Respostas Young Freedman 14e capt01, Exercícios de Física

Baixe Respostas Young Freedman 14e capt01 e outras Exercícios em PDF para Física, somente na Docsity! © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 1-1 1.1. IDENTIFY: Convert units from mi to km and from km to ft. SET UP: 1 in 2 54 cm,. = . 1 km 1000 m,= 12 in 1 ft,. = 1 mi 5280 ft.= EXECUTE: (a) 2 3 5280 ft 12 in 2 54 cm 1 m 1 km1 00 mi (1 00 mi) 1 61 km 1 mi 1 ft 1 in 10 cm 10 m . .⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞. = . = .⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟.⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠ (b) 3 2 310 m 10 cm 1 in 1 ft1 00 km (1 00 km) 3 28 10 ft 1 km 1 m 2 54 cm 12 in ⎛ ⎞⎛ ⎞ .⎛ ⎞⎛ ⎞. = . = . ×⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ . .⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠ EVALUATE: A mile is a greater distance than a kilometer. There are 5280 ft in a mile but only 3280 ft in a km. 1.2. IDENTIFY: Convert volume units from L to 3in .. SET UP: 31 L 1000 cm .= 1 in 2 54 cm. = . EXECUTE: 33 31000 cm 1 in0 473 L 28 9 in 1 L 2 54 cm ⎛ ⎞ .⎛ ⎞. × × = . . .⎜ ⎟ ⎜ ⎟⎜ ⎟ .⎝ ⎠⎝ ⎠ EVALUATE: 31 in. is greater than 31 cm , so the volume in 3in. is a smaller number than the volume in 3cm , which is 3473 cm . 1.3. IDENTIFY: We know the speed of light in m/s. / .t d v= Convert 1.00 ft to m and t from s to ns. SET UP: The speed of light is 83 00 10 m/s.v = . × 1 ft 0 3048 m.= . 91 s 10 ns.= EXECUTE: 98 0 3048 m 1 02 10 s 1 02 ns 3 00 10 m/s t − .= = . × = . . × EVALUATE: In 1.00 s light travels 8 5 53 00 10 m 3 00 10 km 1 86 10 mi.. × = . × = . × 1.4. IDENTIFY: Convert the units from g to kg and from 3cm to 3m . SET UP: 1 kg 1000 g.= 1 m 100 cm.= EXECUTE: 3 4 3 3 g 1 kg 100 cm kg19 3 1 93 10 1000 g 1 mcm m ⎛ ⎞ ⎛ ⎞. × × = . ×⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ EVALUATE: The ratio that converts cm to m is cubed, because we need to convert 3cm to 3m . 1.5. IDENTIFY: Convert volume units from 3in. to L. SET UP: 31 L 1000 cm .= 1 in 2 54 cm.. = . EXECUTE: 3 3 3(327 in ) (2 54 cm/in ) (1L/1000 cm ) 5 36 L. × . . × = . EVALUATE: The volume is 35360 cm . 31 cm is less than 31 in ,. so the volume in 3cm is a larger number than the volume in 3in .. UNITS, PHYSICAL QUANTITIES, AND VECTORS 1 1-2 Chapter 1 © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 1.6. IDENTIFY: Convert 2ft to 2m and then to hectares. SET UP: 4 21 00 hectare 1 00 10 m .. = . × 1 ft 0 3048 m.= . EXECUTE: The area is 22 4 2 43 600 ft 0 3048 m 1 00 hectare(12 0 acres) 4 86 hectares. 1 acre 1 00 ft 1 00 10 m ,⎛ ⎞ . .⎛ ⎞ ⎛ ⎞. = .⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ . . ×⎝ ⎠ ⎝ ⎠⎝ ⎠ EVALUATE: Since 1 ft 0 3048 m,= . 2 2 21 ft (0 3048) m .= . 1.7. IDENTIFY: Convert seconds to years. 1 gigasecond is a billion seconds. SET UP: 91 gigasecond 1 10 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 days ⎛ ⎞⎛ ⎞⎛ ⎞. = . × = .⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠ EVALUATE: The conversion 71 y 3 156 10 s= . × assumes 1 y 365 24 d,= . which is the average for one extra day every four years, in leap years. The problem says instead to assume a 365-day year. 1.8. IDENTIFY: Apply the given conversion factors. SET UP: 1 furlong 0 1250 mi and 1 fortnight 14 days= . = . 1 day 24 h= . EXECUTE: 0 125 mi 1 fortnight 1 day(180 000 furlongs fortnight) 67 mi/h 1 furlong 14 days 24 h , / ⎛ ⎞⎛ ⎞. ⎛ ⎞ =⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠ EVALUATE: A furlong is less than a mile and a fortnight is many hours, so the speed limit in mph is a much 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 mi 3 788 L .⎛ ⎞⎛ ⎞. = . = .⎜ ⎟⎜ ⎟.⎝ ⎠⎝ ⎠ (b) The volume of gas required is 1500 km 64 1 L. 23 4 km/L = . . 64 1 L 1 4 tanks. 45 L/tank . = . EVALUATE: 1 mi/gal 0 425 km/L.= . A km is very roughly half a mile and there are roughly 4 liters in a gallon, so 241 mi/gal km/L,∼ which is roughly our result. 1.10. IDENTIFY: Convert units. SET UP: Use the unit conversions given in the problem. Also, 100 cm 1 m= and 1000 g 1 kg.= EXECUTE: (a) mi 1 h 5280 ft ft60 88 h 3600 s 1 mi s ⎛ ⎞⎛ ⎞⎛ ⎞ =⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠ (b) 2 2 ft 30 48 cm 1 m m32 9 8 1ft 100 cms s ⎛ ⎞.⎛ ⎞ ⎛ ⎞ = .⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠ (c) 3 3 3 3 g 100 cm 1 kg kg1 0 10 1 m 1000 gcm m ⎛ ⎞⎛ ⎞⎛ ⎞. =⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠ EVALUATE: The relations 60 mi/h 88 ft/s= and 3 3 31 g/cm 10 kg/m= are exact. The relation 2 232 ft/s 9 8 m/s= . is accurate to only two significant figures. 1.11. IDENTIFY: We know the density and mass; thus we can find the volume using the relation density mass/volume / .m V= = The radius is then found from the volume equation for a sphere and the result for the volume. SET UP: 3Density 19 5 g/cm= . and critical 60 0 kgm = . . For a sphere 34 3 .V rπ= EXECUTE: 3critical 3 60 0 kg 1000 g/density 3080 cm . 1 0 kg19 5 g/cm V m ⎛ ⎞⎛ ⎞.= = =⎜ ⎟⎜ ⎟⎜ ⎟ .. ⎝ ⎠⎝ ⎠ Units, Physical Quantities, and Vectors 1-5 © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. (b) 1/31/3 4 36 6[1 10 m ] 27 mVd π π ⎛ ⎞×⎛ ⎞= = =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ EVALUATE: Our estimate assumes that each 3cm of air is breathed in only once, where in reality not all the oxygen is absorbed from the air in each breath. Therefore, a somewhat smaller volume would actually be required. 1.21. IDENTIFY: Estimation problem. SET UP: Estimate that the pile is 18 in 18 in 5 ft 8 in.× .× .. Use the density of gold to calculate the mass of gold in the pile and from this calculate the dollar value. EXECUTE: The volume of gold in the pile is 318 in 18 in 68 in 22,000 inV = .× .× . = . . Convert to 3cm : 3 3 3 5 322,000 in (1000 cm /61 02 in ) 3 6 10 cmV = . . . = . × . The density of gold is 319 3 g/cm ,. so the mass of this volume of gold is 3 5 3 6(19 3 g/cm )(3 6 10 cm ) 7 10 gm = . . × = × . The monetary value of one gram is $10, so the gold has a value of 6 7($10/gram)(7 10 grams) $7 10 ,× = × or about 6$100 10× (one hundred million dollars). EVALUATE: This is quite a large pile of gold, so such a large monetary value is reasonable. 1.22. IDENTIFY: Estimate the number of beats per minute and the duration of a lifetime. The volume of blood pumped during this interval is then the volume per beat multiplied by the total beats. SET UP: An average middle-aged (40 year-old) adult at rest has a heart rate of roughly 75 beats per minute. To calculate the number of beats in a lifetime, use the current average lifespan of 80 years. EXECUTE: 9beats 60 min 24 h 365 days 80 yr(75 beats/min) 3 10 beats/lifespan 1 h 1 day yr lifespan N ⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞= = ×⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠ 9 3 7 blood 3 1 L 1 gal 3 10 beats(50 cm /beat) 4 10 gal/lifespan 3 788 L lifespan1000 cm V ⎛ ⎞×⎛ ⎞⎛ ⎞= = ×⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟.⎝ ⎠⎝ ⎠⎝ ⎠ EVALUATE: This is a very large volume. 1.23. IDENTIFY: Estimate the diameter of a drop and from that calculate the volume of a drop, in 3m . Convert 3m to L. SET UP: Estimate the diameter of a drop to be 2 mm.d = The volume of a spherical drop is 3 3 3 34 1 3 6 . 10 cm 1 L.V r dπ π= = = EXECUTE: 3 3 316 (0 2 cm) 4 10 cm .V π −= . = × The number of drops in 1.0 L is 3 5 3 3 1000 cm 2 10 4 10 cm− = × × EVALUATE: Since 3,V d∼ if our estimate of the diameter of a drop is off by a factor of 2 then our estimate of the number of drops is off by a factor of 8. 1.24. IDENTIFY: Draw the vector addition diagram to scale. SET UP: The two vectors A G and B G are specified in the figure that accompanies the problem. EXECUTE: (a) The diagram for = +R A B GG G is given in Figure 1.24a. Measuring the length and angle of R G gives 9 0 mR = . and an angle of 34 .θ = ° (b) The diagram for = −E A B GG G is given in Figure 1.24b. Measuring the length and angle of E G gives 22 mD = and an angle of 250 .θ = ° (c) − − = −( + ),A B A B G G so − −A B G G has a magnitude of 9.0 m (the same as +A B G G ) and an angle with the x+ axis of 214° (opposite to the direction of ).+A B G G (d) − = −( − ),B A A B G GG G so −B A GG has a magnitude of 22 m and an angle with the x+ axis of 70° (opposite to the direction of −A B G G ). EVALUATE: The vector −A G is equal in magnitude and opposite in direction to the vector .A G 1-6 Chapter 1 © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Figure 1.24 1.25. IDENTIFY: Draw each subsequent displacement tail to head with the previous displacement. The resultant displacement is the single vector that points from the starting point to the stopping point. SET UP: Call the three displacements ,A G ,B G and .C G The resultant displacement R G is given by .= + +R A B C G GG G EXECUTE: The vector addition diagram is given in Figure 1.25. Careful measurement gives that R G is 7 8 km, 38 north of east.. D EVALUATE: The magnitude of the resultant displacement, 7.8 km, is less than the sum of the magnitudes of the individual displacements, 2 6 km 4 0 km 3 1 km.. + . + . Figure 1.25 1.26. IDENTIFY: Since she returns to the starting point, the vector sum of the four displacements must be zero. SET UP: Call the three given displacements ,A G ,B G and ,C G and call the fourth displacement .D G 0.+ + + =A B C D G GG G EXECUTE: The vector addition diagram is sketched in Figure 1.26. Careful measurement gives that D G is144 m, 41 south of west° . EVALUATE: D G is equal in magnitude and opposite in direction to the sum .+ +A B C G GG Figure 1.26 Units, Physical Quantities, and Vectors 1-7 © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 1.27. IDENTIFY: For each vector ,V G use that cosxV V θ= and sin ,yV V θ= when θ is the angle V G makes with the x+ axis, measured counterclockwise from the axis. SET UP: For ,A G 270 0 .θ = . ° For ,B G 60 0 .θ = . ° For ,C G 205 0 .θ = . ° For ,D G 143 0 .θ = . ° EXECUTE: 0,xA = 8 00 m.yA = − . 7 50 m,xB = . 13 0 m.yB = . 10 9 m,xC = − . 5 07 m.yC = − . 7 99 m,xD = − . 6 02 m.yD = . EVALUATE: The signs of the components correspond to the quadrant in which the vector lies. 1.28. IDENTIFY: tan ,y x A A θ = for θ measured counterclockwise from the x+ -axis. SET UP: A sketch of ,xA yA and A G tells us the quadrant in which A G lies. EXECUTE: (a) 1 00 mtan 0 500. 2 00 m y x A A θ − .= = = − . . 1tan ( 0 500) 360 26 6 333 .θ −= − . = ° − . ° = ° (b) 1 00 mtan 0 500. 2 00 m y x A A θ .= = = . . 1tan (0 500) 26 6 .θ −= . = . ° (c) 1 00 mtan 0 500. 2 00 m y x A A θ .= = = − . − . 1tan ( 0 500) 180 26 6 153 .θ −= − . = ° − . ° = ° (d) 1 00 mtan 0 500. 2 00 m y x A A θ − .= = = . − . 1tan (0 500) 180 26 6 207θ −= . = ° + . ° = ° EVALUATE: The angles 26 6. ° and 207° have the same tangent. Our sketch tells us which is the correct value of .θ 1.29. IDENTIFY: Given the direction and one component of a vector, find the other component and the magnitude. SET UP: Use the tangent of the given angle and the definition of vector magnitude. EXECUTE: (a) tan32.0 x y A A ° = (9.60 m)tan32.0 6.00 m.xA = ° = 6.00 m.xA = − (b) 2 2 11.3 m.x yA A A= + = EVALUATE: The magnitude is greater than either of the components. 1.30. IDENTIFY: Given the direction and one component of a vector, find the other component and the magnitude. SET UP: Use the tangent of the given angle and the definition of vector magnitude. EXECUTE: (a) tan34.0 x y A A ° = 16.0 m 23.72 m tan34.0 tan34.0 x y A A = = = ° ° 23.7 m.yA = − (b) 2 2 28.6 m.x yA A A= + = EVALUATE: The magnitude is greater than either of the components. 1.31. IDENTIFY: If ,= +C A B G G G then x x xC A B= + and .y y yC A B= + Use xC and yC to find the magnitude and direction of .C G 1-10 Chapter 1 © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 1.35. IDENTIFY: Vector addition problem. ( )− = + − .A B A B G GG G SET UP: Find the x- and y-components of A G and .B G Then the x- and y-components of the vector sum are calculated from the x- and y-components of A G and .B G EXECUTE: cos(60 0 )xA A= . ° (2 80 cm)cos(60 0 ) 1 40 cmxA = . . ° = + . sin (60 0 )yA A= . ° (2 80 cm)sin (60 0 ) 2 425 cmyA = . . ° = + . cos( 60 0 )xB B= − . ° (1 90 cm)cos( 60 0 ) 0 95 cmxB = . − . ° = + . sin ( 60 0 )yB B= − . ° (1 90 cm)sin ( 60 0 ) 1 645 cmyB = . − . ° = − . Note that the signs of the components correspond to the directions of the component vectors. Figure 1.35a (a) Now let = + .R A B GG G 1 40 cm 0 95 cm 2 35 cmx x xR A B= + = + . + . = + . . 2 425 cm 1 645 cm 0 78 cmy y yR A B= + = + . − . = + . . 2 2 2 2(2 35 cm) (0 78 cm)x yR R R= + = . + . 2 48 cmR = . 0 78 cmtan 0 3319 2 35 cm y x R R θ + .= = = + . + . 18 4θ = . ° Figure 1.35b EVALUATE: The vector addition diagram for = +R A B GG G is R G is in the 1st quadrant, with | | | | ,y xR R< in agreement with our calculation. Figure 1.35c Units, Physical Quantities, and Vectors 1-11 © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. (b) EXECUTE: Now let = − .R A B GG G 1 40 cm 0 95 cm 0 45 cmx x xR A B= − = + . − . = + . . 2 425 cm 1 645 cm 4 070 cmy y yR A B= − = + . + . = + . . 2 2 2 2(0 45 cm) (4 070 cm)x yR R R= + = . + . 4 09 cmR = . 4 070 cmtan 9 044 0 45 cm y x R R θ .= = = + . . 83 7θ = . ° Figure 1.35d EVALUATE: The vector addition diagram for ( )= + −R A B GG G is R G is in the 1st quadrant, with | | | |,x yR R< in agreement with our calculation. Figure 1.35e (c) EXECUTE: ( )− = − −B A A B G GG G −B A GG and −A B G G are equal in magnitude and opposite in direction. 4 09 cmR = . and 83 7 180 264θ = . ° + ° = ° Figure 1.35f 1-12 Chapter 1 © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. EVALUATE: The vector addition diagram for ( )= + −R B A GG G is R G is in the 3rd quadrant, with | | | |,x yR R< in agreement with our calculation. Figure 1.35g 1.36. IDENTIFY: The general expression for a vector written in terms of components and unit vectors is ˆ ˆ.x yA A= +A i j G SET UP: ˆ ˆ5 0 5 0(4 6 ) 20 30. = . − = −B i j i j G GG EXECUTE: (a) 5 0,xA = . 6 3yA = − . (b) 11 2,xA = . 9 91yA = − . (c) 15 0,xA = − . 22 4yA = . (d) 20,xA = 30yA = − EVALUATE: The components are signed scalars. 1.37. IDENTIFY: Find the components of each vector and then use the general equation ˆ ˆx yA A= +A i j G for a vector in terms of its components and unit vectors. SET UP: 0,xA = 8 00 m.yA = − . 7 50 m,xB = . 13 0 m.yB = . 10 9 m,xC = − . 5 07 m.yC = − . 7 99 m,xD = − . 6 02 m.yD = . EXECUTE: ˆ( 8 00 m) ;= − .A j G ˆ ˆ(7 50 m) (13 0 m) ;= . + .B i j G ˆ ˆ( 10 9 m) ( 5 07 m) ;= − . + − .C i j G ˆ ˆ( 7 99 m) (6 02 m) .= − . + .D i j G EVALUATE: All these vectors lie in the xy-plane and have no z-component. 1.38. IDENTIFY: Find A and B. Find the vector difference using components. SET UP: Identify the x- and y-components and use = +2 2 x y A A A . EXECUTE: (a) ˆ ˆ4.00 7.00 ;= +A i j G 4.00;xA = + 7.00.yA = + 2 2 2 2(4.00) (7.00) 8.06.x yA A A= + = + = ˆ ˆ5.00 2.00 ;= −B i j G 5.00;xB = + 2.00;yB = − 2 2 2 2(5.00) ( 2.00) 5.39.x yB B B= + = + − = EVALUATE: Note that the magnitudes of A G and B G are each larger than either of their components. EXECUTE: (b) ˆ ˆ ˆ ˆ ˆ ˆ4.00 7.00 (5.00 2.00 ) (4.00 5.00) (7.00 2.00) .− = + − − = − + +A B i j i j i j G G ˆ ˆ1.00 9.00− = − +A B i j G G (c) Let ˆ ˆ1.00 9.00 .− = − +=R A B i j GG G Then 1.00,xR = − 9.00.yR = Units, Physical Quantities, and Vectors 1-15 © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 1.45. IDENTIFY: For all of these pairs of vectors, the angle is found from combining cosAB φ⋅ =A B G G and x x y y z zA B A B A B⋅ = + +A B G G , to give the angleφ as arccos arccos .x x y y A B A B AB AB φ +⎛ ⎞ ⎛ ⎞⋅= =⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎝ ⎠ A B G G SET UP: x x y y z zA B A B A B⋅ = + +A B G G shows how to obtain the components for a vector written in terms of unit vectors. EXECUTE: (a) 22, 40, 13,A B⋅ = − = =A B G G and so 22arccos 165 . 40 13 φ −⎛ ⎞= = °⎜ ⎟ ⎝ ⎠ (b) 60, 34, 136,A B⋅ = = =A B G G 60arccos 28 . 34 136 φ ⎛ ⎞= = °⎜ ⎟ ⎝ ⎠ (c) 0⋅ =A B G G and 90 .φ = ° EVALUATE: If 0,⋅ >A B G G 0 90 .φ≤ < ° If 0,⋅ <A B G G 90 180 .φ° < ≤ ° If 0,⋅ =A B G G 90φ = ° and the two vectors are perpendicular. 1.46. IDENTIFY: The right-hand rule gives the direction and φ× = G G | | sinABA B gives the magnitude. SET UP: 120 0 .φ = . ° EXECUTE: (a) The direction of ×A B G G is into the page (the -directionz− ). The magnitude of the vector product is 2sin (2 80 cm)(1 90 cm)sin120 4 61 cm .AB φ = . . ° = . (b) Rather than repeat the calculations, ×B A GG = – ×A B G G may be used to see that ×B A GG has magnitude 24.61 cm and is in the -directionz+ (out of the page). EVALUATE: For part (a) we could use the components of the cross product and note that the only non- vanishing component is (2 80 cm)cos60 0 ( 1 90 cm)sin60z x y y xC A B A B= − = . . ° − . ° 2 (2 80 cm)sin 60 0 (1 90 cm)cos60 0 4 61 cm .− . . ° . . ° = − . This gives the same result. 1.47. IDENTIFY: ×A D G G has magnitude sin .AD φ Its direction is given by the right-hand rule. SET UP: 180 53 127φ = ° − ° = ° EXECUTE: (a) 2| | (8 00 m)(10 0 m)sin127 63 9 m .× = . . ° = .A D G G The right-hand rule says ×A D G G is in the -directionz− (into the page). ( ) ×b D A GG has the same magnitude as ×A D G G and is in the opposite direction. EVALUATE: The component of D G perpendicular to A G is sin53 0 7 99 m.D D⊥ = . ° = . 2| | 63 9 m ,AD⊥× = = .A D G G which agrees with our previous result. 1.48. IDENTIFY: Apply Eqs. (1.16) and (1.20). SET UP: The angle between the vectors is 20 90 30 140° + ° + ° = °. EXECUTE: (a) cosAB φ⋅ =A B G G gives 2(3 60 m)(2 40 m)cos140 6 62 m⋅ = . . ° = − . .A B G G (b) From φ× = G G | | sinABA B , the magnitude of the cross product is 2(3 60 m)(2 40 m)sin140 5 55 m. . ° = . and the direction, from the right-hand rule, is out of the page (the -directionz+ ). EVALUATE: We could also use x x y y z zA B A B A B⋅ = + +A B G G and the cross product, with the components of A G and .B G 1.49. IDENTIFY: We model the earth, white dwarf, and neutron star as spheres. Density is mass divided by volume. SET UP: We know that density = mass/volume = m/V where 343V rπ= for a sphere. From Appendix B, the earth has mass of 245 97 10 kgm = . × and a radius of 66 37 10 mr = . × whereas for the sun at the end of its lifetime, 301 99 10 kgm = . × and r = 7500 6km 7 5 10 m= . × . The star possesses a radius of r = 10 km = 41 0 10 m. × and a mass of 301 99 10 kgm = . × . 1-16 Chapter 1 © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. EXECUTE: (a) The earth has volume 3 6 3 21 34 43 3 (6 37 10 m) 1 0827 10 mV rπ π= = . × = . × . Its density is 324 3 3 3 3 21 3 2 5 97 10 kg 10 g 1 mdensity (5 51 10 kg/m ) 5 51 g/cm 1 kg1 0827 10 m 10 cm m V ⎛ ⎞. × ⎛ ⎞= = = . × = .⎜ ⎟⎜ ⎟⎜ ⎟. × ⎝ ⎠⎝ ⎠ (b) 3 6 3 21 34 43 3 (7 5 10 m) 1 77 10 mV rπ π= = . × = . × 30 3 9 3 6 3 21 3 3 1 99 10 kg 1 g/cmdensity (1 1 10 kg/m ) 1 1 10 g/cm 1 77 10 m 1000 kg/m m V ⎛ ⎞. ×= = = . × = . ×⎜ ⎟⎜ ⎟. × ⎝ ⎠ (c) 3 4 3 12 34 43 3 (1 0 10 m) 4 19 10 mV rπ π= = . × = . × 30 3 17 3 14 3 12 3 3 1 99 10 kg 1 g/cmdensity (4 7 10 kg/m ) 4 7 10 g/cm 4 19 10 m 1000 kg/m m V ⎛ ⎞. ×= = = . × = . ×⎜ ⎟⎜ ⎟. × ⎝ ⎠ EVALUATE: For a fixed mass, the density scales as 31/r . Thus, the answer to (c) can also be obtained from (b) as 36 6 3 14 3 4 7 50 10 m(1 1 10 g/cm ) 4 7 10 g/cm 1 0 10 m ⎛ ⎞. ×. × = . × .⎜ ⎟⎜ ⎟. ×⎝ ⎠ 1.50. IDENTIFY: Area is length times width. Do unit conversions. SET UP: 1 mi 5280 ft.= 31 ft 7 477 gal.= . EXECUTE: (a) The area of one acre is 21 1 18 80 640mi mi mi ,× = so there are 640 acres to a square mile. (b) 22 21 mi 5280 ft(1 acre) 43,560 ft 640 acre 1 mi ⎛ ⎞ ⎛ ⎞× × =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ (all of the above conversions are exact). (c) (1 acre-foot) 3 53 7 477 gal(43,560 ft ) 3 26 10 gal, 1 ft .⎛ ⎞= × = . ×⎜ ⎟ ⎝ ⎠ which is rounded to three significant figures. EVALUATE: An acre is much larger than a square foot but less than a square mile. A volume of 1 acre- foot is much larger than a gallon. 1.51. IDENTIFY: The density relates mass and volume. Use the given mass and density to find the volume and from this the radius. SET UP: The earth has mass 24E 5 97 10 kgm = . × and radius 6 E 6 37 10 m.r = . × The volume of a sphere is 34 3 .V rπ= 3 31 76 g/cm 1760 km/m .ρ = . = EXECUTE: (a) The planet has mass 25E5 5 3 28 10 kg.m m= . = . × 25 22 3 3 3 28 10 kg 1 86 10 m . 1760 kg/m m V ρ . ×= = = . × 1/31/3 22 3 7 43 3[1 86 10 m ] 1 64 10 m 1 64 10 km 4 4 V r π π ⎛ ⎞. ×⎛ ⎞= = = . × = . ×⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ (b) E2 57r r= . EVALUATE: Volume V is proportional to mass and radius r is proportional to 1/3,V so r is proportional to 1/3.m If the planet and earth had the same density its radius would be 1/3 E E(5 5) 1 8 .r r. = . The radius of the planet is greater than this, so its density must be less than that of the earth. 1.52. IDENTIFY and SET UP: Unit conversion. EXECUTE: (a) 91 420 10 cycles/s,f = . × so 109 1 s 7 04 10 s 1 420 10 −= . × . × for one cycle. (b) 1210 3600 s/h 5 11 10 cycles/h 7 04 10 s/cycle− = . × . × Units, Physical Quantities, and Vectors 1-17 © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. (c) Calculate the number of seconds in 4600 million 9years 4 6 10 y= . × and divide by the time for 1 cycle: 9 7 26 10 (4 6 10 y)(3 156 10 s/y) 2 1 10 cycles 7 04 10 s/cycle− . × . × = . × . × (d) The clock is off by 1 s in 5100,000 y 1 10 y,= × so in 94 60 10 y. × it is off by 9 4 5 4 60 10(1s) 4 6 10 s 1 10 ⎛ ⎞. × = . ×⎜ ⎟⎜ ⎟×⎝ ⎠ (about 13 h). EVALUATE: In each case the units in the calculation combine algebraically to give the correct units for the answer. 1.53. IDENTIFY: Using the density of the oxygen and volume of a breath, we want the mass of oxygen (the target variable in part (a)) breathed in per day and the dimensions of the tank in which it is stored. SET UP: The mass is the density times the volume. Estimate 12 breaths per minute. We know 1 day = 24 h, 1 h = 60 min and 1000 L = 1 m3. The volume of a cube having faces of length l is 3.V l= EXECUTE: (a) ( ) 60 min 24 h12 breaths/min 17,280 breaths/day. 1 h 1 day ⎛ ⎞⎛ ⎞ =⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ The volume of air breathed in one day is 312( L/breath)(17,280 breaths/day) 8640 L 8.64 m .= = The mass of air breathed in one day is the density of air times the volume of air breathed: 3 3(1.29 kg/m )(8.64 m ) 11.1 kg.m = = As 20% of this quantity is oxygen, the mass of oxygen breathed in 1 day is (0.20)(11.1 kg) 2.2 kg 2200 g.= = (b) V = 38.64 m and 3 ,V l= so 1/3 2.1 m.l V= = EVALUATE: A person could not survive one day in a closed tank of this size because the exhaled air is breathed back into the tank and thus reduces the percent of oxygen in the air in the tank. That is, a person cannot extract all of the oxygen from the air in an enclosed space. 1.54. IDENTIFY: Use the extreme values in the piece’s length and width to find the uncertainty in the area. SET UP: The length could be as large as 7.61 cm and the width could be as large as 1.91 cm. EXECUTE: (a) The area is 14.44 ± 0.095 cm2. (b) The fractional uncertainty in the area is 2 2 0.095 cm 0.66%, 14.44 cm = and the fractional uncertainties in the length and width are 0.01 cm 0.13% 7.61 cm = and 0.01 cm 0.53%. 1.9 cm = The sum of these fractional uncertainties is 0.13% 0.53% 0.66%,+ = in agreement with the fractional uncertainty in the area. EVALUATE: The fractional uncertainty in a product of numbers is greater than the fractional uncertainty in any of the individual numbers. 1.55. IDENTIFY: Calculate the average volume and diameter and the uncertainty in these quantities. SET UP: Using the extreme values of the input data gives us the largest and smallest values of the target variables and from these we get the uncertainty. EXECUTE: (a) The volume of a disk of diameter d and thickness t is 2( /2)V d tπ= . The average volume is 2 3(8 50 cm/2) (0 050 cm) 2 837 cmV π= . . = . . But t is given to only two significant figures so the answer should be expressed to two significant figures: 32 8 cmV = . . We can find the uncertainty in the volume as follows. The volume could be as large as 2 3(8 52 cm/2) (0 055 cm) 3 1 cm ,V π= . . = . which is 30 3 cm. larger than the average value. The volume could be as small as 2 3(8 48 cm/2) (0 045 cm) 2 5 cm ,V π= . . = . which is 30 3 cm. smaller than the average value. The uncertainty is 30 3 cm ,± . and we express the volume as 32 8 0 3 cmV = . ± . . 1-20 Chapter 1 © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 1.61. IDENTIFY: Vector addition. Target variable is the 4th displacement. SET UP: Use a coordinate system where east is in the -directionx+ and north is in the -directiony+ . Let ,A G ,B G and C G be the three displacements that are given and let D G be the fourth unmeasured displacement. Then the resultant displacement is = + + + .R A B C D G GG G G And since she ends up back where she started, 0= .R G 0 ,= + + +A B C D G GG G so ( )= − + +D A B C G GG G ( )x x x xD A B C= − + + and ( )y y y yD A B C= − + + EXECUTE: 180 m,xA = − 0yA = cos315 (210 m)cos315 148 5 mxB B= ° = ° = + . sin315 (210 m)sin315 148 5 myB B= ° = ° = − . cos60 (280 m)cos60 140 mxC C= ° = ° = + sin 60 (280 m)sin 60 242 5 myC C= ° = ° = + . Figure 1.61a ( ) ( 180 m 148 5 m 140 m) 108 5 mx x x xD A B C= − + + = − − + . + = − . ( ) (0 148 5 m 242 5 m) 94 0 my y y yD A B C= − + + = − − . + . = − . 2 2 x yD D D= + 2 2( 108 5 m) ( 94 0 m) 144 mD = − . + − . = 94 0 mtan 0 8664 108 5 m y x D D θ − .= = = . − . 180 40 9 220 9θ = ° + . ° = . ° ( D G is in the third quadrant since both xD and yD are negative.) Figure 1.61b The direction of D G can also be specified in terms of 180 40 9 ;φ θ= − ° = . ° D G is 41° south of west. EVALUATE: The vector addition diagram, approximately to scale, is Vector D G in this diagram agrees qualitatively with our calculation using components. Figure 1.61c Units, Physical Quantities, and Vectors 1-21 © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 1.62. IDENTIFY: Find the vector sum of the two displacements. SET UP: Call the two displacements A G and ,B G where 170 kmA = and 230 km.B = .+ =A B R G G G A G and B G are as shown in Figure 1.62. EXECUTE: (170 km)sin 68 (230 km)cos36 343.7 km.x x xR A B= + = ° + ° = (170 km)cos68 (230 km)sin36 71.5 km.y y yR A B= + = ° − ° = − 2 2 2 2(343.7 km) ( 71.5 km) 351 km.x yR R R= + = + − = 71.5 kmtan | | 0 208. 343.7 km y R x R R θ = = = . 11.8 south of east.Rθ = ° EVALUATE: Our calculation using components agrees with R G shown in the vector addition diagram, Figure 1.62. Figure 1.62 1.63. IDENTIFY: We know the resultant of two forces of known equal magnitudes and want to find that magnitude (the target variable). SET UP: Use coordinates having a horizontal x+ axis and an upward y+ axis. Then x x xA B R+ = and 12.8 N.xR = SOLVE: x x xA B R+ = and cos32 sin32 .xA B R° + ° = Since ,A B= 2 cos32 ,xA R° = so 7.55 N.(2)(cos32 ) xRA = = ° EVALUATE: The magnitude of the x component of each pull is 6.40 N, so the magnitude of each pull (7.55 N) is greater than its x component, as it should be. 1.64. IDENTIFY: Solve for one of the vectors in the vector sum. Use components. SET UP: Use coordinates for which x+ is east and y+ is north. The vector displacements are: 2 00 km, 0 of east; 3 50 m, 45 south of east;= . ° = . °A B K K and 5 80 m, 0 east= . °R K EXECUTE: 5 80 km (2 00 km) (3 50 km)(cos45 ) 1 33 km;x x x xC R A B= − − = . − . − . ° = . y y y yC R A B= − − 0 km 0 km ( 3 50 km)(sin 45 ) 2 47 km;= − − − . ° = . 2 2(1 33 km) (2 47 km) 2 81 km;C = . + . = . 1tan [(2 47 km)/(1 33 km)] 61 7 north of eastθ −= . . = . ° . The vector addition diagram in Figure 1.64 shows good qualitative agreement with these values. EVALUATE: The third leg lies in the first quadrant since its x and y components are both positive. Figure 1.64 1-22 Chapter 1 © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 1.65. IDENTIFY: We have two known vectors and a third unknown vector, and we know the resultant of these three vectors. SET UP: Use coordinates for which x+ is east and y+ is north. The vector displacements are: 23.0 km at 34.0° south of east; 46.0 km due north;= =A B K K 32.0 km due west=R K ; C G is unknown. EXECUTE: 32.0 km (23.0 km)cos34.0° 0 51.07 km;x x x xC R A B= − − = − − − = − 0 ( 23.0 km)sin34.0° 46.0 km 33.14 km;y y y yC R A B= − − = − − − = − 2 2 x yC C C= + = 60.9 km Calling θ the angle that C G makes with the –x-axis (the westward direction), we have 33.14tan / 51.07y x C Cθ = = ; θ = 33.0° south of west. EVALUATE: A graphical vector sum will confirm this result. 1.66. IDENTIFY: The four displacements return her to her starting point, so ( ),= − + +D A B C G GG G where ,A G , G B and C G are in the three given displacements and D G is the displacement for her return. SET UP: Let x+ be east and y+ be north. EXECUTE: (a) [(147 km)sin85 (106 km)sin167 (166 km)sin 235 ] 34 3 km.xD = − ° + ° + ° = − . [(147 km)cos85 (106 km)cos167 (166 km)cos235 ] 185 7 km.yD = − ° + ° + ° = + . 2 2( 34 3 km) (185 7 km) 189 km.D = − . + . = (b) The direction relative to north is 34.3 kmarctan 10.5 . 185.7 km φ ⎛ ⎞= = °⎜ ⎟ ⎝ ⎠ Since 0xD < and 0,yD > the direction of D G is 10 5. ° west of north. EVALUATE: The four displacements add to zero. 1.67. IDENTIFY: We want to find the resultant of three known displacement vectors: = + +R A B C G GG G . SET UP: Let x+ be east and y+ be north and find the components of the vectors. EXECUTE: The magnitudes are A = 20.8 m, B = 38.0 m, C = 18.0 m. The components are Ax = 0, Ay = 28.0 m, Bx = 38.0 m, By = 0, Cx = –(18.0 m)(sin33.0°) = –9.804 m, Cy = –(18.0 m)(cos33.0°) = –15.10 m Rx = Ax + Bx + Cx = 0 + 38.0 m + (–9.80 m) = 28.2 m Ry = Ay + By + Cy = 20.8 m + 0 + (–15.10 m) = 5.70 m 2 2 x yR R R= + = 28.8 m is the distance you must run. Calling Rθ the angle the resultant makes with the +x-axis (the easterly direction), we have tan Rθ = Ry/Rx = (5.70 km)/(28.2 km); Rθ = 11.4° north of east. EVALUATE: A graphical sketch will confirm this result. 1.68. IDENTIFY: Let the three given displacements be ,A G B G and ,C G where 40 steps,A = 80 stepsB = and 50 steps.C = .= + +R A B C G GG G The displacement C G that will return him to his hut is .−R G SET UP: Let the east direction be the -directionx+ and the north direction be the -directiony+ . EXECUTE: (a) The three displacements and their resultant are sketched in Figure 1.68. (b) (40)cos45 (80)cos60 11 7xR = ° − ° = − . and (40)sin 45 (80)sin60 50 47 6yR = ° + ° − = . . The magnitude and direction of the resultant are 2 2( 11 7) (47 6) 49,− . + . = 47.6acrtan 76 , 11.7 ⎛ ⎞ = °⎜ ⎟ ⎝ ⎠ north of west. We know that R G is in the second quadrant because 0,xR < 0.yR > To return to the hut, the explorer must take 49 steps in a direction 76° south of east, which is 14° east of south. Units, Physical Quantities, and Vectors 1-25 © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. EXECUTE: (a) 0x x x xA B C D+ + + = gives ( ) [0 (1250 m)sin30 0 (1000 m)cos32 0 ] 223.0 m.x x x xD A B C= − + + = − − . ° + . ° = − 0y y y yA B C D+ + + = gives ( ) [ 825 m (1250 m)cos30 0 (1000 m)sin32 0 ] 787.4 m.y y y yD A B C= − + + = − − + . ° + . ° = − The fourth displacement D G and its components are sketched in Figure 1.73b. 2 2 818.4 m.x yD D D= + = | | 223.0 mtan | | 787.4 m x y D D φ = = and 15.8 .φ = ° You should head 15.8° west of south and must walk 818 m. (b) The vector diagram is sketched in Figure 1.73c. The final displacement D G from this diagram agrees with the vector D G calculated in part (a) using components. EVALUATE: Note that D G is the negative of the sum of ,A G ,B G and C G , as it should be. Figure 1.73 1.74. IDENTIFY: The displacements are vectors in which we want to find the magnitude of the resultant and know the other vectors. SET UP: Calling A G the vector from you to the first post, B G the vector from you to the second post, and C G the vector from the first post to the second post, we have .+A C = B G G G We want to find the magnitude of vector B G . We use components and the magnitude of C G . Let +x be toward the east and +y be toward the north. EXECUTE: Bx = 0 and By is unknown. Cx = –Ax = –(52.0 m)(cos 37.0°) = –41.529 m Ax = 41.53 m 68.0 m,C = so 2 2 –53.8455 m.y xC C C= ± − = We use the minus sign because the second post is south of the first post. By = Ay + Cy = (52.0 m)(sin 37°) + (–53.8455 m) = –22.551 m. Therefore you are 22.6 m from the second post. EVALUATE: By is negative since post is south of you (in the negative y direction), but the distance to you is positive. 1.75. IDENTIFY: We are given the resultant of three vectors, two of which we know, and want to find the magnitude and direction of the third vector. SET UP: Calling C G the unknown vector and A G and B G the known vectors, we have .+ + =A B C R G GG G The components are x x x xA B C R+ + = and .y y y yA B C R+ + = 1-26 Chapter 1 © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. EXECUTE: The components of the known vectors are 12.0 m,xA = 0,yA = sin50.0 21.45 m,xB B= − ° = − cos50.0 18.00 m,yB B= ° = + 0,xR = and 10.0 m.yR = − Therefore the components of C G are 0 12.0 m ( 21.45 m) 9.45 mx x x xC R A B= − − = − − − = and 10.0 m 0 18.0 m 28.0 m.y y y yC R A B= − − = − − − = − Using these components to find the magnitude and direction of G C gives 29.6 mC = and 9.45tan 28.0 θ = and 18.6θ = ° east of south. EVALUATE: A graphical sketch shows that this answer is reasonable. 1.76. IDENTIFY: The displacements are vectors in which we know the magnitude of the resultant and want to find the magnitude of one of the other vectors. SET UP: Calling A G the vector of Ricardo’s displacement from the tree, B G the vector of Jane’s displacement from the tree, and C G the vector from Ricardo to Jane, we have .+ =A C B G G G Let the +x-axis be to the east and the +y-axis be to the north. Solving using components we have x xA C B+ =x and .y y yA C B+ = EXECUTE: (a) The components of A G and B G are (26.0 m)sin60.0 22.52 m,xA = − ° = − (26.0 m)cos60.0 13.0 m,yA = ° = + (16.0 m)cos30.0 13.86 m,xB = − ° = − (16.0 m)sin30.0 8.00 m,yB = − ° = − 13.86 m ( 22.52 m) 8.66 m,x x xC B A= − = − − − = + 8.00 m (13.0 m) 21.0 my y yC B A= − = − − = − Finding the magnitude from the components gives 22.7 m.C = (b) Finding the direction from the components gives 8.66tan 21.0 θ = and 22.4 ,θ = ° east of south. EVALUATE: A graphical sketch confirms that this answer is reasonable. 1.77. IDENTIFY: If the vector from your tent to Joe’s is A G and from your tent to Karl’s is ,B G then the vector from Karl’s tent to Joe’s tent is −A B G G . SET UP: Take your tent’s position as the origin. Let x+ be east and y+ be north. EXECUTE: The position vector for Joe’s tent is ˆ ˆ ˆ ˆ([21 0 m]cos 23 ) ([21 0 m]sin 23 ) (19 33 m) (8 205 m). ° − . ° = . − . .i j i j The position vector for Karl’s tent is ˆ ˆ ˆ ˆ([32 0 m]cos 37 ) ([32 0 m]sin 37 ) (25 56 m) (19 26 m). ° + . ° = . + . .i j i j The difference between the two positions is ˆ ˆ ˆ ˆ(19 33 m 25 56 m) ( 8 205 m 19 25 m) (6 23 m) (27 46 m). − . + − . − . = − . − . .i j i j The magnitude of this vector is the distance between the two tents: 2 2( 6 23 m) ( 27 46 m) 28 2 mD = − . + − . = . EVALUATE: If both tents were due east of yours, the distance between them would be 32 0 m 21 0 m 11 0 m.. − . = . If Joe’s was due north of yours and Karl’s was due south of yours, then the distance between them would be 32 0 m 21 0 m 53 0 m.. + . = . The actual distance between them lies between these limiting values. 1.78. IDENTIFY: Calculate the scalar product and use Eq. (1.16) to determine .φ SET UP: The unit vectors are perpendicular to each other. EXECUTE: The direction vectors each have magnitude 3, and their scalar product is (1)(1) (1)( 1) (1)( 1) 1,+ − + − = − so from Eq. (1.16) the angle between the bonds is 1 1arccos arccos 109 . 33 3 −⎛ ⎞ ⎛ ⎞= − = °⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ EVALUATE: The angle between the two vectors in the bond directions is greater than 90 .° Units, Physical Quantities, and Vectors 1-27 © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 1.79. IDENTIFY: We know the scalar product and the magnitude of the vector product of two vectors and want to know the angle between them. SET UP: The scalar product is cosAB θ⋅A B = G G and the vector product is sinAB .θ×A B = G G EXECUTE: cos 6 00AB .θ⋅ = −A B = G G and sin 9 00AB . .θ× = +A B = G G Taking the ratio gives 9.00tan , 6.00 θ = − so 124 .θ = ° EVALUATE: Since the scalar product is negative, the angle must be between 90° and 180°. 1.80. IDENTIFY: Find the angle between specified pairs of vectors. SET UP: Use cos AB φ ⋅= A B G G EXECUTE: (a) ˆ=A k G (along line ab) ˆ ˆ ˆ= + +B i j k G (along line ad) 1,A = 2 2 21 1 1 3B = + + = ˆ ˆ ˆ ˆ( ) 1⋅ = ⋅ + + =A B k i j k G G So cos 1/ 3; AB φ ⋅= =A B G G 54 7φ = . ° (b) ˆ ˆ ˆ= + +A i j k G (along line ad) ˆ ˆ= +B j k G (along line ac) 2 2 21 1 1 3;A = + + = 2 21 1 2B = + = ˆ ˆ ˆ ˆ ˆ( ) ( ) 1 1 2⋅ = + + ⋅ + = + =A B i j k i j G G So 2 2cos ; 3 2 6AB φ ⋅= = =A B G G 35 3φ = . ° EVALUATE: Each angle is computed to be less than 90 ,° in agreement with what is deduced from the figure shown with this problem in the textbook. 1.81. IDENTIFY: We know the magnitude of two vectors and their scalar product and want to find the magnitude of their vector product. SET UP: The scalar product is cosAB φ⋅ =A B G G and the vector product is φ× = G G | | sinABA B . EXECUTE: cosAB φ⋅ =A B G G = 90.0 m2, which gives 2 2112.0 m 112.0 mcos 0.5833, (12.0 m)(16.0 m)AB φ = = = so 54.31 .φ = ° Therefore 2sin (12 0 m)(16 0 m)(sin54 31 ) 156 mAB . . . .φ× = ° =A B = G G EVALUATE: The magnitude of the vector product is greater than the scalar product because the angle between the vectors is greater than 45º. 1.82. IDENTIFY: The cross product ×A B G G is perpendicular to both A G and .B G SET UP: Use Eq. (1.23) to calculate the components of .×A B G G EXECUTE: The cross product is 6 00 11 00ˆ ˆ ˆ ˆ ˆ ˆ( 13 00) (6 00) ( 11 00) 13 (1 00) . 13 00 13 00 ⎡ . . ⎤⎛ ⎞− . + . + − . = − . + −⎜ ⎟⎢ ⎥. .⎝ ⎠⎣ ⎦ i j k i j k The magnitude of the vector in square brackets is 1 93,. and so a unit vector in this direction is ˆ ˆ ˆ(1 00) (6 00/13 00) (11 00/13 00) . 1 93 ⎡ ⎤− . + . . − . . ⎢ ⎥.⎢ ⎥⎣ ⎦ i j k 1-30 Chapter 1 © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. A2 + 64.00 N2 – 16.00 N A + A2 = 33.99 N2 The quadratic formula gives two solutions: A = 5.00 N and B = 3.00 N or A = 3.00 N and B = 5.00 N. In either case, the larger force has magnitude 5.00 N. (b) Let A = 5.00 N and B = 3.00 N, with the larger vector along the x-axis and the smaller one making an angle of +30.0° with the +x-axis in the first quadrant. The components of the resultant are Rx = Ax + Bx = 5.00 N + (3.00 N)(cos 30.0°) = 7.598 N Ry = Ay + By = 0 + (3.00 N)(sin 30.0°) = 1.500 N = +2 2 x y R R R = 7.74 N EVALUATE: To check our answer, we could use the other resultants and angles given in the table with the problem. 1.89. IDENTIFY: Use the x and y coordinates for each object to find the vector from one object to the other; the distance between two objects is the magnitude of this vector. Use the scalar product to find the angle between two vectors. SET UP: If object A has coordinates ( , )A Ax y and object B has coordinates ( , ),B Bx y the vector ABr G from A to B has x-component B Ax x− and y-component .B Ay y− EXECUTE: (a) The diagram is sketched in Figure 1.89. (b) (i) In AU, 2 2(0 3182) (0 9329) 0 9857. + . = . . (ii) In AU, 2 2 2(1 3087) ( 0 4423) ( 0 0414) 1 3820. + − . + − . = . . (iii) In AU, 2 2 2(0 3182 1 3087) (0 9329 ( 0 4423)) (0 0414) 1 695. − . + . − − . + . = . . (c) The angle between the directions from the Earth to the Sun and to Mars is obtained from the dot product. Combining Eqs. (1.16) and (1.19), ( 0 3182)(1 3087 0 3182) ( 0 9329)( 0 4423 0 9329) (0)arccos 54.6 . (0.9857)(1.695) φ ⎛ ⎞− . . − . + − . − . − . += = °⎜ ⎟ ⎝ ⎠ (d) Mars could not have been visible at midnight, because the Sun-Mars angle is less than 90 .° EVALUATE: Our calculations correctly give that Mars is farther from the Sun than the earth is. Note that on this date Mars was farther from the earth than it is from the Sun. Figure 1.89 Units, Physical Quantities, and Vectors 1-31 © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 1.90. IDENTIFY: Add the vector displacements of the receiver and then find the vector from the quarterback to the receiver. SET UP: Add the x-components and the y-components. EXECUTE: The receiver’s position is ˆ ˆ ˆ ˆ[( 1 0 9 0 6 0 12 0)yd] [( 5 0 11 0 4 0 18 0) yd] (16 0 yd) (28 0 yd) .+ . + . − . + . + − . + . + . + . = . + .i j i j The vector from the quarterback to the receiver is the receiver’s position minus the quarterback’s position, or ˆ ˆ(16 0 yd) (35 0 yd) ,. + .i j a vector with magnitude 2 2(16 0 yd) (35 0 yd) 38 5 yd.. + . = . The angle is 16 0arctan 24 6 35 0 .⎛ ⎞ = . °⎜ ⎟.⎝ ⎠ to the right of downfield. EVALUATE: The vector from the quarterback to receiver has positive x-component and positive y-component. 1.91. IDENTIFY: Draw the vector addition diagram for the position vectors. SET UP: Use coordinates in which the Sun to Merak line lies along the x-axis. Let A G be the position vector of Alkaid relative to the Sun, M G is the position vector of Merak relative to the Sun, and R G is the position vector for Alkaid relative to Merak. 138 lyA = and 77 ly.M = EXECUTE: The relative positions are shown in Figure 1.91. .+ =M R A GG G x x xA M R= + so (138 ly)cos25 6 77 ly 47 5 ly.x x xR A M= − = . ° − = . (138 ly)sin 25 6 0 59 6 ly.y y yR A M= − = . ° − = . 76 2 lyR = . is the distance between Alkaid and Merak. (b) The angle is angle φ in Figure 1.91. 47 5 lycos 76 2 ly xR R θ .= = . and 51 4 .θ = . ° Then 180 129 .φ θ= ° − = ° EVALUATE: The concepts of vector addition and components make these calculations very simple. Figure 1.91 1.92. IDENTIFY: The total volume of the gas-exchanging region of the lungs must be at least as great as the total volume of all the alveoli, which is the product of the volume per alveoli times the number of alveoli. SET UP: V = NValv, and we use the numbers given in the introduction to the problem. EXECUTE: V = NValv = (480 × 106)(4.2 × 106 µm3) = 2.02 × 1015 µm3. Converting to liters gives = × =⎛ ⎞⎜ ⎟ ⎝ ⎠ 3 15 3 6 1 m 2.02 10 m 2.02 L 10 µm V ≈ 2.0 L. Therefore choice (c) is correct. EVALUATE: A volume of 2 L is reasonable for the lungs. 1.93. IDENTIFY: We know the volume and want to find the diameter of a typical alveolus, assuming it to be a sphere. SET UP: The volume of a sphere of radius r is V = 4/3 πr3 and its diameter is D = 2r. EXECUTE: Solving for the radius in terms of the volume gives r = (3V/4π)1/3, so the diameter is D = 2r = 2(3V/4π)1/3 = ( )×⎡ ⎤ ⎢ ⎥ ⎣ ⎦ 6 3 1/ 3 2 3 4.2 10 µm 4π = 200 µm. Converting to mm gives D = (200 µm)[(1 mm)/(1000 µm)] = 0.20 mm, so choice (a) is correct. EVALUATE: A sphere that is 0.20 mm in diameter should be visible to the naked eye for someone with good eyesight. 1-32 Chapter 1 © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 1.94. IDENTIFY: Draw conclusions from a given graph. SET UP: The dots lie more-or-less along a horizontal line, which means that the average alveolar volume does not vary significantly as the lung volume increases. EXECUTE: The volume of individual alveoli does not vary (as stated in the introduction). The graph shows that the volume occupied by alveoli stays constant for higher and higher lung volumes, so there must be more of them, which makes choice (c) the correct one. EVALUATE: It is reasonable that a large lung would need more alveoli than a small lung because a large lung probably belongs to a larger person than a small lung.