Statistics Fourth Edition Freedman Solution Manual

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Solution manual for university physics with modern physics 14th edition young freedman.1.1UNITS, PHYSICAL QUANTITIES, AND VECTORSDownload Full Solution manual for university physics with modern physics 14thedition young freedmanIDENTIFY: Convert unitsfrommi to km and fromkm to ft.SET UP: 1 in. = 2.54 cm, 1 km = 1000 m, 12 in. = 1 ft, 1 mi = 5280 ft.5280 ft 12 in. 2.54 cm1 m 1 kmEXECUTE: (a) 1.00 mi = (1.

00 mi) = 1.61 km1 in. 1031 mi 1 ft 102 cm m103 m 102 cm 1 in. 1 ft 3(b) 1.00 km = (1.00 km) = 3. 28 ×10 ft1 m 12 in.1 km 2. 54 cmEVALUATE: A mile is a greater distance than a kilometer. There are 5280 ft in a mile but only 3280 ft ina km.1.2.

IDENTIFY: Convert volume units from L to in.3.SET UP: 1 L = 1000 cm3. = 2.54 cmEXECUTE:1000 cm3 1 in.

9 in.3.0.473 L × ×1 L 2.54 cmEVALUATE: 1 in.3 is greater than 1 cm3, so the volume in in.3 is a smaller number than the volume incm3, which is 473 cm3.1.3. IDENTIFY: We knowthe speed oflight in m/s. Convert 1.00 ft to m and t from s to ns.SET UP: The speed oflight is v = 3.00 ×108 m/s.

1 ft = 0.3048 m. 1 s = 109 ns.EXECUTE: t =0.3048 m= 1.02× 10−9s = 1.02 ns3. 00 ×108m/sEVALUATE: In 1.00 s light travels 3. 00 × 108 m = 3. 00 × 105 km = 1. 86 ×105 mi.1.4.

IDENTIFY: Convert the units fromg to kg and from cm3 to m3.SET UP: 1 kg = 1000 g. 1 m = 100 cm.EXECUTE: 19.

3g 1 kg 100 cm 34 kg× × = 1.93×10cm31 m m31000 gEVALUATE: The ratio that converts cm to m is cubed, because we need to convert cm3 to m3.1.5. IDENTIFY: Convert volume units fromin.3 to L. SETUP: 1 L = 1000 cm3.

= 2.54 cm.EXECUTE: (327 in.3 ) × (2.54 cm/in.)3 × (1L/1000 cm3) = 5.36 LEVALUATE: The volume is 5360 cm3. 1 cm3 is less than 1 in.3, so thevolume in cm3 is a larger number than thevolume in in.3.© Copyright 2016Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as theycurrentlyexist. Noportion ofthis material maybe reproduced, in anyform orby any means, without permissionin writingfromthe publisher.1-1.1-2 Chapter 11.6. IDENTIFY: Convert ft2 to m2 and then to hectares.SET UP: 1.00 hectare = 1. 1 ft = 0.3048 m.EXECUTE: Thearea is43,600 ft2 0.3048 m 2 1.00 hectare(12.0 acres) = 4.86 hectares.4 21 acre 1.

00 ×10 m1.00 ftEVALUATE: Since 1 ft = 0.3048 m, 1 ft21.7. IDENTIFY: Convert seconds to years. 1 gigasecond is a billion seconds.SET UP: 1 gigasecond = 1×109 s. 1 day = 24 h. 1 h = 3600 s.EXECUTE: 9 1 h 1 day 1 y1.00 gigasecond = (1.00 × 10 s) = 31.7 y.3600 s 24 h 365 daysEVALU A TE: The conversion 1 y = 3.156 × 107 s assumes 1 y = 365.24 d, which is the average foroneextra day every fouryears,in leap years.The problemsays insteadto assume a 365-day year.1.8. IDENTIFY: Apply the givenconversionfactors.SET UP: 1 furlong = 0.

1250 mi and 1 fortnight = 14 days. 1 day = 24 h.0.125 mi 1 fortnight 1 dayEXECUTE: (180, 000 furlongs/fortnight) = 67 mi/h14 days1 furlong 24 hEVALUATE: A furlong is less than a mile and a fortnight is many hours,sothe speedlimit in mph is amuch smaller number.1.9.

IDENTIFY: Convert miles/gallon to km/L.SET UP: 1 mi = 1.609 km. 1 gallon = 3.788 L.EXECUTE: (a)1.609 km 1 gallon55. 0 miles/gallon = (55.0 miles/gallon) = 23.4 km/L.1 mi3.788 L(b) The volume ofgas required is1500 km= 64.1 L.64.1 L= 1.4 tanks.23.4 km/L 45 L/tankEVALUATE: 1 mi/gal = 0.425 km/L. A km is very roughly halfa mile and there are roughly 4liters in agallon,so 1 mi/gal ∼ 24 km/L, which is roughly ourresult.1.10. IDENTIFY: Convert units.SET UP: Use the unit conversionsgiven in the problem.Also,100 cm = 1 m and 1000 g = 1 kg.EXECUTE: (a)mi 1 h 5280 ft= 88ft60h 1 mi s3600 s(b) 32ft 30.48 cm 1 m= 9.8m1ft s2s2 100 cm(c)g 100 cm 3 1 kg 3 kg1.

0 = 10cm31 m m31000 gEVALUATE: The relations 60mi/h = 88 ft/s and 1 g/cm3 = 103 kg/m3 are exact. The relation32 ft/s2 = 9.8 m/s2 is accurate to only two significantfigures.1.11. IDENTIFY: We knowthe density andmass; thus we can find the volume using the relationdensity = mass/volume = m/V. The radius is then foundfromthe volume equation fora sphere andthe result forthe volume.SET UP:EXECUTE:Density = 19.5 g/cm3 and m = 60.0 kg.

For a sphere V =4 π r3.critical 3V = m 60.0 kg 1000 g = 3080 cm3./density =3critical= (0.3048)2 m2.19.5 g/cm 1.0 kg© Copyright 2016Pearson Education, Inc. All rights reserved.

This material is protectedunder all copyright laws as they currentlyexist.No portion ofthis material maybe reproduced, in anyformor by anymeans, without permission in writingfrom thepublisher.Units, Physical Quantities,and Vectors 1-3r = 33V= 33(3080 cm3) = 9.0 cm.4π 4πEVALUATE: The density is very large, so the 130-pound sphereis small in size.1.12. IDENTIFY: Convert units.SET UP: We know theequalities 1 mg = 10−3 g, 1 µg 10−6 g, and 1 kg = 103g.10−3g 1 μg 5EXECUTE: (a) (410 mg/day) = 4.10 ×10 μg/day.1 mg 10−6g(b) (12 mg/kg)(75 kg)10−3 g= (900 mg) = 0.900 g.1 mg10−3g(c) The mass of each tablet is (2.0 mg) = 2.0 ×10−3g. The numberoftablets required eachday is1 mgthe numberofgrams recommended perday divided by the numberofgrams pertablet:0.0030 g/day= 1.5 tablet/day. Take2 tablets each day.2.0 ×10−3 g/tablet1 mg(d) (0.000070 g/day) = 0.070 mg/day.−310 gEVALUATE: Quantities in medicine and nutrition are frequently expressed in a wide variety of units.1.13.

IDENTIFY: Modelthebacteria as spheres. Usethe diameter to find theradius, then find the volume and surfacearea using the radius.SET UP: From Appendix B, thevolume V of a sphere in terms of its radius is V = 43 π r3 while its surfacearea A is A = 4π r2. The radius is one-halfthe diameterorr = d/2= 1.0 μ m.

Finally, the necessaryequalities forthis problemare: 1 μ m = 10−6 m; 1 cm = 10−2 m; and 1 mm = 10−3 m.EXECUTE: V = 4 π r3=4 π (1.0 μ m)3 10−6 m 3 1 cm 3= 4.2×10−12 cm3and−23 31 μ m 10 mA = 4π r2= 4π (1.0 μ m)2 10−6 m 2 1 mm 2= 1.3×10−5 mm2−31 μ m 10 mEVALUATE: On a human scale, the results are extremely small. This is reasonable because bacteria are notvisible without a microscope.1.14. IDENTIFY: When numbers are multiplied or divided, the number of significant figures in the result can be nogreater than in the factor with the fewest significant figures. When we add or subtract numbers it is the locationof the decimal that matters.SET UP: 12 mm has two significant figures and 5.98 mm has three significant figures.EXECUTE: (a) (12 mm) × (5.98 mm) = 72 mm2 (two significant figures)(b)5.98 mm= 0.50 (also two significantfigures) 12 mm(c) 36 mm (to the nearest millimeter)(d) 6 mm(e) 2.0 (two significant figures)EVALUATE: The length of therectangle is known only to thenearest mm, so theanswers in parts (c) and(d) are known only to the nearest mm.1.15.

IDENTIFY: Useyour calculator to display π ×107. Compare that number to the number of seconds in a year.SET UP: 1 yr= 365.24 days,1day = 24 h, and 1 h = 3600 s.© Copyright 2016Pearson Education, Inc. All rights reserved.

This material is protectedunder all copyright laws as they currentlyexist.No portion ofthis material maybe reproduced, in anyformor by anymeans, without permission in writingfrom thepublisher.1-4 Chapter 1EXECUTE: 24 h 3600 s 7π × 10 7s = 3. 14159×10 7s(365.

24 days/1 yr) = 3. 15567×10 s;1 day 1 hThe approximate expression is accurate to two significant figures.The percenterroris 0.45%.EVALUATE: The close agreement is a numerical accident.1.16. IDENTIFY: To asses theaccuracy of the approximations, we must convert them to decimals.SET UP: Use a calculator to calculate thedecimal equivalent of each fraction and then round thenumeral to thespecified number of significant figures. Compare to π rounded to the same number of significant figures.EXECUTE: (a) 22/7 = 3.14286 (b) 355/113 = 3.14159 (c) Theexact value of π rounded to six significant figuresis 3.14159.EVALUATE: We see that 355/113 is a much better approximation to π than is 22/7.1.17.

IDENTIFY: Express 200 kg in pounds. Express each of 200 m, 200 cm and 200 mm in inches.

Express 200months in years.SET UP: A mass of 1 kg is equivalent to a weight of about 2.2 lbs.1 in. 1 y = 12 months.EXECUTE: (a) 200 kg is a weight of 440 lb. This is much larger than thetypicalweight of a man.(b) 200 m = (2. 00 × 104 1 in.

9 ×103inches. This is much greater than the height of a person.cm)2.54 cm(c) 200 cm = 2.00 m = 79 inches = 6.6 ft.

Some peopleare this tall, but not an ordinary man.(d) 200 mm = 0.200 m = 7.9 inches. This is much too short.1 y(e) 200 months = (200 mon) = 17 y. This is the age of a teenager; a middle-aged man is much12 monolderthan this.EVALUATE: None are plausible. When specifying the value of a measured quantity it is essential to give theunits in which it is being expressed.1.18. IDENTIFY: Estimatethe number of peopleand then use the estimates given in theproblem to calculate the numberof gallons.SET UP: Estimate 3×108 people, so 2 ×108 cars.EXECUTE: (Number of cars × miles/car day)/(mi/gal) = gallons/day(2 × 108 cars × 10000 mi/yr/car × 1 yr/365 days)/(20mi/gal) = 3×108 gal/dayEVALUATE: The number of gallons of gas used each day approximately equals the population of the U.S.1.19. IDENTIFY: Estimatethenumber of blinks per minute.

Convert minutes to years. Estimatethetypical lifetimein years.SET UP: Estimate that we blink 10 times per minute.1 y = 365 days. 1 day = 24 h, 1 h = 60 min. Use80 yearsfor the lifetime.EXECUTE: Thenumber of blinks is60 min 24 h 365 days (80 y/lifetime) = 4 ×108(10 per min)1 y1 h 1 dayEVALUATE: Our estimate of the number of blinks per minute can be off by a factor of two but ourcalculation is surely accurate to a power of 10.1.20. IDENTIFY: Approximate the number of breaths per minute. Convert minutes to years and cm3 to m3 to findthe volume in m3 breathed in a year.24 h 60 min = 5.3×105 min.

102SET UP: Assume 10 breaths/min. 1 y = (365 d) cm = 1 m so1 d 1 h106 cm3 = 1 m3. The volume of a sphere is V = 43 π r 3 = 16 π d 3, where r is the radius and d is thediameter. Don’t forget to accountforfourastronauts.5minEXECUTE: (a) The volumeis (4)(10 breaths/min)(500 × 10−6m3) 5. 3×10= 1× 104m3/yr.1 y© Copyright 2016Pearson Education, Inc. All rights reserved.

This material is protectedunder all copyright laws as they currentlyexist.No portion ofthis material maybe reproduced, in anyformor by anymeans, without permission in writingfrom thepublisher.Units, Physical Quantities,and Vectors 1-5(b) d =6V 1/3 61×104m3 1/3= = 27 mπ πEVALUATE: Our estimate assumes that each cm3 of air is breathed in only once, where in reality not all theoxygen is absorbed from the air in each breath. Therefore, a somewhat smaller volume would actually berequired.1.21. IDENTIFY: Estimation problem.SET UP: Estimate that the pile is 18 in. Use the density of gold to calculate the mass of goldin the pile and from this calculate thedollar value.EXECUTE: Thevolume of gold in the pile is V = 18 in. = 22,000 in.3. Convert to cm3:V = 22,000 in.3 (1000 cm3 /61. 02 in.3 ) = 3.

6 × 105 cm3.The density of gold is 19.3 g/cm3, so the mass of this volume of gold is m =(19. 6 × 105 cm3 ) = 7 × 106 g.The monetary value ofone gramis $10, so the gold has a value of($10/gram)(7 × 106 grams)= $7 ×107,or about $100 ×106 (one hundred million dollars).EVALUATE: This is quite a large pile of gold, so such a large monetary value is reasonable.1.22. IDENTIFY: Estimatethenumber of beats per minute and theduration of a lifetime.

Thevolume of blood pumpedduring this interval is then the volume per beat multiplied by the totalbeats.SET UP: An average middle-aged (40 year-old) adult at rest has a heart rateof roughly 75 beats per minute.To calculate the number of beats in a lifetime, use the current average lifespan of 80 years.EXECUTE:60 min 24 h 365 days 80 yr = 3×109 beats/lifespanNbeats = (75 beats/min)yr1 h 1 day lifespanV = (50 cm3 /beat) 1 L 1 galb lo o d1000 cm3 3.788 LEVALUATE: This is a very large volume.3×109 beats= 4 ×107gal/lifespanlifespan1.23. IDENTIFY: Estimate thediameter of a drop and from that calculate the volume of a drop, in m3.

Convertm3 to L.SET UP: Estimate the diameter of a drop to be d = 2 mm. The volume of a spherical drop is V =43 π r 3 = 16 π d 3. 103 cm3 = 1 L.π (0.2 cm)3 = 4 ×10−3 cm3.

Thenumber of drops in 1.0 L is1000 cm3EXECUTE: V =1= 2 ×1056 4 ×10−3 cm3EVALUATE: Since V ∼ d 3, if ourestimate of the diameterofa drop is off by a factorof2 then ourestimate ofthe numberofdrops is offby a factorof8.1.24. IDENTIFY: Draw the vector addition diagram to scale.SET UP: The two vectors A and B are specified in the figure that accompanies the problem.EGXEC U TE: (a) The diagram forGR = A + B is given in Figure 1.24a. Measuring the length and angle ofR gives R = 9.0 m and an angle of θ = 34°.(b) Thediagram forG G GE = A − B is given in Figure 1.24b.

Measuring the length and angle of E givesD = 22 m and an angle of θ = 250°.G GG G(c) − A − B = −(A + B), so − A − B has a magnitude of 9.0 m (the same as A + B ) and an angle with the+ x axis of 214° (oppositeto thedirection of A + B).G G G G G+ x axis of 70° (opposite(d) B − A = −( A − B), so B − A has a magnitude of 22 m and an angle with theG Gto the direction ofA − B ). GEVALUATE: The vector− A is equalin magnitude and oppositein direction to the vector A.© Copyright 2016Pearson Education, Inc. All rights reserved. This material is protectedunder all copyright laws as they currentlyexist.No portion ofthis material maybe reproduced, in anyformor by anymeans, without permission in writingfrom thepublisher.1-6 Chapter 1Figure 1.241.25.

IDENTIFY: Draw each subsequent displacement tail to head with the previous displacement. The resultantdisplacement is the single vector that points fromthe starting point to the stoppingpoint.SET UP: Call the three displacements A,B, and C. The resultant displacement R is given byG G G GR  A  B  C.

GEXECUTE: The vectoradditiondiagramis given in Figure 1.25. Careful measurement gives that R is78 km, 38Dnorth ofeast.EVALUATE: Themagnitude of the resultant displacement, 7.8 km, is less than thesum of the magnitudes of theindividual displacements, 2 6 km  4 0 km  31 km.Figure 1.251.26. IDENTIFY: Since she returns to the starting point, the vector sum of the four displacements must be zero.SET U P: Call thethree given displacementsGA, B, and C, and call the fourth displacement D.G G G GA  B  C  D  0. GEXECUTE: The vectoradditiondiagramis sketchedin Figure 1.26. Careful measurement gives that Dis144 m, 41 southofwestGEVALU A TE:GD is equal in magnitude and oppositein direction to thesum A  B  C.Figure 1.26© Copyright 2016Pearson Education, Inc.

All rights reserved. This material is protectedunder all copyright laws as they currentlyexist.No portion ofthis material maybe reproduced, in anyformor by any means, without permission in writingfrom thepublisher.Units, Physical Quantities,and Vectors 1-71.27. IDENTIFY: For each vector V, use that Vx = V cosθ and V y = V sinθ, whenwith the + x axis, measured counterclockwise fromthe axis.GGSET UP: For A, θ = 270.0°.

For B, θ = 60.0°. For C, θ = 205.0°. For D,θ is the angle V makesθ = 143.0°.EXECUTE: Ax = 0, Ay = −8.00 m.

Bx = 7.50 m, By = 13.0 m. Cx = −10.9 m, Cy = −5.07 m.Dx = −7.99 m, Dy = 6.02 m.EVALUATE: The signs of the components correspond to thequadrant in which the vector lies.1.28. IDENTIFY: tanθ = Ay, for θ measured counterclockwise from the + x -axis.AxSET UP: A sketch of Ax, Ay and A tells us the quadrant in which A lies.EXECUTE:Ay(a) tan θ = =− 1.00 m= −0.500. Θ = tan−1(− 0. 500) = 360° − 26.6° = 333°.A 2.00 mx(b) tan θ = Ay = 1.00 m = 0.500. Θ = tan−1(0.500) = 26. 6°.A 2.00 mx(c) tan θ =Ay=1.00 m= −0.500.

Θ = tan−1(− 0. 500) = 180° − 26.6° = 153°.A − 2.00 mx(d) tan θ =Ay=−1.00 m= 0.500. Θ = tan−1(0.500) = 180° + 26.

6° = 207°A −2.00 mxEVALUATE: Theangles 26.6° and 207° have the same tangent. Our sketch tells us which is the correct valueof θ.1.29. IDENTIFY: Given the direction and one component of a vector, find the other component and themagnitude.SET UP: Use the tangent of the given angle and thedefinition of vector magnitude.EXECUTE: (a) tan32.0° =AxAyAx = (9.60 m)tan32.0° = 6.00 m. Ax = −6.00 m.(b) A = Ax2+ Ay2=11.3 m.EVALUATE: The magnitude is greater than either of thecomponents.1.30. IDENTIFY: Given the direction and one component of a vector, find the other component and themagnitude.SET UP: Use the tangent of the given angle and thedefinition of vector magnitude.EXECUTE: (a) tan34.0° =AxAyA = Ax = 16.0 m = 23.72 mytan34.0° tan34.0°Ay = −23.7 m.(b) A = A2+ A2 = 28.6 m.x yEVALUATE: The magnitude is greater than either of thecomponents.1.31.

IDENTIFY:G G GIf C= A + B, then C x = Ax + Bx and C y = Ay + By. Use Cx and Cy to find the magnitude andGdirection ofC.© Copyright 2016PearsonEducation, Inc. All rights reserved.

This material is protectedunder all copyright laws as theycurrently exist. Noportion ofthis material maybe reproduced, in anyform orby any means, without permissionin writingfromthe publisher.1-8 Chapter 1SET UP: From Figure E1.24 in the textbook, Ax = 0, Ay = −8.00 m and B x = + Bsin30.0° = 7.50 m,B y = + B cos30.

0° = 13.0 m.G G GEXECU T E: (a) C = A + B so C x = Ax + Bx = 7.50 m and C y = Ay + By = +5.00 m. C = 9.01 m.tanθ =Cy=5.00 mand θ = 33. 7°.Cx 7.50 mG G G(b)G G GB + A = A + B, so B + A has magnitude 9.01 m and direction specified by 33.

7°.G G G= Ax − Bx = − 7.50 m and D y = Ay − By = − 21.0 m. D = 22.3 m.(c) D = A − B so DxDy − 21.0 m rdtanφ = = − 7.50 m and φ = 70.

D is in the 3 quadrant and the angle θ counterclockwise from theDx+ x axis is 180° + 70. 3°.(d)G G G G G GB − A = − ( A − B), so B − A has magnitude 22.3 m and direction specified by θ = 70.3°.EVALUATE: These results agree with those calculated from a scale drawing in Problem 1.24.1.32. IDENTIFY: Find thevector sum of the three given displacements.SET UP: Use coordinates for which + x is east and + y is north. The driver’s vector displacements are:K K KA = 2. 6 km, 0° of north;B = 4.0 km, 0° of east; C = 3.

1 km, 45° north of east.EXECUTE: R x = Ax + B x + Cx = 0 + 4.0 km + (3. 1 km)cos(45°) = 6.2 km; R y = Ay + B y + Cy =2. 6 km + 0 + (3. 1 km)(sin 45°) = 4.8 km; R = R x2 + Ry2 = 7.8 km; θ = tan−1(4.

2 km) =38°;KR = 7. 8 km, 38° north of east. This result is confirmed by thesketch in Figure 1.32.EVALUATE: Both Rx and Ry are positiveand R is in the first quadrant.Figure 1.321.33.

IDENTIFY: Vector addition problem. We are given themagnitude and direction of three vectors and are askedto find their sum.SET UP:A = 3.25 kmB = 2.20 kmC = 1.50 kmFigure 1.33a.© Copyright 2016Pearson Education, Inc.

All rights reserved. This material is protectedunder all copyright laws as they currentlyexist.No portion ofthis material maybe reproduced, in anyformor by anymeans, without permission in writingfrom thepublisher.Units, Physical Quantities,and Vectors 1-9Select a coordinate systemwhere + x is east and + y is north.LetGA, B, and C be the threeG G G Gdisplacementsofthe professor.Then the resultantdisplacement R is given by R = A + B + C.

By themethod of components, Rx = Ax + Bx + Cx and Ry  Ay  B y  Cy. Find the x and y components ofeachvector; add themto find the components ofthe resultant.Then the magnitude anddirection oftheresultant can be foundfromits x and y components that we have calculated.As always it is essentialtodraw a sketch.EXECUTE:Ax  0, Ay 3.25 kmBx = −2.20 km, By = 0C x = 0, Cy = −1.50 kmRx = Ax + Bx + CxRx = 0 − 2.20 km + 0 = −2.20 kmRy  Ay  B y  CyRy = 3.25 km + 0 − 1.50 km = 1.75 kmFigure 1.33bR = R 2+ R2= (−2.20 km)2+ (1.75 km)2x yR = 2.81 kmtanθ =Ry=1.75 km= −0.800R −2.20 kmxθ = 141.5°Figure 1.33cThe angle θ measured counterclockwise fromthe + x-axis.

In terms of compass directions,the resultantdisplacement is 38.5° N of W.EVALU A TE: Rx 0, so R is in the 2nd quadrant.This agrees with the vectoradditiondiagram.1.34. IDENTIFY: Use A  A 2  A2 and tanθ  Ay to calculate the magnitude and directionofeach ofthex yAxgiven vectors.SET UP: A sketch of Ax, Ay and A tells us the quadrant in which A lies.2 2 5.20EXECUTE: (a) (−8.

20 cm) = 10.0 cm, arctan = 148.8° (which is180° − 31. 2° ).−8.60(b) (− 9.7 m)2 + (− 2.45 m)2 −2.45= 14° + 180° = 194°.= 10.0 m, arctan−9.7(c) (7.75 km)2 + (− 2. 70 km)2 = 8.21 km,−2.7= 340.8° (which is 360° − 19.

2° ).arctan7.75EVALUATE: In each case the angle is measured counterclockwise from the + x axis. Our results for θ agreewith our sketches.© Copyright 2016Pearson Education, Inc. All rights reserved. This material is protectedunder all copyright laws as they currentlyexist.No portion ofthis material maybe reproduced, in anyformor by anymeans, without permission in writingfrom thepublisher.1-10 Chapter 1G G1.35.

IDENTIFY: Vector addition problem. A − B = A + ( − B).SET UP: Find the x- and y-components of A and B. Then the x- and y-components of the vector sum arecalculated fromthe x- and y-components ofA and B.EXECUTE:Ax = Acos(60.

80 cm)cos(60. 0°) = + 1.40cm Ay = Asin (60.0°)Ay = (2. 80 cm)sin (60.0°) = +2.425 cmB x = B cos(− 60.

0°)Bx = (1.90 cm)cos(−60.0°)= +0.95cm B y = Bsin(−60.0°)By = (1.90 cm)sin (−60.0°) = −1.645 cmNotethat the signs of thecomponents correspond tothe directions of the component vectors.Figure 1.35aG G G(a) Now let R = A + B.R x = Ax + Bx = +1. 40 cm + 0.95 cm = + 2.35 cm.R y = Ay + By = +2.425 cm − 1. 78 cm.R = R x2 + Ry2 = (2.35 cm)2 + (0.78 cm)2R = 2.48 cmtanθ =Ry =+0.78 cm= +0.3319R +2.35 cmxθ = 18. 4°Figure 1.35bGEVALUATE: The vector addition diagram for R = A + B isR is in the 1st quadrant, with Ry 0, so R is in the 2nd quadrant.1.39. IDENTIFY: Use trigonometry to find the components ofeach vector.Use R  A  B  andR y  Ay  By  to find the components ofthe vectorsum.The equationxG xˆ x ˆexpresses aA = Axi + Ay jvectorin terms of its components.SET UP: Use the coordinates in the figure that accompanies theproblem.EXECUTE: G ˆ ˆ ˆ ˆ(a) A = (3. 23 m)i + (3.38 m) jG ˆ ˆ ˆ ˆB = − (2. 40 m)sin30.0° j = (−2.08 m)i + (−1.20 m) jG G− (4.00) G ˆ ˆ ˆ ˆ(b) C = (3.00) A B = (3.00)(1.23 m)i + (3.00)(3.38 m) j − (4.00)(−2.08 m)i − (4.00)(−1.20 m) jG ˆ ˆC = (12.01 m)i + (14.94 m) j(c) From A  A 2  A2 and tanθ Ay,x yAx14.94 mC = (12.01 m)2+ (14.94 m)2= 19.17 m, arctan = 51.2°12.01 mEVALU A TE: Cx and Cyare both positive,so θ is in the first quadrant.1.40.

IDENTIFY: We use the vector components and trigonometry to find the angles. SETUP: Usethe fact that tanθ  Ay / Ax.EXECUTE: (a) tanθ  A / A  6.00. Θ = 117° with the +x-axis.y x−3.00(b) tanθ  B / B  2.00. Θ = 15.9°.y x7.00(c) First find thecomponents of C.

Cx = Ax + Bx = -3.00 + 7.00 = 4.00,Cy = Ay + By = 6.00 + 2.00 = 8.00tanθ  C / C 8.00  2.00. Θ = 63.4°y x4.00EVALU A TE: Sketching each of the three vectors to scale will show that the answers are reasonable.1.41. IDENTIFY:G G GGA and B are given in unit vector form.

Find A, B and the vector difference A − B.G G G G GSET UP: A = −2.00 i + 3. 00 j + 4.00 k, B = 3.00 i + 1.00 j− 3.

00kUse A = Ax2 + Ay2 + Az2 to findthe magnitudes of the vectors.EXECUTE: (a) A = A 2 + A 2 + A2 = (−2.00)2 + (3.00)2 + (4.00)2 = 5.38x y z© Copyright 2016Pearson Education, Inc. All rights reserved. This material is protectedunder all copyright laws as they currentlyexist.No portion ofthis material maybe reproduced, in anyformor by anymeans, without permission in writingfrom thepublisher.1-14 Chapter 1B = B 2 + B 2 + B2= (3. 00) 2 + ( − 3.

36Gx y z(b) G ˆ ˆ ˆ ˆ ˆ ˆA − B = (−2. 00i + 3.00 j + 4. 00k ) − (3.00i + 1.00 j − 3.00k)G G ˆ ˆ ˆ ˆ ˆ ˆA − B = ( −2. 00 − 3.00)i + (3. Install forticlient without antivirus protection.

00) j + (4.00 − (−3. 00i + 2.00 j + 7. 00k.G G G(c) Let C = A − B, so Cx = −5.00, Cy = + 2.00, Cz = +7.00G G G GC = C x2 + C y2 + Cz2 = (− 5.00)2 + (2. 00)2 = 8.83soG G G− A have thesame magnitude but oppositedirections.B − A = − ( A − B), A − B and BEVALUATE: A, B, and C are each larger than any of their components.1.42. IDENTIFY: Target variables areG GA⋅ B and the angle φ between the two vectors.SET UP: We are givenG GA and B in unit vector form andcan take the scalar product usingG G= Ax Bx + Ay B y + Az Bz. The angle φ can then be found from A ⋅ B = AB cosφ.A⋅ BEXECUTE: G ˆ ˆ G ˆ ˆ A = 8.06, B = 5.39.(a) A = 4.00i + 7.00 j, B = 5.00i − 2.00 j;G G ˆ ˆ ˆ ˆA ⋅ B = (4.00i + 7.0 0 j ) ⋅ (5.00i − 2.00 j) = (4.00)(5.00) + (7.00)(− 2.00) = 20.0 − 14.0 = +6.00.G G(b) cosφ =A⋅ B = 6.00 = 0.1382; φ = 82.1 °.(8.06)(5.39)AB GEVALU A TE: The component of along A is in the same direction as A, so the scalar product isBpositive and theangle φ is less than 90°.1.43.

IDENTIFY:G GA ⋅ B = AB cosφSET UP: ForG Gφ= 150. For B and C, φ = 145. For A and C, φ = 65. 0°.A and B,EXECUTE:G G= (8. 0 m)cos150.0° = −104 m2(a) A⋅B(b)G G= (15.0 m)(12.0 m)cos145. 0° = −148 m2B⋅C(c)G GA ⋅ C = (8.00 m)(12. 0 m)cos65.0° = 40.6 m2EVALU A TE: When φ 90° the scalar product is negative.1.44.

IDENTIFY: Target variable is the vector A × B expressed in terms ofunit vectors.SET UP: We are givenG Gin unit vectorformand can take the vectorproductusingA and Bˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆi × i = j × j = 0, i × j = k, and j × i = −k.EXECUTE: G ˆ ˆ G = ˆ ˆA = 4.00i + 7.00 j, B 5.00i − 2.00 j.G G ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ = 0A × B = (4.00i + 7.00 j) × (5.00i − 2.00 j) = 20.0i × i − 8.00i × j + 35.0 j × i − 14.0 j × j. But i × i = j × jand ˆ ˆ ˆ ˆ ˆ ˆ, so G G ˆ ˆ ˆ Gi × j = k, j × i = −k A × B = −8.00k + 35.0(− k ) = −43.0k. The magnitude of A × B is 43.0.EVALU A TE: Sketch the vectorsGA and B in a coordinate systemwhere the xy-plane is in theplane of thepaperand the z-axis is directed out towardyou.By the right-handrule A × B is directed into the plane ofthe paper,in the − z-direction.This agrees with the above calculation that usedunit vectors.Figure 1.44.© Copyright 2016Pearson Education, Inc.

All rights reserved. This material is protectedunder all copyright laws as they currentlyexist.No portion ofthis material maybe reproduced, in any formor by anymeans, without permission in writingfrom thepublisher.Units, Physical Quantities,and Vectors 1-151.45.

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