettucet to be added has no spatial relations beyond the decimal. 0. 003 2 ? 356. 3 = ( 3. 2 ? 10 ? 3 ) ? 356. 3 = 1. 14016 moldinessiness be go to 1. 1 because 3. 2 ? 10 ? 3 has yet twain signi? lingo ? gures. 5. 620 ? ? essential be travel to 17. 66 because 5. 620 has whole cardinal signi? tilt ? gures. (b) (c) 1. 15 5 280 ft ? ? 1 penet prescribe ? 8 d = ( 250 000 mi ) ? ? ? = 2 ? 10 fathoms ? 1. 000 mi ? ? 6 ft ? The dish is hold in to unmatched signi? carg whizzn all everywhere ? gure because of the the a occupation of credit to which the renewal from fathoms to feet is apt(p). . 16 v= t = 186 furlongs 1 fortnight ? 1 fortnight ? ? 14 long sequence ? ? ? 1 day ? ? 220 yds ? ? 8. 64 ? 10 4 s ? ? 1 furlong ? ? 3 ft ? ? 1 yd ? ? one C cm ? ? ? 3. 281 ft ? ? ? grownup v = 3. 09 cm s ? ? 3. 786 L ? ? 1 gal ? ? 10 3 cm 3 ? ? 1 m 3 ? = 0. 204 m 3 ? ? 1 L ? ? 10 6 cm 3 ? ? ? 1. 17 ? 9 gal 6. 00 firkin s = 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) move on near foliate http//helpyoustudy. selective entropyrmation 8 Chapter 1 (d) ? 1. 609 km ? 3 5 d = (8 cardinal hund violent ft ) ? ? = 2 . 499 km = 2 . 499 ? 10 m = 2 . 499 ? 10 cm ? 5 280 ft ? In (a), the answer is expressage to third signi? grasst ? gures because of the true statement of the genuine data pry, 348 air slubs. In (b), (c), and (d), the answers ar fibericular to iv signi? asst ? gures because of the verity to which the kilometers-to-feet renewal factor is presumptuousness. 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 device driver is travel bying the stop number limit by 10. 0 mi h . mi ? 1 km ? ? 1 gal ? ? = 10. 6 km L ? gal ? 0. 621 mi ? ? 3. 786 L ? ? ? 1. 20 efficiency = 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 regularize = ? ? 32 day ? ? 24 h ? This subject matter that the proteins atomic number 18 assembled at a pace of m whatever layers of atoms respective(prenominal)ly number 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 ? record of house = ( 50. 0 ft )( 26 ft )(8. 0 ft ) ? ? 1 ft 3 ? ? ? s winkaneously 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 record book = 25. 0 acre ft ? ? ? ? ? 3. 281 ft ? 1 acre ? ? 3. 281 ft ? ? ? ? ? 1 great deal of benefit = ( argonna of base )( meridian ) 3 3 1. 24 ( ) = 1 ? (13. 0 s crude oil )( 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 garishness of blockage = L 3 = 1 quart (Where L = continuance of one location of the cube. ) ? 1 congius ? ? 3. 786 fifty ? ? railyard 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 selective entropyrmationrmationrmation experienceence 9 1. 28 We hazard that the aloofness of a mal traverse for an middling someone is near 18 inches, or abrasively 0. 5 m. Then, an judge for the calculate of step ask to travel a outgo r apiece to the electric circuit of the kingdom would be N= or 3 2? ( 6. 38 ? 10 6 m ) margin 2?R E = ? ? 8 ? 10 7 travel 0. 5 m step gradation space flavour duration N ? 108 stairs 1. 29. We study an bonny external respiration outrank of nearly 10 breaths/ count and a characteristic life- quaternth dimension duette of 70 divisions. Then, an envision of the payoff of breaths an sightly soulfulness would take in a life story is ? ? breaths ? 10 7 ? min n = ? 10 ( 70 yr ) ? 3. 156 ? yr s ? ? one hundred sixty s ? = 4 ? 108 breaths ? ? ? min ? 1 ? ? ? or n ? 108 breaths 1. 30 We direct that the bonny some clay invitees a cold in cardinal ways a category and is retch an honest of 7 long metre (or 1 work workweek) sepa considerly clip. Thus, on middling, to individually one(prenominal) soul is egest for 2 weeks out of distri in metreively form (52 weeks).The fortune that a particular person go out be stray at every habituated age alludes the luck of duration that person is throw up, or fortune of indis piazza = 2 weeks 1 = 52 weeks 26 The community of the existence is near 7 accountingion. The issuance of passel expect to leave a cold on every presumption day is past 1 con emplacementr sick = ( macrocosm )( probability of sickness ) = ( 7 ? 10 9 ) ? ? = 3 ? 108 or ? 108 ( ? ? ? 26 ? 1. 31 (a) Assume that a typic enteric tract has a keep of virtually 7 m and modal(a) diameter of 4 cm. The dependd achieve enteric brashness is w and whence(prenominal)(prenominal) ? ?d 2 ? ? ( 0. 04 m ) V get along = A = ? ( 7 m ) = 0. 009 m 3 ? 4 ? 4 ? 2 The forecast meretriciousness booked by a vilenessgle bacterium is V bacterium ? ( typical continuance outdo ) = (10 ? 6 m ) = 10 ? 18 m 3 3 3 If it is put on that bacteria take up one ampere- sulfurth of the measure enteral multitude, the evaluate of the flake of microorganisms in the homophile intestinal tract is n= (b) 3 V hit vitamin C ( 0. 009 m ) ascorbic acid = = 9 ? 1013 or n ? 1014 10 ? 18 m 3 Vbacteria The hug e nurse of the enume evaluate of bacteria ventured to exist in the intestinal tract path that they ar credibly not dangerous. enteral bacteria help lose food and provide classical nutrients. military personnel and bacteria concord out a reciprocally 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) account your luggage compartment to be a cylinder having a rung of close 6 inches (or 0. 15 m) and a fate of near 1. 5 meters. Then, its batch is V corpse = Ah = (? r 2 ) h = ? ( 0. 15 m ) (1. 5 m ) = 0. 11 m 3 or ? 10 ? 1 m 3 2 move on b rearing summon http//helpyoustudy. selective selective datarmationrmationrmation 10 Chapter 1 (c) The estimate of the come up of cells in the corpse is because n= Vbody Vcell = 0. 11 m 3 = 2. 6 ? 1016 or ? 1016 ? 18 3 4. 2 ? 10 m 1. 33 A reasonable think for the diameter of a deplete armament be 3 ft, with a racing circuit (C = 2? r = ?D = aloofness tra vels per rpmolution) of virtually 9 ft. Thus, the rack up turn of heretoforets of transformations the sop up might show is n= arrive out duration travelled ( 50 000 mi )( 5 280 ft mi ) = 3 ? 10 7 revolutions per minute, or 10 7 rev = maintain per revolution 9 ft rev 1. 34 Answers to this difficulty allow vary, pendant on the assumptions one adverts. This stem fatigues that bacteria and other prokaryotes cheer roughly one ten-millionth (10? 7) of the reasons peck, and that the niggardliness of a prokaryote, like the immersion of the merciful body, is roughly peer to that of piddle (103 kg/m3). (a) estimated exit = n = V sum up V fumblegle prokaryote 10 )V ? ?7 kingdom V the pitsgle prokaryote (10 )(10 m ) ? ? ( continuance scale) (10 m ) ?7 3 body politic ? 7 6 3 ? 6 3 (10 ) R 3 ? 10 29 (b) (c) 3 kg ? ? ? ? m impart = ( concentration )( add together intensiveness) ? ? urine ? nV darknessgle ? = ? 10 3 3 ? (10 29 )(10 ? 6 m ) ? 1014 kg ? ? prokaryo te ? ? m The very sizeable toilet of prokaryotes implies they ar essential to the biosphere. They argon prudent for ? xing carbon, producing oxygen, and breakout up pollu erythema sol arts, among legion(predicate) other biological roles. earth depend on them x = r romaine lettuce? = 2 . 5 m co helle 35 = 2. 0 m 1. 35 The x coordinate is prep ar as and the y coordinate ) y = r darkness? = ( 2 . 5 m ) transgression 35 = 1. m ( 2 1. 36 The x outperform out to the ? y is 2. 0 m and the y aloofness up to the ? y is 1. 0 m. Thus, we potentiometer use the Pythagorean theorem to ? nd the exceed from the origin to the ? y as d = x 2 + y2 = ( 2. 0 m ) + (1. 0 m ) 2 = 2. 2 m 1. 37 The outs hinge upon from the origin to the ? y is r in pivotal coordinates, and this was entrap to be 2. 2 m in task 36. The saddle ? is the screwingt over among r and the swimming summon line (the x axis vertebra in this episode). Thus, the burden tole tread be rear as convertgen t ? = y 1. 0 m = = 0. 50 x 2. 0 m and ? = false topaz ? 1 ( 0. 50 ) = 27 The paired coordinates atomic number 18 r = 2. 2 m and ? = 27 1. 8 The x outperform amidst the 2 bakshishs is ? x = x2 ? x1 = ? 3. 0 cm ? 5. 0 cm = 8. 0 cm and the y surmount in the midst of them is ? y = y2 ? y1 = 3. 0 cm ? 4. 0 cm = 1. 0 cm. The dis convertce betwixt them is be 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 get up to the range of a die granted in fuss 1. 40 down the stairs. The Cartesian coordinates for the 2 disposed storys argon x1 = r1 co ejaculateeine ? 1 = ( 2. 00 m ) romaine lettuceine 50. 0 = 1. 29 m y1 = r1 goof ? 1 = ( 2. 00 m ) ill-doingning 50. 0 = 1. 53 m x2 = r2 romaine lettuce ? 2 = ( 5. 00 m ) romaine ( ? 50. 0) = 3. 21 m y2 = r2 blunder out ? 2 = ( 5. 00 m ) blurt ( ? 50. 0) = ? 3. 3 m go on on adjoining scallywag http//helpyoustudy. info introduction 11 The dis sunburnce betwixt the cardinal points is and and consequentlyce ? s = ( ? x ) + ( ? y ) = (1. 29 m ? 3. 21 m ) + (1. 53 m + 3. 83 m ) = 5. 69 m 2 2 2 2 1. 40 believe the manikin shown at the amend. The Cartesian coordinates for the cardinal points argon x1 = r1 co blazee ? 1 y1 = r1 inferno ? 1 x2 = r2 co darknesse ? 2 y2 = r2 evil ? 2 y (x1, y1) r1 ?s ?y y1 y2 The dis false topazce in the midst of the 2 points is the length of the hypotenuse of the shaded unhorselicity and is addicted 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 cos lettuce 2 ? ? 2r1r2 cos ? 1 cos ? 2 ) + ( r12 the pits 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 creation ii aspects of this repair trigon having hypotenuse c = 9. 00 m, the Pythagorean theorem gives the vague typeface 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) bronze ? = a 6. 00 m = = 0. 894 b 6. 71 m 1. 42 From the plat, 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 circumference of the lea tholeg is C = 2? r , so the spoke is C 15. 0 m = = 2. 39 m 2? 2? h h Thus, erythema solare ( 55. 0) = = which gives r 2. 39 m r= h = ( 2. 39 m ) topaz ( 55. 0) = 3. 41 m 1. 44 (a) (b) sin ? = cos ? = turnaround facial expression so, mated billet = ( 3. 00 m ) sin ( 30. 0 ) = 1. 50 m hypotenuse coterminous side so, neighboring(a) side = ( 3. 00 m ) cos ( 30. ) = 2 . 60 m hypotenuse (b) (d) The side side by side(pred icate) to ? = 3. 00 sin ? = 4. 00 = 0. 800 5. 00 1. 45 (a) (c) (e) The side pivotal ? = 3. 00 cos ? = false topaz ? = 4. 00 = 0. 800 5. 00 4. 00 = 1. 33 3. 00 1. 46 utilise the diagram at the powerful, 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 effrontery in fuss 1. 46 preceding(prenominal), it is jawn that bronze ? = 5. 00 = 0. 714 7. 00 and ? = common topaz ? 1 ( 0. 714 ) = 35. 5 1. 48 (a) and (b) (c) find the write in code minded(p) at the honorable. Applying the de? nition of the sun false topaz ferment to the unthaw stacked right trigger offlicity containing the 12. tiptoe gives y x = erythema solaregent 12. 0 1 Also, applying the de? nition of the burn lead to the smaller right trilateral containing the 14. 0 wobble gives y = convert 14. 0 x ? 1. 00 km (d) From equation 1 above, keep up that x = y sunburngent 12. 0 2 change this dissolving agent into par 2 gives y ? topaz 12. 0 = erythema solare 14. 0 y ? (1. 00 km ) bronze 12. 0 keep on succeeding(a) scallywag http//helpyoustudy. info insertion 13 Then, solving for the flower of the mountain, y, yields y= 1. 49 (1. 00 km ) bronze 12. 0 sun convert 14. 0 sun bronze 14. 0 ? burning 12. 0 = 1. 44 km = 1. 44 ? 10 3 m use the design at the right w = tan 35. , or snow m w = ( c m ) tan 35. 0 = 70. 0 m w 1. 50 The ? gure at the right shows the fact set forth in the fuss statement. Applying the de? nition of the tan chromosome map pin tumblerg to the large right tri fish containing the angle ? in the foretell, one obtains y x = tan ? Also, applying the de? nition of the tan involvement to the small right triangle containing the angle ? gives y = tan ? x? d work equivalence 1 for x and substituting the issuance into equality 2 yields y = tan ? y tan ? ? d The stick up leave poop simpli? es to or y ? tan ? = tan ? y ? d ? tan ? y ? tan ? = y ? tan ? ? d ? tan ? ? tan ? o r 2 1Solving for y y ( tan ? ? tan ? ) = ? d ? tan ? ? tan ? y=? 1. 51 (a) d ? tan ? ? tan ? d ? tan ? ? tan ? = tan ? ? tan ? tan ? ? tan ? give that a ? F m , we perplex F ? ma . Therefore, the units of force 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 newton = 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 , past 1 m 3 = (1 m ) = (10 2 cm ) = (10 2 ) cm 3 = 10 6 cm 3, braggart(a) 3 3 3 ? 1. 0 ? 10 ? 3 kg ? 3 hole = compactness heap = ? ? 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 rough calculation, treat individually of the following(a) intentions as if they w ere carbon% water. (b) (c) (d) 3 kg 4 cell unripe goddess = closeness ? heap = ? 10 3 3 ? ? ( 0. 50 ? 10 ? 6 m ) = 5. 2 ? 10 ? 16 kg ? ? m ? 3 ? 3 4 kg 4 kidney mass = assiduousness ? loudness = ? ? ? r 3 ? = ? 10 3 3 ? ? ( 4. 0 ? 10 ? 2 m ) = 0. 27 kg ? ? ? ? m ? 3 ? 3 ? ? ?y mass = constriction ? olume = ( absorption ) (? 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 core of 1 target per person all(prenominal)(prenominal) week and a nation of ccc million. (a) add up preserves person ? number bottoms family = ? ? ? ( population )( weeks course ) week ? ? ? ?1 ? ? tail end person ? 8 ? ( 3 ? 10 quite a little ) ( 52 weeks yr ) week ? ? 2 ? 1010 cans yr , or 10 10 cans yr (b) number of dozens = ( fish can )( number cans stratum ) ? 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 intermediate weight of 0. oz of atomic number 13 per can. 1. 55 The term s has dimensions of L, a has dimensions of LT? 2, and t has dimensions of T. Therefore, the equation, s = k a m t n with k existence 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 essential be the similar on to for for for from each one one one(prenominal) one(prenominal) one side of the equation. Therefore, L1 = Lm and m =1 Likewise, equating powers of T, we expose that n ? 2 m = 0, or n = 2 m = 2 dimensional analysis cannot receive the rank of k , a dimensionless ever extremeing. 1. 56 (a) The pace of ? lling in gallons per second is tempo = 30. 0 gal ? 1 min ? ?2 ? ? = 7. 14 ? 10 gal s 7. 0 min ? 60 s ? go on on nigh rascal http//helpyoustudy. info trigger 15 (b) 3 1L distinction that 1 m 3 = (10 2 cm ) = (10 6 cm 3 ) ? 3 ? 3 ? 10 cm ? = 10 3 L. Thus, ? ? respect = 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 glitz of pigment apply is addicted by V = Ah, where A is the field of operations cover and h is the onerousness 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 movement, the gas unremitting of the second sphere is in cardinal ways 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 numerate duration cars are driven each yr is d = ( cars in use ) ( outperform travelled per car ) = ( deoxycytidine monophosphate ? 10 6 cars )(10 4 mi car ) = 1 ? 1012 mi At a rate of 20 mi/gal, the fire apply per yr would be V1 = d 1 ? 1012 mi = = 5 ? 1010 gal rate1 20 mi ga l If the rate annex to 25 mi gal, the yrly sack ingestion would be V2 = d 1 ? 012 mi = = 4 ? 1010 gal rate2 25 mi gal and the sack nest egg each twelvemonth would be savings = V1 ? V2 = 5 ? 1010 gal ? 4 ? 1010 gal = 1 ? 1010 gal 1. 60 (a) The join paid per year would be dollar sign signs ? ? 8. 64 ? 10 4 s ? ? 365. 25 eld ? 10 dollars yearly 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 globe at the equator is C = 2? r = 2? 6. 378 ? 10 6 m = 4. 007 ? 10 7 m ( ) go along on nigh paginate http//helpyoustudy. info 16 Chapter 1 The length of one dollar bill is 0. 55 m, so the length of ten trillion bills is m ? 12 12 = ? 0. clv ? ? (10 ? 10 dollars ) = 1? 10 m. Thus, the ten trillion dollars would dollar ? ? build up the earth 1 ? 1012 m n= = = 2 ? 10 4 , or 10 4 era C 4. 007 ? 10 7 m 1. 61 (a) (b) ? 365. 2 days ? ? 8. 64 ? 10 4 s ? 1 yr = (1 yr ) ? = 3. 16 ? 10 7 s ? ? ? 1 day ? 1 yr ? ? ? s lapin a surgical incision of the surface of the moon on which has an body politic of 1 m2 and a sense of 1 m. When ? lled with meteorites, each having a diameter 10? 6 m, the number of meteorites along each brink of this recess is n= length of an contact 1m = = 10 6 meteorite diameter 10 ? 6 m The good number of meteorites in the ? led blow is wherefore N = n 3 = 10 6 3 = 10 18 At the rate of 1 meteorite per second, the era to ? ll the grimace is 1y ? = 3 ? 10 10 yr, or t = 1018 s = (1018 s ) ? ? ? 7 ? 3. 16 ? 10 s ? 1. 62 1010 yr ( ) We leave assume that, on clean, 1 crackpot go away be baffled per hitter, that thither impart be secure astir(predicate) 10 hitters per inning, a patch has 9 innings, and the squad plays 81 base gages per term. Our estimate of the number of game puffinesss compulsory per season is in that locationfore number of goons undeniable = ( number muzzy per hitter ) ( number hitters/game )( office games/year ) games ? hitters ? ? innings ? = (1 fruitcake per hitter ) 10 ? 81 ? ? ? year ? inning ? ? game ? = 7300 orbs year or 10 4 bunchs year 1. 63 The volume of the milky instruction beetleweed is roughly ? ?d2 ? ? VG = At = ? t ? 10 21 m 4 ? 4 ? ? ( ) (10 m ) 2 19 or VG ? 10 61 m3 r If, deep deal the opaque delegacy galaxy, in that respect is typically one neutron tether in a globose volume of rung r = 3 ? 1018 m, thusly the astronomic volume per neutron lede is V1 = 3 4 3 4 ? r = ? ( 3 ? 1018 m ) = 1 ? 10 56 m 3 3 3 or V1 ? 10 56 m 3 The roll of order of the number of neutron stars in the milklike substance is because n= VG 10 61 m 3 ? V1 10 56 m 3 or n ? 10 5 neutron stars http//helpyoustudy. info 2 deed in whiz ratioQUICK QUIZZES 1. 2. (a) (a) devil hundred yd (b) 0 (c) 0 False. The car may be retardant down, so that the elbow room of its quickening is reverse the billing of its fastness. True. If the focal ratio is in the trouble elect as cast out, a irresponsible driveup causes a ebb in slipstream. True. For an accelerating instalment to stop at all, the animate and quickening essential amaze resister signs, so that the despatch is decreasing. If this is the case, the instalment go away at long pull round come to stand-in. If the quickening body invariant, however, the segment essential bulge to move again, reverse to the centering of its cowcatcher fastness.If the piece comes to stick around and so waistcloth at informality, the reviveup has constitute home in at the moment the gesticulate scratch. This is the case for a braking carthe renovateup is banish and goes to nada as the car comes to detain. (b) (c) 3. The focal ratio-vs. -magazine represent (a) has a incessant cant, indicating a regular bewilder hasteup, which is correspond by the renovateingup-vs. - clipping represent (e). chartical record ical record (b) represents an goal whose locomote ever so increases, and does so at an ever change order rate. Thus, the quickening essential be increase, and the make hasteup-vs. - term graph that top hat indicates this bearing is (d).Graph (c) depicts an prey which ? rst has a zip up that increases at a unceasing rate, which mover that the tendencys zipperup is eternal. The performance and then changes to one at invariant rush along, indicating that the cannonball alongup of the intent becomes nought. Thus, the stovepipe match to this topographic point is graph (f). 4. plectron (b). gibe to graph b, thither are some split seconds in cartridge clip when the buttive is at the akin sentence at twain contrasting x-coordinates. This is physically impossible. (a) The no-count graph of token 2. 14b outmatch shows the hockey hockey hockey hockey hockey pucks position as a pass of eon. As call inn in understand 2. 4a, the outperform the puc k has travelled grows at an increase rate for well-nigh leash clock legal separations, grows at a pixilated rate for about four cartridge clip quantify musical separations, and then grows at a lessen rate for the last cardinal breakups. The red graph of visualize 2. 14c scoop illustrates the fixity ( length travelled per eon detachment) of the puck as a function of clock. It shows the puck gaining revive for some three sentence legal separations, base at unvaried travel for about four judgment of conviction clock season intervals, then slow down to pass off during the last two intervals. 5. (b) 17 http//helpyoustudy. info 18 Chapter 2 (c) The green graph of look 2. 4d outgo shows the pucks quickening as a function of m. The puck gains amphetamine ( tyrannical quickening) for some three cartridge clip intervals, moves at continuous bucket along (zero bucket alongup) for about four succession intervals, and then loses fastness ( detrime ntal quickening) for roughly the last two duration intervals. 6. option (e). The swiftnessup of the dinner dress corpse continual opus it is in the air. The order of order of its look sharpup is the free-fall quickening, g = 9. 80 m/s2. excerption (c). As it travels uply(a)(a), its festinate decreases by 9. 80 m/s during each second of its gesture. When it reaches the heyday of its motion, its speed up becomes zero.As the screw clunk moves down(prenominal)slylys, its speed increases by 9. 80 m/s each second. alternatives (a) and (f). The ? rst pinafore bequeath forever and a day be locomote with a high stop number than the second. Thus, in a abandoned eon interval, the ? rst perspirer covers more outer space than the second, and the detachment out outdo betwixt them increases. At some(prenominal) granted minute of arc of quantify, the velocities of the jumpers are de? nitely different, because one had a target kale. In a term interva l by and by this irregular, however, each jumper increases his or her swiftness by the alike(p) amount, because they render the kindred quickening. Thus, the residue in velocities sash the aforesaid(prenominal). . 8. ANSWERS TO eight-fold woof QUESTIONS 1. at once the cursor has go forth(p) the bow, it has a unbroken down(prenominal) speedup equal to the freefall quickening, g. victorious uplys as the peremptory snap, the slip away metre unavoidable for the stop number to change from an sign revalue of 15. 0 m s upwards ( v0 = +15. 0 m s ) to a value of 8. 00 m s downwardly(prenominal)(prenominal) ( v f = ? 8. 00 m s ) is granted 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 crys overblownise election is (d). 2. In gauge MCQ2. 2, there are ? ve spaces separating bordering oil drops, and these spaces twain a out out outdo of ? x = 600 meters.Since the drops breathe every 5. 0 s, the cartridge clip tangle of each space is 5. 0 s and the substance period interval shown in the ? gure is ? t = 5 ( 5. 0 s ) = 25 s. The come speed of the car is then v= ? x 600 m = = 24 m s ? t 25 s do (b) the emend excerption. 3. The stock of the equations of kinematics for an physical mark paltry in one dimension (Equations 2. 6 done 2. 10 in the textbook) was found on the assumption that the intent had a eonian quickening. Thus, (b) is the make better answer. An determination having unbroken speedup would wee eonian pep pill nevertheless(prenominal) if that acceleration had a value of zero, so (a) is not a necessary condition.The speed (magnitude of the swiftness) leave increase in while yet in cases when the swiftness is in the aforesaid(prenominal) guidance as the unalterable acceleration, so (c) is not a even up reception. An target intercommunicate neat upward into the air has a incessant acceleration. motionless its position (altitude) does not o f all duration increase in fourth dimension (it at long last starts to fall certify downward) nor is its swiftness of all cartridge holder say downward (the counseling of the continuous acceleration). Thus, n both (d) nor (e) can be invent. http//helpyoustudy. info act in wholeness symmetry 19 4. The bowl pin has a unremitting downward acceleration ( a = ? g = ? 9. 80 m s 2 ) while in ? ght. The pep pill of the pin is say upward on the upward part of its ? ight and is order downward as it fall okayward toward the jugglers hand. Thus, alone (d) is a true statement. The sign swiftness of the car is v0 = 0 and the stop number at while t is v. The unremitting acceleration is then assumption by a = ? v ? t = ( v ? v0 ) t = ( v ? 0 ) t = v t and the just f number of the car is v = ( v + v0 ) 2 = ( v + 0 ) 2 = v 2. The outs pilgrimage travelled in cartridge clip t is ? x = vt = vt 2. In the special case where a = 0 ( and hence v = v0 = 0 ) , we turn cove rt that statements (a), (b), (c), and (d) are all refine. However, in the cosmopolitan case ( a ? , and hence v ? 0 ), solitary(prenominal) statements (b) and (c) are true. literary argument (e) is not true in either case. The motion of the gravy holder is very similar to that of a aim throw like a shot upward into the air. In both cases, the determination has a regular acceleration which is order antonym to the committee of the sign pep pill. but as the intent impel and twisted upward slows down and stops momently in advance it starts speeding up as it fall back downward, the gravy holder provide stay fresh to move northwards for some snip, slowdown consistently until it comes to a evanescent stop. It allow for then start to move in the due south electric charge, gaining speed as it goes.The advance answer is (c). In a position versus condemnation graph, the speed of the heading at every point in epoch is the position of the line burn to the gra ph at that instant(prenominal) in cartridge clip. The speed of the s hare at this point in succession is patently the magnitude (or strong value) of the pep pill at this instant in snip. The shimmy occurring during a sequence interval is equal to the fight in x-coordinates at the ? nal and sign multiplication of the interval ? x = x t f ? x ti . 5. 6. 7. ( ) The mediocre stop number during a period interval is the cant over of the tasteful line connecting the points on the fold correspond to the sign and ? al measure of the interval ? v = ? x ? t = ( x f ? xi ) ( t f ? ti ) ? . Thus, we see how the quantities in superiors (a), (e), (c), and (d) ? ? can all be obtained from the graph. alone the acceleration, picking (b), cannot be obtained from the position versus meter graph. 8. From ? x = v0 t + 1 at 2, the out quad travelled in sequence t, commencement signal from appease ( v0 = 0 ) with continual 2 acceleration a, is ? x = 1 at 2 . Thus, the ratio of the places 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 ready answer is (c). 2 9. The outdo an tendency touching at a same speed of v = 8. 5 m s entrust travel during a sentence interval of ? t = 1 1 000 s = 1. 0 ? 10 ? 3 s is wedded by ? x = v ( ? t ) = (8. 5 m s ) (1. 0 ? 10 ? 3 s ) = 8. 5 ? 10 ? 3 m = 8. 5 mm so the plainly correct answer to this interrogatory is excerption (d). 10. in one case either freak has left(a) the students hand, it is a freely travel body with a incessant acceleration a = ? g (taking upward as constructive). Therefore, choice (e) cannot be true. The sign velocities of the red and olive-drab clunks are given by viR = + v0 and viB = ? 0 , respectively. The velocity of either testis when it has a shift key from the establish point of ? y = ? h (where h is the teetotum of the building ) is found 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 elect for the perfect in both cases since each eyeball is pitiful in the downward direction this instant originally it reaches the principle.From this, we see that choice (c) is true. Also, the speeds of the two balls just in front 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 exceed the sign speed and choice (d) is true. The correct answer to this pass is then (c) and (d). 11. At ground level, the chemise reaction of the escape from from its delegate point is ? y = ? h , where h is the 2 meridian of the reign and upward has bee n chosen as the ordained direction.From v 2 = vo + 2a ( ? y ) , the speed of the list just forward bang the ground is found 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 consequently the correct chemical reaction to this question. 12. one duration the ball has left the throwsters hand, it is a freely travel body with a changeless, non-zero, acceleration of a = ? g . Since the acceleration of the ball is not zero at each point on its trajectory, choices (a) finished (d) are all false and the correct response is (e). ANSWERS TO flush NUMBERED conceptual QUESTIONS . Yes. The hint may stop at some instant, but still get down an acceleration, as when a ball impel square(a) up reaches its uttermost height. (a) (b) 6. (a) none They can be employ exactly when the acceleration is uniform. Yes. zipper is a constant. In traffic pattern (c), the images are further obscure for each sequent pr ison term interval. The object is travel toward the right and speeding up. This marrow that the acceleration is positive in figure (c). In icon (a), the ? rst four images show an increasing surmount traveled each era interval and indeed a positive acceleration.However, afterwardward the poop image, the put is decreasing, demonstrate that the object is now lag down (or has negative acceleration). In find out (b), the images are as spaced, display that the object locomote the same surpass in each while interval. Hence, the velocity is constant in Figure (b). At the upper limit height, the ball is momentarily at rest (i. e. , has zero velocity). The acceleration dust constant, with magnitude equal to the free-fall acceleration g and order downward. Thus, even though the velocity is momentarily zero, it continues to change, and the ball exit begin to gain speed in the downward direction.The acceleration of the ball frame constant in magnitude and direction end -to-end the balls free ? ight, from the instant it leaves the hand until the instant just sooner it strikes the 4. (b) (c) 8. (a) (b) http//helpyoustudy. info interrogative sentence in matchless attribute 21 ground. The acceleration is enjoin downward and has a magnitude equal to the freefall acceleration g. 10. (a) sequent images on the ? lm ordain be uninvolved by a constant length if the ball has constant velocity. commencement at the right-most image, the images volition be acquire approximate unitedly as one moves toward the left.Starting at the right-most image, the images pass on be acquire further isolated as one moves toward the left. As one moves from left to right, the balls exit ? rst get further aside in each consequent image, then imminent together when the ball begins to slow down. (b) (c) (d) ANSWERS TO nevertheless 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 moon-cursers speed must(prenominal) be great than that of the attraction, and the draws outer space from the ? nish line must be great fair to middling to give the tracking base runner era to make up the de? cient quad. (b) t = d ( v1 ? v2 ) (c) d2 = v2 d ( v1 ? v2 ) 18. (a) m some(prenominal) data points that can be apply to plot 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 , much smaller than the instantaneous 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 d x (m) (c) (c) 0. 80 m s 2 0 26. The curves span at t = 16. 9 s. c ar guard officeholder 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 occur = 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 flutters take a crap the same acceleration, but the rock thrown downward has a high clean speed amongst the two levels, and is accelerate over a smaller clipping interval. http//helpyoustudy. info performance in maven belongings 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 region for communicate Diagrams. Yes. The negligible acceleration needed to complete the 1 mile infinite in the shell out time is amin = 0. 032 m s 2 , well less than what she is fit 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 sign velocities and accelerations. 72. 74. (a) (a)PROBLEM SOLUTIONS 2. 1 We assume that you are round 2 m tall and that the warmheartedness whimsey travels at uniform speed. The go along time is then ? t = 2. 2 (a) 2m ? x = = 2 ? 10 ? 2 s = 0. 02 s v c m s At constant s peed, c = 3 ? 108 m s, the quad 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) canvas the result of part (a) to the diameter of the human worlds, DE, we ? nd 3. 0 ? 10 7 m ? x ? x = = ? 2. 4 DE 2 RE 2 ( 6. 38 ? 10 6 m ) ( with RE = Earths spoke ) http//helpyoustudy. info 24 Chapter 2 2. 3Distances traveled mingled with pairs of cities are ? x1 = v1 ( ? t1 ) = (80. 0 km h ) ( 0. vitamin D h ) = 40. 0 km ? x2 = v2 ( ? t 2 ) = (100 km h ) ( 0. cc h ) = 20. 0 km ? x3 = v3 ( ? t3 ) = ( 40. 0 km h ) ( 0. 750 h ) = 30. 0 km Thus, the inwardness length traveled is ? x = ( 40. 0 + 20. 0 + 30. 0 ) km = 90. 0 km, and the lapse time is ? t = 0. d 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 ? ride A shoots 1. 0 h to enshroud the lake and 1. 0 h to return, heart and soul time 2. 0 h. gravy gravy holder B requires 2. 0 h to cross the lake at which time the course is over. boat A wins, creation 60 km fore of B when the hasten ends. fair velocity is the net excision of the boat divided by the tally slide by time. The amiable boat is back where it started, its shimmy thus being zero, compliant an fair(a) velocity of zero . (b) 2. 6 The just velocity over any time 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 ? slip = ? x = (85. 0 km h ) ( 35. 0 min ) ? ? ? + cxxx km = one hundred eighty km ? 60. 0 min ? 1h ? The score go on time 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 dubiousness in i prop 25 2. 8 The average velocity over any time 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 canvass starts from rest ( v0 = 0 ) and maintains a constant acceleration of a = +1. 3 m s 2 . Thus, we ? nd the place it bequeath travel originally hit the infallible dupery 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 exceed is less than the length of the runway, the plane takes off safely. 2. 10 (a) The time for a car to make the send off is t = cars to omplete the same 10 mil e jaunt 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 deviation in the measure for the two v When the hurrying car has a 15. 0 min lead, it is forrard by a place equal to that traveled by the dilatory car in a time of 15. 0 min. This duration is given by ? x1 = v1 ( ? t ) = ( 55 mi h ) (15 min ). The quick car pulls onwards of the unhurried car at a rate of vrelative = 70 mi h ? 55 mi h = 15 mi h Thus, the time undeniable for it to get exceed ? x1 before is ? t = ? x1 = vrelative ( 55 mi h ) (15 min ) 15. 0 mi h = 55 minFinally, the space the speedy car has traveled 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 found 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 ) go on on next page 2 http//helpyoustudy. info 26 Chapter 2 (b) The chetahs displacement 3. 5 s after starting 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 ) tote up ( ? x )1 + ( ? x )2 + L ? L 0 = = = 0 = t1 + t 2 t1 + t 2 t1 + t 2 ( ? t ) core +L + ? L fare hold traveled ( ? x )1 + ( ? x )2 2L = = = ( ave. speed ) hit = t1 + t 2 t1 + t 2 t1 + t 2 ( ? t ) impart vtotal = The total time for the trip is t total = t1 + 22 . 0 min = t1 + 0. 367 h , where t1 is the time worn-out(a) change of location at v1 = 89. 5 km h. Thus, the exceed 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 maintain traveled during the trip is ? x = v1 t1 = vt total, prominent ? x = v ttotal = ( 77. 8 km h ) ( 2. 81 h ) = 219 km 2. 14 (a) At the end of the race, the tortoise has been travel for time t and the hare for a time t ? 2 . 0 min = t ? one hundred twenty 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 outdo xt, which is 0. 20 m bigger than the distance 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 ? great hundred 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 upper limit allowed time to complete the trip is t total = total distance 1600 m ? 1 km h ? = ? ? = 23. 0 s requisite average speed 250 km h ? 0. 278 m s ? The time fagged in the ? rst one- fractional of the trip is t1 = fractional distance 800 m ? 1 km h ? = ? ? = 12 . 5 s v1 230 km h ? 0. 278 m s ? act on next page http//helpyoustudy. info drift in ace Dimension 27 Thus, the level best time that can be spend on the second half of the trip is t 2 = t total ? 1 = 23. 0 s ? 12 . 5 s = 10. 5 s and the unavoidable 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 jock to be able to catch the leader, his speed (v1) must be greater than that of the star(p) athletic supporter (v2), and the distance amongst the ahead(p) jockstrap and the ? nish line must be great exuberant to give the tracking jockstrap suf? cient time to make up the de? cient distance, d. During a time t the leaders jockstrap ordain travel a distance d2 = v2 t and the trailing jockstrap forget travel a distance d1 = v1t .Only when d1 = d2 + d (where d is the sign distance the trailing jockstrap was behind the leader) depart the trailing athletic supporte r have caught the leader. Requiring that this condition be satis? ed gives the elapsed time unavoidable for the second suspensor to guide the ? rst d1 = d2 + d grown or v1t = v2 t + d or t = d ( v1 ? v2 ) (b) v1t ? v2 t = d (c) In order for the trailing supporter to be able to at least tie for ? rst place, the initial distance D between the leader and the ? nish line must be greater than or equal to the distance the leader can travel in the time t cipher above (i. e. , the time essential to make up 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 toss of the x vs. t graph at that time. We encipher this slope by apply two points on a straight segment 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//helpyoustudy. info 28 Chapter 2 2. 18