Solution manual for university physics with modern physics 14th edition by young

I DENTIFY : We know the density and mass; thus we can find the volume using the relation density mass/volume= =m V/.. I DENTIFY : Since she returns to the starting point, the vector sum

1.1 I DENTIFY : Convert units from mi to km and from km to ft

1.3 I DENTIFY : We know the speed of light in m/s t d v= / Convert 1.00 ft to m and t from s to ns

S ET U P : The speed of light is v= ×3 00 10 m/s.8 1 ft 0 3048 m.= 1 s 10 ns.= 9

E XECUTE : 0 3048 m8 1 02 10 s 1 02 ns9

3 00 10 m/s

E VALUATE : In 1.00 s light travels 3 00 10 m 3 00 10 km 1 86 10 mi × 8 = × 5 = × 5

1.4 I DENTIFY : Convert the units from g to kg and from cm to 3 m 3

E VALUATE : The ratio that converts cm to m is cubed, because we need to convert cm to 3 m 3

1.5 I DENTIFY : Convert volume units from in.3 to L

1.6 I DENTIFY : Convert ft to 2 m and then to hectares 2

S ET U P : 1 00 hectare 1 00 10 m = × 4 2 1 ft 0 3048 m.=

E XECUTE : The area is

2 2

1.8 I DENTIFY : Apply the given conversion factors

S ET U P : 1 furlong 0 1250 mi and 1 fortnight 14 days= = 1 day 24 h=

E XECUTE : (180 000 furlongs fortnight) 0 125 mi 1 fortnight 1 day 67 mi/h

1 mi/gal∼ km/L, which is roughly our result

1.10 I DENTIFY : Convert units

S ET U P : Use the unit conversions given in the problem Also, 100 cm 1 m= and 1000 g 1 kg.=

32 ft/s = 9 8 m/s is accurate to only two significant figures

1.11 I DENTIFY : 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

S ET U P : Density 19 5 g/cm= 3 and mcritical= 60 0 kg For a sphere 4 3

E VALUATE : The density is very large, so the 130-pound sphere is small in size

1.12 I DENTIFY : Convert units

S ET U P : We know the equalities 1 mg =10 g,− 3 1 µg 10 g,−6 and 1 kg =10 g.3

E XECUTE : (a)

3

5 6

the number of grams recommended per day divided by the number of grams per tablet:

E VALUATE : Quantities in medicine and nutrition are frequently expressed in a wide variety of units

1.13 I DENTIFY : Model the bacteria as spheres Use the diameter to find the radius, then find the volume and

surface area using the radius

S ET U P : From Appendix B, the volume V of a sphere in terms of its radius is 4 3

3

V= πr while its surface

area A is A=4πr2 The radius is one-half the diameter or r d= /2 1 0 m= μ Finally, the necessary equalities for this problem are: 1 m 10 m;μ = −6 1 cm 10 m;= −2 and 1 mm 10 m= −3

E VALUATE : On a human scale, the results are extremely small This is reasonable because bacteria are not

visible without a microscope

1.14 I DENTIFY : 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

S ET U P : 12 mm has two significant figures and 5.98 mm has three significant figures

E XECUTE : (a) (12 mm) (5 98 mm) 72 mm× = 2(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)

E VALUATE : 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

1.15 I DENTIFY : Use your calculator to displayπ×10 7 Compare that number to the number of seconds in a year

S ET U P : 1 yr 365 24 days,= 1 day 24 h,= and 1 h 3600 s=

E XECUTE : (365 24 days/1 yr) 24 h 3600 s 3 15567 10 s;7

The approximate expression is accurate to two significant figures The percent error is 0.45%

E VALUATE : The close agreement is a numerical accident

1.16 I DENTIFY : To asses the accuracy of the approximations, we must convert them to decimals

S ET U P : Use a calculator to calculate the decimal equivalent of each fraction and then round the numeral

to the specified number of significant figures Compare to π rounded to the same number of significant figures

E XECUTE : (a) 22/7 = 3.14286 (b) 355/113 = 3.14159 (c) The exact value of π rounded to six significant figures is 3.14159

E VALUATE : We see that 355/113 is a much better approximation to π than is 22/7

1.17 I DENTIFY : Express 200 kg in pounds Express each of 200 m, 200 cm and 200 mm in inches Express

200 months in years

S ET U P : A mass of 1 kg is equivalent to a weight of about 2.2 lbs.1 in = 2 54 cm 1 y 12 months.=

E XECUTE : (a) 200 kg is a weight of 440 lb This is much larger than the typical weight of a man

(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

S ET U P : Estimate 3 10× 8people, so 2 10× 8cars

E XECUTE : (Number of cars miles/car day)/(mi/gal) gallons/day× =

(2 10 cars 10000 mi/yr/car 1 yr/365 days)/(20 mi/gal) 3 10 gal/day× × × = ×

E VALUATE : The number of gallons of gas used each day approximately equals the population of the U.S

1.19 I DENTIFY : Estimate the number of blinks per minute Convert minutes to years Estimate the typical

1.20 I DENTIFY : Approximate the number of breaths per minute Convert minutes to years and cm to 3 m to 3

find the volume in m breathed in a year 3

S ET U P : Assume 10 breaths/min.1 y (365 d) 24 h 60 min 5 3 10 min.5

V= πr = πd where r is the radius and d is the diameter

Don’t forget to account for four astronauts

E XECUTE : (a) The volume is (4)(10 breaths/min)(500 10 m )6 3 5 3 10 min5 1 10 m /yr.4 3

1 y

1.21 I DENTIFY : Estimation problem

S ET U P : 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

E XECUTE : The volume of gold in the pile is V=18 in 18 in 68 in.× × =22,000 in .3 Convert to cm : 3

The monetary value of one gram is $10, so the gold has a value of ($10/gram)(7 10 grams) $7 10 ,× 6 = × 7

or about $100 10× 6 (one hundred million dollars)

E VALUATE : This is quite a large pile of gold, so such a large monetary value is reasonable

1.22 I DENTIFY : 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

S ET U P : 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

E XECUTE : beats (75 beats/min) 60 min 24 h 365 days 80 yr 3 10 beats/lifespan9

E VALUATE : This is a very large volume

1.23 I DENTIFY : Estimate the diameter of a drop and from that calculate the volume of a drop, in m Convert 3

1.24 I DENTIFY : Draw the vector addition diagram to scale

S ET U P : The two vectors AGand BGare specified in the figure that accompanies the problem

E XECUTE : (a) The diagram for R A BG= +G Gis given in Figure 1.24a Measuring the length and angle of

RGgives 9 0 R= mand an angle of θ= °34

(b) The diagram for EG= −A BG G is given in Figure 1.24b Measuring the length and angle of EG gives

22 m

D= and an angle of θ=250 °

(c) − − = −( + ),A BG G A B so − −A BG G has a magnitude of 9.0 m (the same asA BG+ G) and an angle with the

x

+ axis of 214°(opposite to the direction of A BG+ G)

(d) B AG− = −( − ),G A BG G so B AG− Ghas a magnitude of 22 m and an angle with the +x axis of 70° (opposite

to the direction ofA BG− G)

E VALUATE : The vector −AGis equal in magnitude and opposite in direction to the vector AG.

Figure 1.24 1.25 I DENTIFY : 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

S ET U P : Call the three displacements ,AG BG, and CG. The resultant displacement RG is given by

E VALUATE : 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 I DENTIFY : Since she returns to the starting point, the vector sum of the four displacements must be zero

S ET U P : Call the three given displacements ,AG BG, and ,CG and call the fourth displacement DG.

1.27 I DENTIFY : For each vector ,VG use that V x=Vcosθ and V y=Vsin ,θ when θ is the angle VG makes

with the +x axis, measured counterclockwise from the axis

S ET U P : For ,AG 270 0 θ= ° For ,BG θ = °60 0 For ,CG θ =205 0 ° For ,DG θ =143 0 °

θ = for θ measured counterclockwise from the +x-axis

S ET U P : A sketch of A x, A and A y G tells us the quadrant in which AG lies

E XECUTE :

2 00 m

y x

A A

A A

1tan (0 500) 26 6

θ= − = °

2 00 m

y x

A A

A A

S ET U P : Use the tangent of the given angle and the definition of vector magnitude

E XECUTE : (a) tan 32.0 x

y

A A

E VALUATE : The magnitude is greater than either of the components

1.30 I DENTIFY : Given the direction and one component of a vector, find the other component and the

magnitude

S ET U P : Use the tangent of the given angle and the definition of vector magnitude

E XECUTE : (a) tan 34.0 x

y

A A

° =

16.0 m

23.72 mtan 34.0 tan 34.0

x y

E VALUATE : The magnitude is greater than either of the components

1.31 I DENTIFY : If ,CG= +A BG G then C x=A x+B xand C y=A y+B y Use C and x C to find the magnitude and y

direction of CG

S ET U P : From Figure E1.24 in the textbook,A x=0, A y= − 8 00 mand sin 30 0B x= +B ° = 7 50 m,

7 50 m

y x

C C

7 50 m

y x

D D

φ= =−

− and 70 3 φ= ° DGis in the 3 quadrant and the angle rd θcounterclockwise from the

x

+ axis is 180° + ° =70 3 250 3 °

(d) B AG− = − −G (A BG G),so B AG− Ghas magnitude 22.3 m and direction specified by θ = °70 3

E VALUATE : These results agree with those calculated from a scale drawing in Problem 1.24

1.32 I DENTIFY : Find the vector sum of the three given displacements

S ET U P : 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

E VALUATE : Both R and x R are positive and y RG is in the first quadrant

Figure 1.32 1.33 I DENTIFY : Vector addition problem We are given the magnitude and direction of three vectors and are

asked to find their sum

Select a coordinate system where x+ is east and y+ is north Let ,AG B and CG, G be the three

displacements of the professor Then the resultant displacement RG is given by R A B CG= + +G G G By the method of components, R x=A x+B x+C xand R y=A y+B y+C y. 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

R R

S ET U P : A sketch of A x, A and y AGtells us the quadrant in which AGlies

E XECUTE : (a) ( 8 60 cm)− 2+ (5 20 cm)2= 10 0 cm, arctan 5.20 148.8

E VALUATE : In each case the angle is measured counterclockwise from the +x axis Our results for θ

agree with our sketches

1.35 I DENTIFY : Vector addition problem A B AG− = + − G G ( BG)

S ET U P : Find the x- and y-components of AG and BG. Then the x- and y-components of the vector sum are calculated from the x- and y-components of AG and BG

Note that the signs of the components correspond

to the directions of the component vectors

R R

E VALUATE : The vector addition diagram for R A BG= +G G is

RGis in the 1st quadrant, with |R y| | | ,< R x

in agreement with our calculation

Figure 1.35c

(b) E XECUTE : Now let R A BG= − G G

R R

E VALUATE : The vector addition diagram for R AG= + −G ( BG) is

RGis in the 1st quadrant, with | | |R x < R y|,

in agreement with our calculation

Figure 1.35e (c) E XECUTE :

E VALUATE : The vector addition diagram for R BG= + −G ( AG) is

RG is in the 3rd quadrant, with | | |R x < R y|,

in agreement with our calculation

Figure 1.35g 1.36 I DENTIFY : The general expression for a vector written in terms of components and unit vectors is

E VALUATE : The components are signed scalars

1.37 I DENTIFY : Find the components of each vector and then use the general equation AG=A x iˆ+A yˆj for a

vector in terms of its components and unit vectors

E VALUATE : All these vectors lie in the xy-plane and have no z-component

1.38 I DENTIFY : Find A and B Find the vector difference using components

S ET U P : Identify the x- and y-components and use = 2+ 2

R R

E VALUATE : R x<0 and R y>0, so RG is in the 2nd quadrant

1.39 I DENTIFY : Use trigonometry to find the components of each vector Use R x = A x+B x+" and

= + +"

R A B to find the components of the vector sum The equation AG=A x iˆ+A yˆj expresses a vector in terms of its components

S ET U P : Use the coordinates in the figure that accompanies the problem

E XECUTE : (a) AG= (3 60 m)cos70 0 ° + iˆ (3 60 m)sin 70 0 ° = ˆj (1 23 m)iˆ+ (3 38 m)ˆj

E VALUATE : C and x C are both positive, so y θis in the first quadrant

1.40 I DENTIFY : We use the vector components and trigonometry to find the angles

S ET U P : Use the fact thattanθ = A A y/ x

E XECUTE : (a) θ =

−

3.00tan A A y / x θ = 117° with the +x-axis

E VALUATE :Sketching each of the three vectors to scale will show that the answers are reasonable

1.41 I DENTIFY : AG and BG are given in unit vector form Find A, B and the vector difference A BG− G

S ET U P : AG= − 2 00iG+ 3 00jG+ 4 00 ,kG BG= 3 00iG+ 1 00Gj− 3 00kG

Use A= A2x+A2y+A z2 to find the magnitudes of the vectors

E XECUTE : (a) A= A x2+A2y+A z2 = − ( 2 00)2+ (3 00)2+ (4 00)2 = 5 38

B AG G A BG G so A BG−G and B AG− G have the same magnitude but opposite directions

E VALUATE : A, B, and C are each larger than any of their components

1.42 I DENTIFY : Target variables are A BG G⋅ and the angle φ between the two vectors

S ET U P : We are given AG and BG in unit vector form and can take the scalar product using

S ET U P : For AG and ,BG φ=150 0 ° For BG and ,CG φ=145 0 ° For AG and ,CG φ= °65 0

E XECUTE : (a) A BG G⋅ = (8 00 m)(15 0 m)cos150 0 ° = −104 m2

(b) B CG⋅ =G (15 0 m)(12 0 m)cos145 0 ° = −148 m2

(c) A CG G⋅ = (8 00 m)(12 0 m)cos65 0 ° = 40 6 m2

E VALUATE : When 90φ< ° the scalar product is positive and when φ> °90 the scalar product is negative

1.44 I DENTIFY : Target variable is the vector A BG× G expressed in terms of unit vectors

S ET U P : We are given AG and BG in unit vector form and can take the vector product using

and ˆ ˆi× =j k ˆ ˆˆ, j i× = −kˆ, so A BG× = −G 8.00kˆ+35.0(− = −kˆ) 43.0 kˆ The magnitude of A BG×G is 43.0

E VALUATE : Sketch the vectors AG and BG in a coordinate system where the xy-plane is in the plane of the paper and the z-axis is directed out toward you By the right-hand rule A BG×G is directed into the plane of the paper, in the −z-direction This agrees with the above calculation that used unit vectors

Figure 1.44

1.45 I DENTIFY : For all of these pairs of vectors, the angle is found from combining A BG G⋅ =ABcosφ and

4.61 cm and is in the +z-direction (out of the page)

E VALUATE : For part (a) we could use the components of the cross product and note that the only vanishing component is C z=A B x y−A B y x= (2 80 cm)cos60 0 ( 1 90 cm)sin 60 ° − °

non-2 (2 80 cm)sin 60 0 (1 90 cm)cos60 0− ° ° = − 4 61 cm This gives the same result

1.47 I DENTIFY : A DG× G has magnitude ADsin φ Its direction is given by the right-hand rule

S ET U P : φ=180° − ° =53 127°

E XECUTE : (a)|A DG×G| (8 00 m)(10 0 m)sin127= ° = 63 9 m 2 The right-hand rule says A DG× G is in the -direction

z

( )b D AG×G has the same magnitude as A DG× G and is in the opposite direction

E VALUATE: The component of DG perpendicular to AG is D⊥=Dsin 53 0 ° = 7 99 m

2

|A DG×G|=AD⊥= 63 9 m , which agrees with our previous result

1.48 I DENTIFY : Apply Eqs (1.16) and (1.20)

S ET U P : The angle between the vectors is 20° + ° + ° =90 30 140°

E XECUTE : (a) cosA BG G⋅ =AB φ gives A BG G⋅ = (3 60 m)(2 40 m)cos140 ° = − 6 62 m2

(b) From |A BG× G| =ABsinφ, the magnitude of the cross product is(3 60 m)(2 40 m)sin140 ° = 5 55 m2and the direction, from the right-hand rule, is out of the page (the +z-direction)

E VALUATE : We could also use A BG G⋅ =A B x x+A B y y+A B z z and the cross product, with the components

4

1 0 10 m × and a mass of m= ×1 99 10 kg30

E XECUTE : (a) The earth has volume 4 3 4 6 3 21 3

E XECUTE : (a) The area of one acre is 1 1 1 2

8 mi×80 mi=640 mi ,so there are 640 acres to a square mile

(b)

2 2

(all of the above conversions are exact)

(c) (1 acre-foot) (43,560 ft )3 7 477 gal3 3 26 10 gal,5

1 ft

E VALUATE : 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 I DENTIFY : The density relates mass and volume Use the given mass and density to find the volume and

from this the radius

S ET U P : The earth has mass mE= ×5 97 10 kg24 and radius rE= ×6 37 10 m.6 The volume of a sphere is 3

1.52 I DENTIFY and S ET U P : Unit conversion

E XECUTE : (a) f = 1 420 10 cycles/s,× 9 so 1 9 s 7 04 10 10 s