HOLT - Physics is Beautiful

1 PhysicsSolutions ManualHOLTISBN-13: 978-0-03-099807-2 ISBN-10: 0-03-099807-71 2 3 4 5 6 7 082 11 10 09 08 07 Copyright by holt , Rinehart and WinstonAll rights reserved. No part of this publication may be reproduced or transmittedin any form or by any means, electronic or mechanical, including photocopy,recording, or any information storage and retrieval system, without permission inwriting from the using holt Physics may photocopy complete pages in sufficient quantities for classroom use only and not for the Owl Design are trademarks licensed to holt , Rinehart andWinston, registered in the United States of America and/or other in the United States of AmericaHolt PhysicsTeacher s Solutions ManualIf you have received these materials as examination copies free of charge, holt ,Rinehart and Winston retains title to the materials and they may not be of examination copies is strictly of this publication in print format does not entitle users to convertthis publication, or any portion of it, into electronic I Student Edition SolutionsChapter 1 The Science of Physics .

2 I-1-1 Chapter 2 Motion in One Dimension.. I-2-1 Chapter 3 Two-Dimensional Motion and Vectors.. I-3-1 Chapter 4 Forces and the Laws of Motion.. I-4-1 Chapter 5 Work and Energy.. I-5-1 Chapter 6 Momentum and Collisions.. I-6-1 Chapter 7 Circular Motion and Gravitation.. I-7-1 Chapter 8 Fluid Mechanics.. I-8-1 Chapter 9 Heat.. I-9-1 Chapter 10 Thermodynamics.. I-10-1 Chapter 11 Vibrations and Waves.. I-11-1 Chapter 12 Sound.. I-12-1 Chapter 13 Light and Reflection.. I-13-1 Chapter 14 Refraction.. I-14-1 Chapter 15 Interference and Diffraction.. I-15-1 Chapter 16 Electric Forces and Fields.. I-16-1 Chapter 17 Electrical Energy and Current.. I-17-1 Chapter 18 Circuits and Circuit Elements.. I-18-1 Chapter 19 Magnetism.. I-19-1 Chapter 20 Electromagnetic Induction.. I-20-1 ContentsCopyright by holt , Rinehart and Winston. All rights by holt , Rinehart and Winston. All rights 21 Atomic Physics .. I-21-1 Chapter 22 Subatomic Physics .

3 I-22-1 Appendix IAdditional Practice.. I-Apx I-1 Section II Problem Workbook SolutionsChapter 1 The Science of Physics .. II-1-1 Chapter 2 Motion in One Dimension .. II-2-1 Chapter 3 Two-Dimensional Motion and Vectors.. II-3-1 Chapter 4 Forces and the Laws of Motion.. II-4-1 Chapter 5 Work and Energy.. II-5-1 Chapter 6 Momentum and Collisions.. II-6-1 Chapter 7 Circular Motion and Gravitation.. II-7-1 Chapter 8 Fluid Mechanics.. II-8-1 Chapter 9 Heat.. II-9-1 Chapter 10 Thermodynamics.. II-10-1 Chapter 11 Vibrations and Waves.. II-11-1 Chapter 12 Sound.. II-12-1 Chapter 13 Light and Reflection.. II-13-1 Chapter 14 Refraction.. II-14-1 Chapter 15 Interference and Diffraction.. II-15-1 Chapter 16 Electric Forces and Fields.. II-16-1 Chapter 17 Electrical Energy and Current.. II-17-1 Copyright by holt , Rinehart and Winston. All rights 18 Circuits and Circuit Elements.. II-18-1 Chapter 19 Magnetism.. II-19-1 Chapter 20 Electromagnetic Induction.

4 II-20-1 Chapter 21 Atomic Physics .. II-21-1 Chapter 22 Subatomic Physics .. II-22-1 Section III Study Guide Worksheets AnswersIII-1solutionsStudent EditionSolutionsISectionHolt PhysicsIICopyright by holt , Rinehart and Winston. All rights Science of PhysicsStudent Edition = mg 1 11m0 g3g 1 11k0g3g =time =3 10 10 9s 1 11m0 s3s =distance = km 1 11k0m3m = = = = = = = + = = 104m3 10 10 6kgThe Science of Physics , Section 2 ReviewThe Science of Physics , Practice =50 m50 m 1 11 0m 6m = =1 s1 s 1 11 0s 6s = =10 nm 1 11n0m 9m = 10 8m 1 11m0 m3m = 10 8m 1 11 0m 6m = = 1011m 1 11T0m12m = 1011m 1 11k0m3m = = 106g 1 11k0g3g = 10-1Tm1 10 2 m1 10 5mm1 10 8m1 10 6s5 10 5mSection One Student Edition SolutionsI Ch. 1 1 ICopyright by holt , Rinehart and Winston. All rights Science of Physics , Chapter dm 1 11d0m 1m 1 11m0 m3m =2 h 10 h 601mhin =120 min120 min +10 min =130 min130 min 16m0isn =16 g 1 11 0g 6g = km 1 11k0m3m 1 11c0m 2m = mg 1 11m0 g3g =462 m 1 11 0m 6m 1 11c0m 2m =35 km/hg.

5 35hkm 3610h0s 1 11k0m3m = rations 11d0e1kraartaiotinosn =2000 mockingbirds =10 6phones 101 6pphhoonnees =10 9goats 101 n9ggooaatts = 1011E8mmiinneerrs = of light = s108m 3610h0s 1 h 1 11k0m3m = 108m/s t=1 ton = 103kg 18p5erksgon =mass/person =85 kgNote that the numerical answer, people, must be rounded downto 11 109km1 examiner1 nanogoat1 microphone2 kilomockingbirds1 kmockingbirds 1 103mockingbirds1 10 10 107 103s2 102mmHolt Physics Solution ManualI Ch. 1 2 Copyright by holt , Rinehart and Winston. All rights One Student Edition SolutionsI Ch. 1 g + g + g + g = g =b. = m/s = mm p= mm = s s = s = cm, cm + cm = cm = mw= m + m + m + m = m = (a+b+c) 2r= r, a, b, c ,and sall have units of = = = (l en gt h )2 =lengthThus, the equation is dimensionally Substitute the proper dimensions into the = = (t im e) 2 =timeThus, the dimensions are (m/s)2 m/s2 sm2/s2 m/sThe dimensions are not one breath every 5 years 3615ydeaarys 124dahy 3610h0s 1b5resath = one heart beat per day 124dahy 3610h0s 1bseat = will years 3615ydeaarys 124dahy 3610h0s = a tire s radius to be 000 mi 110k3mm 2p1( ) =4 108s9 104beats4 108breathslength [length/(time)2]L ag(length)3 lengthlength length length length(s a)(s b)(s c) m/s797 gGivensSolutionsCopyright by holt , Rinehart and Winston.

6 All rights 30 balls lost per games 310gbamalles = 14 lb per burger and 800 lb per head of 1010burgers =1 1010lb2 103ballsGivensSolutionsI5 1010burgers 180h0ealbd = =8 million peopleEstimate 5 people per family. 58pemoiplllieopneprefoampleily =2 million familiesEstimate that 1/5 of families have a of pianos =(2 million families) 15 =400,000 pianosEstimate 3 tunings per day per tuner, with 200 work days per of pianos tuned each year (per tuner) =(3)(200) =600 Number of tuners == = cmFind the number of balls that can fit along the length and mw=4 mh=3m4 m =100 ballsFind the number that can be stacked to the m =80 ballsMultiply all three figures to find the number of balls that can fit in the balls 100 balls 80 balls =A rough estimate: divide the volume of the room by the volume of a ( cm) =A=pr2=p( cm)2=r= ( cm) =A=pr2=p( cm)2= 109bills 11bsill 3610h0s 114dahy 3615ydeaarys =Take the $5000.

7 It would take 272 years to count 5 billion $1 cm38 cm222 cm8 105balls7 102tuners400,000 pianos 600 pianos/year per tuner2 107head of cattleHolt Physics Solution ManualI Ch. 1 4 Copyright by holt , Rinehart and Winston. All rights One Student Edition SolutionsI Ch. 1 quart u1a0r ts3m3 = 10 4m3V=L3L= 3V = 10 4m 3 = 10 10 6mnumber of micrometeorites per side:l= m om10e t6eomrite = 106micrometeoritesnumber of micrometeorites needed to cover the moon to a depth of m:( 106micrometeorites)3= 1018micrometeorites 1 micro1mseteorite 3610h0s 124dahy 3615ydeaarys =Note that a rougher estimate can be made by dividing the volume of the m3boxby the volume of a cm3 .01c0m 33kg (1 11c0m 23m)3 m3=m= 10 1010 = 10 7kgdensity =918 kg/m3r= cmarea =pr2volume = dmenassisty = k1g0/ m73kg = 10 10m3diameter = voalruemae = (0 .41108 m10)m23 = 10 cubit = mVark=300 cubits 50 cubits 30 cubitsVark=(300 cubits)(50 cubits)(30 cubits) 3 Vark=Estimate the average size of a house to be 2000 ft2and 10 ft (2000 ft2)(10 ft) 3 Vhouse= VVhaorukse = 66 110042mm33 = 1006 102m36 104m3 IGivensSolutionsCopyright by holt , Rinehart and Winston.

8 All rights Physics Solution ManualI Ch. 1 = r= 103kg/m3diameter = ml= mmdiameter =2r= mmdensity =r= 103 43 pr3= 43 p 120 6m 3= 10 19m3m=rV=( 103kg/m3)( 10 19m3)( ) = ( 10 3m) (p) 120 3m 2= 10 8m3m=rV=( 103kg/m3)( 10 8m3)( ) =1 10 5kg5 10 107mm= 43 pr3density = mV = 43pmr3 density = (43p)(( 1100276mk)g3) 10k3gg 1012mcm 3density = area =4pr2=4p( 107m)2surface area = ly =9 500 000 000 000 km= 1012km 110k3mm = 1015mThe Science of Physics , Standardized Test PrepSection One Student Edition SolutionsI Ch. 2 1 Copyright by holt , Rinehart and Winston. All rights In One DimensionStudent Edition m/s east x=vavg t=( m/s)(34 min)(60 s/min) t=34 min x = 103m =2. t=15 min x=vavg t=( km/h)(15 min) vavg= km/h south x =3. t= min x=vavg t=( m/s) ( min)(60 s/min)vavg= m/s north x = km/h east t= v avxg = = x=144 km km/h east t= v avxg = = h x=144 km easttime saved = h h = h = h680 m km1h 60 km eastMotion In One Dimension, Practice AGivensSolutions6.

9 X1=280 km southvavg,1= 88 km/h south t2=24 minvavg,2= 0 km/h x3=210 km southvavg,3= 75 km/h southa. ttot= t1+ t2+ t3 = v avxg1,1 + t2+ v avxg3,3 ttot= +(24 min) 601mhin + 72510kmkm/h ttot= h + h + h = , tot= xtttoott = xt11++ xt22++ tx33 x2=vavg,2 t2=(0 km/h)(24 min) 601mhin =0 kmvavg, tot= = =77 km/h south280 km +0 km +210 km h =6 h 24 min280 km 88 mm/s x= cm t= vx = =24 sMotion In One Dimension, Section 1 m/s x= m t= vx = = sHolt Physics Solution ManualI Ch. 2 2 Copyright by holt , Rinehart and Winston. All rights x1= m south t1= s x2= m north t2= m/sv2= m/s x=780 m t1 t2=( min)(60 s/min) = ,1= xt11 = = ,2= xt22 = = xtot= x1+ x2=( m) +( m) = m ttot= t1+ t2= s + s = svavg= xtttoott = = m/s m/s southa. t1= v1x = =870 s t2= v2x = =410 s t1 t2=870 s 410 s =b. x1=v1 t1 x2=v2 t2 x1= x2v1 t1=v2 t2v1[ t2+( 102s)] =v2 t2v1 t2+ v1( 102s) =v2 t2 t2(v1 v2) = v1( 102s) t2= v1( v2102s) == t2= 102s t1= t2+( 102s) =( 102s) +( 102s) =630 s x1=v1 t1=( m/s)(630 s) = x2=v2 t2=( m/s)( 102s) =570 m570 m ( m/s)( 102s) m/s ( m/s)( 102s) m/s m/s460 m/s2 t= a avvg = = = =vi= m/svf= m/s2 t= a avvg == = =vi= m/svf= m/s2 t= vafa vgvi == =vi= m/s svf vi m/s m/s m/s2vf vi aavgMotion In One Dimension, Practice BSection One Student Edition SolutionsI Ch.

10 2 3 Copyright by holt , Rinehart and Winston. All rights m/saavg= vf tvi ===vf= m/s t=25 10 3m/s2a. v=aavg t =( 10 3m/s2)( min)(60 s/min) = t= v+vi= m/s + m/s =vi= m/s 10 3m/s2 m/s 1500 s m/s ( m/s) (25 min)(60 s/min) m/s x= 12 (vi+vf) t= 12 ( m/s + m/s)( s) =vf= m/s t= m/s x= 12 (vi+vf) t= 12 ( m/s + m/s)( s) =vf= m/s t= m/s t= v2i +xvf == x=99 mvf= m/svf= 2 tx vi= m/s = 101m/s m/s = x= km t= min24 m/s(2)( 103m) ( min)(60 s/min) s(2)(99 m) m/s+ m21 mMotion In One Dimension, Practice m/sa= m/s2 t= svf=vi+a t= m/s +( m/s2)( s)vf= m/s + m/s = x=vi t+ 12 a t2 x=( m/s)( s) + 12 ( m/s2)( s)2 x=23 m + m =29 m/sMotion In One Dimension, Practice m/sa= m/s2 t= svf=vi+a t= m/s +( m/s2)( s)vf= m/s + m/s = x=vi t+ 12 a t2 x=( m/s)( s) + 12 ( m/s2)( s)2 x= m + m = m/sHolt Physics Solution ManualI Ch. 2 4 Copyright by holt , Rinehart and Winston.