1. University Physics Solution Manual Download
2. University Physics 13e Solutions Manual

University physics 13th edition solution manual. 1. © Copyright 2012 Pearson Education, Inc.

Access University Physics 13th Edition Chapter 6 solutions now. Our solutions are written by Chegg experts so you can be assured of the highest quality! Solutions to university physics westfall manual was found in our database. Modern physics 13th edition solutions manual, mastering physics solutions manual.

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. 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 km 1 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 ft 1 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. IDENTIFY: Convert volume units from L to 3 in. SET UP: 3 1 L 1000 cm.= 1 in 2 54 cm.

EXECUTE: 33 31000 cm 1 in 0 473 L 28 9 in 1 L 2 54 cm ⎛ ⎞.⎛ ⎞. × × =.⎜ ⎟ ⎜ ⎟⎜ ⎟.⎝ ⎠⎝ ⎠ EVALUATE: 3 1 in. Is greater than 3 1 cm, so the volume in 3 in. Is a smaller number than the volume in 3 cm, which is 3 473 cm. 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 8 3 00 10 m/s.v =. × 1 ft 0 3048 m.=. 9 1 s 10 ns.= EXECUTE: 9 8 0 3048 m 1 02 10 s 1 02 ns 3 00 10 m/s t.

× 2 EVALUATE: In 1.00 s light travels 8 5 5 3 00 10 m 3 00 10 km 1 86 10 mi. IDENTIFY: Convert the units from g to kg and from 3 cm to 3 m. SET UP: 1 kg 1000 g.= 1 m 1000 cm.= EXECUTE: 3 4 3 3 g 1 kg 100 cm kg 19 3 1 g 1 mcm m ⎛ ⎞ ⎛ ⎞.

University Physics Solution Manual Download

×⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ EVALUATE: The ratio that converts cm to m is cubed, because we need to convert 3 cm to 3 m. IDENTIFY: Convert volume units from 3 in. SET UP: 3 1 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 3 5360 cm. 3 1 cm is less than 3 1 in.

So the volume in 3 cm is a larger number than the volume in 3 in. IDENTIFY: Convert 2 ft to 2 m and then to hectares. SET UP: 4 2 1 00 hectare 1 00 10 m. × 1 ft 0 3048 m.=. UNITS, PHYSICAL QUANTITIES AND VECTORS 1. 1-2 Chapter 1 © Copyright 2012 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 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 2 1 ft (0 3048) m.=. IDENTIFY: Convert seconds to years.

SET UP: 9 1 billion seconds 1 10 s.= × 1 day 24 h.= 1 h 3600 s.= EXECUTE: 9 1 h 1 day 1 y 1 00 billion seconds (1 00 10 s) 31 7 y. 3600 s 24 h 365 days ⎛ ⎞⎛ ⎞ ⎛ ⎞. × =.⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ EVALUATE: The conversion 7 1 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. IDENTIFY: Apply the given conversion factors. SET UP: 1 furlong 0 1250 mi and 1 fortnight 14 days=.

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. 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 gallon 55 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. 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 2 4 1 mi/gal km/L,∼ which is roughly our result. 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 1h 5280 ft ft 60 88 h 3600 s 1mi s ⎛ ⎞⎛ ⎞⎛ ⎞ =⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠ (b) 2 2 ft 30 48 cm 1 m m 32 9 8 1ft 100 cms s ⎛ ⎞.⎛ ⎞ ⎛ ⎞ =.⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠ (c) 3 3 3 3 g 100 cm 1 kg kg 1 0 10 1 m 1000 gcm m ⎛ ⎞⎛ ⎞⎛ ⎞. =⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠ EVALUATE: The relations 60 mi/h 88 ft/s= and 3 3 3 1 g/cm 10 kg/m= are exact. The relation 2 2 32 ft/s 9 8 m/s=. Is accurate to only two significant figures. 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: 3 Density 19 5 g/cm=. And critical 60 0 kgm =. For a sphere 34 3.V rπ= EXECUTE: 3 critical 3 60 0 kg 1000 g /density 3080 cm. 1 0 kg19 5 g/cm V m ⎛ ⎞⎛ ⎞. = = =⎜ ⎟⎜ ⎟⎜ ⎟. ⎝ ⎠⎝ ⎠ 33 3 3 3 (3080 cm ) 9 0 cm. 4 4 V r π π = = =.

EVALUATE: The density is very large, so the 130-pound sphere is small in size. Units, Physical Quantities and Vectors 1-3 © Copyright 2012 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. IDENTIFY: Convert units. SET UP: We know the equalities 3 1 mg 10 g,− = 1 µg 6 10 g,− and 3 1 kg 10 g.= EXECUTE: (a) 3 5 6 10 g 1 g (410 mg/day) 4.10 10 g/day.

1 mg 10 g µ µ − − ⎛ ⎞⎛ ⎞ = ×⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ (b) 3 10 g (12 mg/kg)(75 kg) (900 mg) 0.900 g. 1 mg −⎛ ⎞ = =⎜ ⎟⎜ ⎟ ⎝ ⎠ (c) The mass of each tablet is 3 310 g (2.0 mg) 2.0 10 g/day. 1 mg − −⎛ ⎞ = ×⎜ ⎟ ⎝ ⎠ The number of tablets required each day is the number of grams recommended per day divided by the number of grams per tablet: 3 0.0030 g/day 1.5 tablet/day. 2.0 10 g/tablet− = × Take 2 tablets each day. (d) 3 1 mg (0.000070 g/day) 0.070 mg/day. 10 g− ⎛ ⎞ =⎜ ⎟⎜ ⎟ ⎝ ⎠ EVALUATE: Quantities in medicine and nutrition are frequently expressed in a wide variety of units.

IDENTIFY: The percent error is the error divided by the quantity. SET UP: The distance from Berlin to Paris is given to the nearest 10 km. EXECUTE: (a) 3 3 10 m 1 1 10 890 10 m − =. ×, (b) Since the distance was given as 890 km, the total distance should be 890,000 meters. We know the total distance to only three significant figures. EVALUATE: In this case a very small percentage error has disastrous consequences.

IDENTIFY: When numbers are multiplied or divided, the number of significant figures in the result can be no greater than in the factor with the fewest significant figures. When we add or subtract numbers it is the location of the decimal that matters. SET UP: 12 mm has two significant figures and 5.98 mm has three significant figures. EXECUTE: (a) 2 (12 mm) (5 98 mm) 72 mm×.

= (two significant figures) (b) 5 98 mm 0 50 12 mm. (also two significant figures) (c) 36 mm (to the nearest millimeter) (d) 6 mm (e) 2.0 (two significant figures) EVALUATE: The length of the rectangle is known only to the nearest mm, so the answers in parts (c) and (d) are known only to the nearest mm.

IDENTIFY: Use your calculator to display 7 10.π × Compare that number to the number of seconds in a year. SET UP: 1 yr 365 24 days,=.

1 day 24 h,= and 1 h 3600 s=. EXECUTE: 724 h 3600 s (365 24 days/1 yr) 3 15567 10 s; 1 day 1 h ⎛ ⎞⎛ ⎞.

University Physics 13e Solutions Manual

×⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ 7 7 10 s 3 14159 10 sπ × =. × The approximate expression is accurate to two significant figures.

The percent error is 0.45%. EVALUATE: The close agreement is a numerical accident. IDENTIFY: Estimate the number of people and then use the estimates given in the problem to calculate the number of gallons. SET UP: Estimate 8 3 10× people, so 8 2 10× cars. EXECUTE: (Number of cars miles/car day)/(mi/gal) gallons/day× = 8 8 (2 10 cars 10000 mi/yr/car 1 yr/365 days)/(20 mi/gal) 3 10 gal/day× × × = × EVALUATE: The number of gallons of gas used each day approximately equals the population of the U.S. IDENTIFY: Express 200 kg in pounds. Express each of 200 m, 200 cm and 200 mm in inches.

Express 200 months in years. 1-4 Chapter 1 © Copyright 2012 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. SET UP: A mass of 1 kg is equivalent to a weight of about 2.2 lbs.1 in 2 54 cm. 1 y 12 months.= EXECUTE: (a) 200 kg is a weight of 440 lb. This is much larger than the typical weight of a man. (b) 4 31 in 200 m (2 00 10 cm) 7 9 10 inches. 2 54 cm.⎛ ⎞ =.

×⎜ ⎟.⎝ ⎠ This is much greater than the height of a person. (c) 200 cm 2 00 m 79 inches 6 6 ft.=. Some people are this tall, but not an ordinary man. (d) 200 mm 0 200 m 7 9 inches.=.

This is much too short. (e) 1 y 200 months (200 mon) 17 y.

12 mon ⎛ ⎞ = =⎜ ⎟ ⎝ ⎠ This is the age of a teenager; a middle-aged man is much older than this. EVALUATE: None are plausible. When specifying the value of a measured quantity it is essential to give the units in which it is being expressed. IDENTIFY: The number of kernels can be calculated as bottle kernel/N V V=. SET UP: Based on an Internet search, Iowa corn farmers use a sieve having a hole size of 0.3125 in.

≅ 8 mm to remove kernel fragments. Therefore estimate the average kernel length as 10 mm, the width as 6 mm and the depth as 3 mm.

We must also apply the conversion factors 3 1 L 1000 cm and 1 cm 10 mm= =. EXECUTE: The volume of the kernel is: 3 kernel (10 mm)(6 mm)(3 mm) 180 mm.V = = The bottle’s volume is: 3 3 3 6 3 bottle (2 0 L)(1000 cm )/(1 0 L)(10 mm) /(1 0 cm) 2 0 10 mm.V =. × The number of kernels is then 6 3 3 kernels bottle kernels/ (2 0 10 mm )/(180 mm ) 11 000 kernels.N V V,= ≈. × = EVALUATE: This estimate is highly dependent upon your estimate of the kernel dimensions. And since these dimensions vary amongst the different available types of corn, acceptable answers could range from 6,500 to 20,000. IDENTIFY: Estimate the number of pages and the number of words per page.

SET UP: Assuming the two-volume edition, there are approximately a thousand pages, and each page has between 500 and a thousand words (counting captions and the smaller print, such as the end-of-chapter exercises and problems). EXECUTE: An estimate for the number of words is about 6 10. EVALUATE: We can expect that this estimate is accurate to within a factor of 10. IDENTIFY: Approximate the number of breaths per minute. Convert minutes to years and 3 cm to 3 m to find the volume in 3 m breathed in a year. SET UP: Assume 10 breaths/min. 524 h 60 min 1 y (365 d) 5 3 10 min.

1 d 1 h ⎛ ⎞⎛ ⎞ = =. ×⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ 2 10 cm 1 m= so 6 3 3 10 cm 1 m.= The volume of a sphere is 3 34 1 3 6,V r dπ π= = where r is the radius and d is the diameter. Don’t forget to account for four astronauts. EXECUTE: (a) The volume is 5 6 3 4 35 3 10 min (4)(10 breaths/min)(500 10 m ) 1 10 m /yr. × × = ×⎜ ⎟⎜ ⎟ ⎝ ⎠ (b) 1/31/3 4 3 6 61 10 m 27 m V d π π ⎛ ⎞×⎛ ⎞ = = =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ EVALUATE: Our estimate assumes that each 3 cm 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. IDENTIFY: Estimate the number of blinks per minute.

Convert minutes to years. Estimate the typical lifetime in years. SET UP: Estimate that we blink 10 times per minute.1 y 365 days.= 1 day 24 h,= 1 h 60 min.= Use 80 years for the lifetime. Units, Physical Quantities and Vectors 1-5 © Copyright 2012 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 number of blinks is 860 min 24 h 365 days (10 per min) (80 y/lifetime) 4 10 1 h 1 day 1 y ⎛ ⎞⎛ ⎞⎛ ⎞ = ×⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠ EVALUATE: Our estimate of the number of blinks per minute can be off by a factor of two but our calculation is surely accurate to a power of 10. 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: 9 beats 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.

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 3 18 in 18 in 68 in 22,000 inV =.×.×. Convert to 3 cm: 3 3 3 5 3 22,000 in (1000 cm /61 02 in ) 3 6 10 cmV =. The density of gold is 3 19 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.

IDENTIFY: Estimate the diameter of a drop and from that calculate the volume of a drop, in 3 m. Convert 3 m 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 31 6 (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. IDENTIFY: Estimate the number of students and the average number of pizzas eaten by each student in a school year. SET UP: Assume a school of a thousand students, each of whom averages ten pizzas a year (perhaps an underestimate) EXECUTE: They eat a total of 4 10 pizzas. EVALUATE: The same answer applies to a school of 250 students averaging 40 pizzas a year each.

IDENTIFY: The displacements must be added as vectors and the magnitude of the sum depends on the relative orientation of the two displacements. SET UP: The sum with the largest magnitude is when the two displacements are parallel and the sum with the smallest magnitude is when the two displacements are antiparallel.

EXECUTE: The orientations of the displacements that give the desired sum are shown in Figure 1.26. EVALUATE: The orientations of the two displacements can be chosen such that the sum has any value between 0.6 m and 4.2 m. 1-6 Chapter 1 © Copyright 2012 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 1.26 1.27. 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,B and.C The resultant displacement R is given by.= + +R A B C EXECUTE: The vector addition diagram is given in Figure 1.27. Careful measurement gives that R is 7 8 km, 38 north of east.

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 1.27 1.28. 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. EXECUTE: (a) The diagram for = +C A B is given in Figure 1.28a. Measuring the length and angle of C gives 9 0 mC =.

And an angle of 34.θ = ° (b) The diagram for = −D A B is given in Figure 1.28b. Measuring the length and angle of D gives 22 mD = and an angle of 250.θ = ° (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° (opposite to the direction of ).+A B (d) − = −( − ),B A A B so −B A has a magnitude of 22 m and an angle with the x+ axis of 70° (opposite to the direction of −A B ). EVALUATE: The vector −A is equal in magnitude and opposite in direction to the vector.A. Units, Physical Quantities and Vectors 1-7 © Copyright 2012 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 1.28 1.29. 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,B and,C and call the fourth displacement.D 0.+ + + =A B C D EXECUTE: The vector addition diagram is sketched in Figure 1.29.

Careful measurement gives that D is144 m, 41 south of west°. EVALUATE: D is equal in magnitude and opposite in direction to the sum.+ +A B C Figure 1.29 1.30. IDENTIFY: tan, y x A A θ = for θ measured counterclockwise from the x+ -axis. SET UP: A sketch of,xA yA and A tells us the quadrant in which A lies. EXECUTE: (a) 1 00 m tan 0 500. 2 00 m y x A A θ −. 1 tan ( 0 500) 360 26 6 333.θ − = −.

° = ° (b) 1 00 m tan 0 500. 2 00 m y x A A θ. 1 tan (0 500) 26 6.θ − =. ° (c) 1 00 m tan 0 500.

2 00 m y x A A θ. = = = −.2 1 tan ( 0 500) 180 26 6 153.θ − = −.

° = ° (d) 1 00 m tan 0 500. 2 00 m y x A A θ −. 1 tan (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 1.31. IDENTIFY: For each vector,V use that cosxV V θ= and sin,yV V θ= when θ is the angle V makes with the x+ axis, measured counterclockwise from the axis. 1-8 Chapter 1 © Copyright 2012 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.

SET UP: For,A 270 0.θ =. ° For,B 60 0.θ =. ° For,C 205 0.θ =. ° For,D 143 0.θ =. ° EXECUTE: 0,xA = 8 00 m.yA = −. 10 9 m,xC =.2 5 07 m.yC = −. 7 99 m,xD = −.

EVALUATE: The signs of the components correspond to the quadrant in which the vector lies. 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. 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 ° = (13.0 m)tan32.0 8.12 m.xA = ° = 8.12 m.xA = − (b) 2 2 15.3 m.x yA A A= + = EVALUATE: The magnitude is greater than either of the components. IDENTIFY: Find the vector 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: 2 6 km, 0 of north; 4 0 km, 0 of east; 3 1 km, 45 north of east.=. °A B C EXECUTE: 0 4 0 km (3 1 km)cos(45 ) 6 2 km;x x x xR A B C= + + = +.

Y y y yR A B C= + + = 2 6 km 0 (3 1 km)(sin45 ) 4 8 km. 2 2 7 8 km;x yR R R= + =. 1 tan (4 8 km)/(6 2 km) 38;θ − =. = ° 7 8 km, 38 north of east=. °.R This result is confirmed by the sketch in Figure 1.34. EVALUATE: Both xR and yR are positive and R is in the first quadrant.

Figure 1.34 1.35. IDENTIFY: If,= +C A B then x x xC A B= + and.y y yC A B= + Use xC and yC to find the magnitude and direction of.C SET UP: From Figure E1.28 in the textbook, 0,xA = 8 00 myA = −. And sin30 0 7 50 m,xB B= +.

Cos30 0 13 0 m.yB B= +. Units, Physical Quantities and Vectors 1-9 © Copyright 2012 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) = +C A B so 7 50 mx x xC A B= + =. And 5 00 m.y y yC A B= + = +.

5 00 m tan 7 50 m y x C C θ. And 33 7.θ =. ° (b),+ = +B A A B so +B A has magnitude 9.01 m and direction specified by 33 7. ° (c) = −D A B so 7 50 mx x xD A B= − = −. And 21 0 m.y y yD A B= − =.2 22 3 m.D =. 21 0 m tan 7 50 m y x D D φ. 2 2 and 70 3.φ =.

° D is in the rd 3 quadrant and the angle θ counterclockwise from the x+ axis is 180 70 3 250 3.° +. ° (d) ( ),− = − −B A A B so −B A has magnitude 22.3 m and direction specified by 70 3.θ =.

° EVALUATE: These results agree with those calculated from a scale drawing in Problem 1.28. IDENTIFY: Use Equations (1.7) and (1.8) to calculate the magnitude and direction of each of the given vectors. SET UP: A sketch of,xA yA and A tells us the quadrant in which A lies. EXECUTE: (a) 2 2 ( 8 60 cm) (5 20 cm) 10 0 cm,−.

5.20 arctan 148.8 8.60 ⎛ ⎞ = °⎜ ⎟ −⎝ ⎠ (which is 180 31 2° −. (b) 2 2 ( 9 7 m) ( 2 45 m) 10 0 m,−. 2.45 arctan 14 180 194. 9.7 −⎛ ⎞ = ° + ° = °⎜ ⎟ −⎝ ⎠ (c) 2 2 (7 75 km) ( 2 70 km) 8 21 km. 2.7 arctan 340.8 7.75 −⎛ ⎞ = °⎜ ⎟ ⎝ ⎠ (which is 360 19 2° −. EVALUATE: In each case the angle is measured counterclockwise from the x+ axis.

Our results for θ agree with our sketches. IDENTIFY: Vector addition problem. We are given the magnitude and direction of three vectors and are asked to find their sum.

SET UP: A = 3.25 km B = 2.90 km C = 1.50 km Figure 1.37a Select a coordinate system where x+ is east and + y is north. Let,A B and C be the three displacements of the professor. Then the resultant displacement R is given by.= + +R A B C By the method of components, Rx = Ax + Bx + Cx and Ry = Ay + By + Cy. Find the x and y components of each vector; add them to find the components of the resultant.

Then the magnitude and direction of the resultant can be found from its x and y components that we have calculated. As always it is essential to draw a sketch. 1-10 Chapter 1 © Copyright 2012 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: 0, 3.25 kmx yA A= = + Bx = −2.90 km, By = 0 0, 1.50 kmx yC C= = − Rx = Ax + Bx + Cx Rx = 0 − 2.90 km + 0 = −2.90 km Ry = Ay + By + Cy 3.25 km 0 1.50 km 1.75 kmyR = + − = Figure 1.37b 2 2 2 2 ( 2.90 km) (1.75 km)x yR R R= + = − + R = 3.39 km tanθ = Ry Rx = 1.75 km −2.90 km = −0.603 148.9θ = ° Figure 1.37c The angle θ measured counterclockwise from the +x-axis. In terms of compass directions, the resultant displacement is 31.1 N° of W.

EVALUATE: Rx 0, so R is in 2nd quadrant. This agrees with the vector addition diagram. IDENTIFY: We know the vector sum and want to find the magnitude of the vectors. Use the method of components. SET UP: The two vectors A and B and their resultant C are shown in Figure 1.38.

Let y+ be in the direction of the resultant.A B= EXECUTE:.y y yC A B= + 372 N 2 cos43 0A=. ° and 254 N.A = EVALUATE: The sum of the magnitudes of the two forces exceeds the magnitude of the resultant force because only a component of each force is upward.

Figure 1.38. Units, Physical Quantities and Vectors 1-11 © Copyright 2012 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. 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 are calculated from the x- and y-components of A and.B 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 1.39a (a) Now let = +.R A B 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= + =. 0 78 cm tan 0 3319 2 35 cm y x R R θ +.

° Figure 1.39b EVALUATE: The vector addition diagram for = +R A B is R is in the 1st quadrant, with ,y xR R so R is in the 2nd quadrant. IDENTIFY: Use trig to find the components of each vector. (1.11) to find the components of the vector sum. (1.14) expresses a vector in terms of its components. SET UP: Use the coordinates in the figure that accompanies the problem.

EXECUTE: (a) ˆ ˆ ˆ ˆ(3 60 m)cos70 0 (3 60 m)sin70 0 (1 23 m) (3 38 m)=. +.A i j i j ˆ ˆ ˆ ˆ(2 40 m)cos30 0 (2 40 m)sin30 0 ( 2 08 m) ( 1 20 m)= −. + −.B i j i j ˆ ˆ ˆ ˆ( ) (3 00) (4 00) (3 00)(1 23 m) (3 00)(3 38 m) (4 00)( 2 08 m) (4 00)( 1 20 m) ˆ ˆ(12 01 m) (14 94) =. B C A B i j i j i j. 1-14 Chapter 1 © Copyright 2012 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) From Equations (1.7) and (1.8), 2 2 14 94 m (12 01 m) (14 94 m) 19 17 m, arctan 51 2 12 01 m C.⎛ ⎞ =. EVALUATE: xC and yC are both positive, so θ is in the first quadrant. IDENTIFY: A unit vector has magnitude equal to 1. SET UP: The magnitude of a vector is given in terms of its components by Eq.

EXECUTE: (a) 2 2 2ˆ ˆ ˆ 1 1 1 3 1+ + = + + = ≠i j k so it is not a unit vector. (b) 2 2 2.x y zA A A= + +A If any component is greater than 1+ or less than 1,− 1,A so it cannot be a unit vector. A can have negative components since the minus sign goes away when the component is squared.

(c) 1=A gives 2 2 2 2 (3 0) (4 0) 1a a. = and 2 25 1.a = 1 0 20. 5 0 a = ± = ±.

EVALUATE: The magnitude of a vector is greater than the magnitude of any of its components. IDENTIFY: cosAB φ⋅ =A B SET UP: For A and,B 150 0.φ =. ° For B and,C 145 0.φ =. ° For A and,C 65 0.φ =.

° EXECUTE: (a) 2 (8 00 m)(15 0 m)cos150 0 104 m⋅ =. ° =A B 2 (b) 2 (15 0 m)(12 0 m)cos145 0 148 m⋅ =. ° = −B C (c) 2 (8 00 m)(12 0 m)cos65 0 40 6 m⋅ =. ° =.A C EVALUATE: When 90φ ° the scalar product is negative. IDENTIFY: Target variables are ⋅A B and the angle φ between the two vectors. SET UP: We are given A and B in unit vector form and can take the scalar product using Eq. The angle φ can then be found from Eq.

EXECUTE: (a) ˆ ˆ4.00 7.00,= +A i j ˆ ˆ5.00 2.00;= −B i j A = 8.06, 5.39.B = ˆ ˆ ˆ ˆ(4.00 7.00 ) (5.00 2.00 ) (4.00)(5.00) (7.00)( 2.00)⋅ = + ⋅ − = + − =A B i j i j 20.0 − 14.0 = +6.00. (b) 6.00 cos 0.1382; (8.06)(5.39)AB φ ⋅ = = = A B φ = 82.1°. EVALUATE: The component of B along A is in the same direction as,A so the scalar product is positive and the angle φ is less than 90°. IDENTIFY: For all of these pairs of vectors, the angle is found from combining Eqs. (1.18) and (1.21), to give the angleφ as arccos arccos.x x y yA B A B AB AB φ +⎛ ⎞ ⎛ ⎞⋅ = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ A B SET UP: Eq. (1.14) shows how to obtain the components for a vector written in terms of unit vectors. EXECUTE: (a) 22, 40, 13,A B⋅ = − = =A B and so 22 arccos 165.

40 13 φ −⎛ ⎞ = = °⎜ ⎟ ⎝ ⎠ (b) 60, 34, 136,A B⋅ = = =A B 60 arccos 28. 34 136 φ ⎛ ⎞ = = °⎜ ⎟ ⎝ ⎠ (c) 0⋅ =A B and 90.φ = ° EVALUATE: If 0,⋅ A B 0 90.φ≤ the direction of D is 10 5.

° west of north. EVALUATE: The four displacements add to zero. IDENTIFY: The sum of the vector forces on the beam sum to zero, so their x components and their y components sum to zero. Solve for the components of.F SET UP: The forces on the beam are sketched in Figure 1.75a. Choose coordinates as shown in the sketch. The 100-N pull makes an angle of 30 0 40 0 70 0. ° with the horizontal.

F and the 100-N pull have been replaced by their x and y components. EXECUTE: (a) The sum of the x-components is equal to zero gives (100 N)cos70 0 0xF +. ° = and 34 2 N.xF = −.

The sum of the y-components is equal to zero gives (100 N)sin70 0 124 N 0yF +. ° − = and 30 0 N.yF = +. F and its components are sketched in Figure 1.75b. 2 2 45 5 N.x yF F F= + =. 30 0 N tan 34 2 N y x F F φ.

And 41 3.φ =. ° F is directed at 41 3. ° above the x− -axis in Figure 1.75a. 1-24 Chapter 1 © Copyright 2012 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) The vector addition diagram is given in Figure 1.75c. F determined from the diagram agrees with F calculated in part (a) using components. EVALUATE: The vertical component of the 100 N pull is less than the 124 N weight so F must have an upward component if all three forces balance. Figure 1.75 1.76.

IDENTIFY: Let the three given displacements be,A B and,C where 40 steps,A = 80 stepsB = and 50 steps.C =.= + +R A B C The displacement C that will return him to his hut is.−R 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 1.76. (b) (40)cos45 (80)cos60 11 7xR = ° − ° = −. And (40)sin45 (80)sin60 50 47 6yR = ° + ° − =. The magnitude and direction of the resultant are 2 2 ( 11 7) (47 6) 49,−. = 47.6 acrtan 76, 11.7 ⎛ ⎞ = °⎜ ⎟⎝ ⎠ north of west. We know that R is in the second quadrant because 0,xR To return to the hut, the explorer must take 49 steps in a direction 76° south of east, which is 14° east of south.

EVALUATE: It is useful to show,xR yR and R on a sketch, so we can specify what angle we are computing. Figure 1.76. Units, Physical Quantities and Vectors 1-25 © Copyright 2012 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. IDENTIFY and SET UP: The vector A that connects points 1 1(, )x y and 2 2(, )x y has components 2 1xA x x= − and 2 1.yA y y= − EXECUTE: (a) Angle of first line is 1 200 20 tan 42 210 10 θ − −⎛ ⎞ = = °.⎜ ⎟⎝ ⎠− Angle of second line is 42 30 72° + ° = °. Therefore 10 250cos72 87,X = + ° = 20 250sin72 258Y = + ° = for a final point of (87,258). (b) The computer screen now looks something like Figure 1.77. The length of the bottom line is 2 2 (210 87) (200 258) 136− + − = and its direction is 1 258 200 tan 25 210 87 − −⎛ ⎞ = °⎜ ⎟⎝ ⎠− below straight left. EVALUATE: Figure 1.77 is a vector addition diagram.

The vector first line plus the vector arrow gives the vector for the second line. Figure 1.77 1.78. IDENTIFY: Vector addition. One vector and the sum are given; find the second vector (magnitude and direction). SET UP: Let x+ be east and y+ be north. Let A be the displacement 285 km at 40 0.

° north of west and let B be the unknown displacement. + =A B R where 115 km,=R east = −B R A,x x xB R A= − y y yB R A= − EXECUTE: cos40 0 218 3 km,xA A= −. ° =.2 sin40 0 183 2 kmyA A= +. 115 km, 0x yR R= = Then 333 3 km,xB =. 183 2 kmyB =.2 2 2 380 km;x yB B B= + = tan / (183 2 km)/(333 3 km)y xB Bα = =.

° south of east Figure 1.78 EVALUATE: The southward component of B cancels the northward component of.A The eastward component of B must be 115 km larger than the magnitude of the westward component of.A 1.79. IDENTIFY: Vector addition. One force and the vector sum are given; find the second force.

SET UP: Use components. Let y+ be upward. 1-26 Chapter 1 © Copyright 2012 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 is the force the biceps exerts. Figure 1.79a E is the force the elbow exerts.,+ =E B R where 132 5 NR =. And is upward.,x x xE R B= − y y yE R B= − EXECUTE: sin43 158 2 N,xB B= − ° = −. Cos43 169 7 N,yB B= + ° = +. 0,xR = 132 5 NyR = +.

Then 158 2 N,xE = +. 37.2 N.yE = − 2 2 160 N;x yE E E= + = tan / 37 2/158 2y xE Eα = =. 13,α = ° below horizontal Figure 1.79b EVALUATE: The x-component of E cancels the x-component of.B The resultant upward force is less than the upward component of,B so yE must be downward. IDENTIFY: Find the vector sum of the four displacements. SET UP: Take the beginning of the journey as the origin, with north being the y-direction, east the x-direction, and the z-axis vertical. The first displacement is then ˆ( 30 m),− k the second is ˆ( 15 m),− j the third is ˆ(200 m),i and the fourth is ˆ(100 m).j EXECUTE: (a) Adding the four displacements gives ˆ ˆ ˆ ˆ ˆ ˆ ˆ( 30 m) ( 15 m) (200 m) (100 m) (200 m) (85 m) (30 m)− + − + + = + −.k j i j i j k (b) The total distance traveled is the sum of the distances of the individual segments: 30 m 15 m 200 m 100 m 345 m+ + + =.

The magnitude of the total displacement is: 2 2 2 2 2 2 (200 m) (85 m) ( 30 m) 219 mx y zD D D D= + + = + + − =. EVALUATE: The magnitude of the displacement is much less than the distance traveled along the path. IDENTIFY: The sum of the force displacements must be zero. Use components.

SET UP: Call the displacements,A,B C and,D where D is the final unknown displacement for the return from the treasure to the oak tree. Vectors,A,B and C are sketched in Figure 1.81a. 0+ + + =A B C D says 0x x x xA B C D+ + + = and 0.y y y yA B C D+ + + = 825 m,A = 1250 m,B = and 1000 m.C = Let x+ be eastward and y+ be north.

EXECUTE: (a) 0x x x xA B C D+ + + = gives ( ) (0 1250 msin30 0 1000 mcos40 0 ) 141 m.x x x xD A B C= − + + = − −. ° = − 0y y y yA B C D+ + + = gives ( ) ( 825 m 1250 mcos30 0 1000 msin40 0 ) 900 m.y y y yD A B C= − + + = − − +. ° = − The fourth. Units, Physical Quantities and Vectors 1-27 © Copyright 2012 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. Displacement D and its components are sketched in Figure 1.81b. 2 2 911 m.x yD D D= + = 141 m tan 900 m x y D D φ = = and 8 9.φ =. ° You should head 8 9. ° west of south and must walk 911 m.

(b) The vector diagram is sketched in Figure 1.81c. The final displacement D from this diagram agrees with the vector D calculated in part (a) using components. EVALUATE: Note that D is the negative of the sum of,A,B and.C Figure 1.81 1.82.

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 the vector from you to the first post, B the vector from you to the second post, and C the vector from the first to the second post, we have.+ +A C B Solving using components and the magnitude of C gives x xA C B+ =x and.y y yA C B+ = EXECUTE: 0,xB = 41.53 mxA = and 41.53 m.x x xC B A= − = − 80.0 m,C = so 2 2 68.38 m.y xC C C= ± − = ± The post is 37.1 m from you. EVALUATE: 37.1 myB = − (negative) since post is south of you (in the negative y direction).

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.

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