Solution manual for university physics 13th 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 : Given the direction and one component of a vector, find

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 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

E XECUTE : The area is

2 2

1.7 I DENTIFY : Convert seconds to years

S ET U P : 1 billion seconds 1 10 s.= × 9 1 day 24 h.= 1 h 3600 s.=

E XECUTE : 1 00 billion seconds (1 00 10 s)9 1 h 1 day 1 y 31 7 y

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 furlong 14 days 24 h

, / ⎛⎜ ⎞⎛⎟⎜ ⎞⎟⎜⎛ ⎞⎟=

E VALUATE : 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 I DENTIFY : Convert miles/gallon to km/L

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

10 g 1 g(410 mg/day) 4.10 10 g/day

(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

E VALUATE : In this case a very small percentage error has disastrous consequences

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 : Estimate the number of people and then use the estimates given in the problem to calculate the

number of gallons

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 1.17 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

(e) 200 months (200 mon) 1 y 17 y

E VALUATE : None are plausible When specifying the value of a measured quantity it is essential to give

the units in which it is being expressed

1.18 I DENTIFY : The number of kernels can be calculated as N V= bottle/Vkernel

S ET U P : 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 1 L 1000 cm and 1 cm 10 mm= 3 =

E XECUTE : The volume of the kernel is: Vkernel=(10 mm)(6 mm)(3 mm) 180 mm = 3 The bottle’s volume is: Vbottle= (2 0 L)[(1000 cm )/(1 0 L)][(10 mm) /(1 0 cm) ] 2 0 10 mm 3 3 3 = × 6 3 The number of kernels is then Nkernels=Vbottle/Vkernels≈ ×(2 0 10 mm )/(180 mm ) 11 000 kernels.6 3 3 = ,

E VALUATE : 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

1.19 I DENTIFY : Estimate the number of pages and the number of words per page

S ET U P : 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)

E XECUTE : An estimate for the number of words is about 10 6

E VALUATE : We can expect that this estimate is accurate to within a factor of 10

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

E XECUTE : The number of blinks is (10 per min) 60 min 24 h 365 days (80 y/lifetime) 4 108

E VALUATE : 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

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 : 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

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

E XECUTE : They eat a total of 10 pizzas 4

E VALUATE : The same answer applies to a school of 250 students averaging 40 pizzas a year each 1.26 I DENTIFY : The displacements must be added as vectors and the magnitude of the sum depends on the

relative orientation of the two displacements

S ET U P : 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

E XECUTE : The orientations of the displacements that give the desired sum are shown in Figure 1.26

E VALUATE : The orientations of the two displacements can be chosen such that the sum has any value

between 0.6 m and 4.2 m

Figure 1.26

1.27 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 B and G, 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 1.27 1.28 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 CG= +A B is given in Figure 1.28a Measuring the length and angle of G G

CGgives 9 0 C= mand an angle of θ= °34

(b) The diagram for D A BG= −G G is given in Figure 1.28b Measuring the length and angle of DG gives

22 m

D= and an angle of θ =250 °

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

x

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

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

to the direction of A BG− G)

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

Figure 1.28 1.29 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.

0

+ + + =

G G G G

A B C D

E XECUTE : The vector addition diagram is sketched in Figure 1.29 Careful measurement gives that DG

is144 m, 41 south of west°

E VALUATE : DG is equal in magnitude and opposite in direction to the sum A B CG+ +G G

Figure 1.29 1.30 I DENTIFY : tan y,

x

A A

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 : (a) tan 0 500.1 00 m

2 00 m

y x

A A

A A

A A

θ= = . = −

2

1tan ( 0 500) 180 26 6 153

θ = − − = ° − ° = °

(d) tan 0 500.1 00 m

2 00 m

y x

A A

−

1tan (0 500) 180 26 6 207

θ= − = ° + ° = °

E VALUATE : The angles 26 6 °and 207°have the same tangent Our sketch tells us which is the correct value of θ

1.31 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 ,A 270 0 G θ= ° For ,BG θ = °60 0 For ,CG θ =205 0 ° For ,DG θ =143 0 °

E XECUTE : A x=0, A y= − 8 00 m B x= 7 50 m, B y= 13 0 m.C x=210 9 m, C y= − 5 07 m

7 99 m,

x

D = − D y= 6 02 m

E VALUATE : The signs of the components correspond to the quadrant in which the vector lies

1.32 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

1.33 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 32.0 x

y

A A

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

1.34 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

RK This result is confirmed by the sketch in Figure 1.34

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

Figure 1.34 1.35 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 y

E XECUTE : (a) CG= +A B so G G C x=A x+B x= 7 50 mandC y=A y+B y= + 5 00 m C= 9 01 m.

5 00 mtan

7 50 m

y x

C C

7 50 m

y x

D D

(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 1.28

1.36 I DENTIFY : Use Equations (1.7) and (1.8) to calculate the magnitude and direction of each of the given

vectors

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

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

asked to find their sum

R x = 1.75 km

−2.90 km = −0.603148.9

R y >0, so RG is in 2nd quadrant This agrees with the vector addition diagram

1.38 I DENTIFY : We know the vector sum and want to find the magnitude of the vectors Use the method of

components

S ET U P : The two vectors AGand BGand their resultant CGare shown in Figure 1.38 Let y+ be in the direction of the resultant A B=

E XECUTE : C y=A y+B y.372 N 2 cos 43 0= A ° and A=254 N

E VALUATE : 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

1.39 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

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

in agreement with our calculation

R R

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

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

in agreement with our calculation

Figure 1.39e (c) E XECUTE :

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

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

in agreement with our calculation

Figure 1.39g

1.40 I DENTIFY : The general expression for a vector written in terms of components and unit vectors is

E VALUATE : The components are signed scalars

1.41 I DENTIFY : Find the components of each vector and then use Eq (1.14)

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

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

S ET U P : Deduce the x- and y-components and use Eq (1.8)

R x <0 and R y>0, so R is in the 2nd quadrant G

1.43 I DENTIFY : Use trig to find the components of each vector Use Eq (1.11) to find the components of the

vector sum Eq (1.14) 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

(c) From Equations (1.7) and (1.8),

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

1.44 I DENTIFY : A unit vector has magnitude equal to 1

S ET U P : The magnitude of a vector is given in terms of its components by Eq (1.12)

E XECUTE : (a) |iˆ ˆ+ + =j kˆ| 12+ + =12 12 3 1≠ so it is not a unit vector

(b) | |AG = A x2+A2y+A z2 If any component is greater than 1+ or less than 1,− | | 1,AG > so it cannot be a unit vector AG can have negative components since the minus sign goes away when the component is squared

(c) | | 1AG = gives a2(3 0) 2+a2(4 0) 2 =1 and a2 25 1.= 1 0 20

5 0

a= ± = ±

E VALUATE : The magnitude of a vector is greater than the magnitude of any of its components

1.45 I DENTIFY : A BG G⋅ =ABcosφ

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 ° =2104 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

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

S ET U P : We are given A and G B in unit vector form and can take the scalar product using Eq (1.19) G

The angle φ can then be found from Eq (1.18)

E VALUATE : The component of B along G A is in the same direction as ,G A so the scalar product is G

positive and the angle φ is less than 90°

1.47 I DENTIFY : 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 A B x x A B y y

S ET U P : Eq (1.14) shows how to obtain the components for a vector written in terms of unit vectors

E XECUTE : (a) A BG G⋅ = −22, =A 40, B= 13, and so arccos 22 165

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

S ET U P : We are given A and G B in unit vector form and can take the vector product using Eq (1.24) G

E XECUTE : AG=4.00iˆ+7.00 ,ˆj BG=5.00iˆ−2.00 ˆj

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ(4.00 7.00 ) (5.00 2.00 ) 20.0 8.00 35.0 14.0

Figure 1.48 1.49 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

− (into the page)

( )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

1.50 I DENTIFY : The right-hand rule gives the direction and Eq (1.22) gives the magnitude

This gives the same result

1.51 I DENTIFY : Apply Eqs (1.18) and (1.22)

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

E XECUTE : (a) Eq (1.18) gives A BG G⋅ = (3 60 m)(2 40 m)cos140 ° = − 6 62 m2

(b) From Eq (1.22), the magnitude of the cross product is(3 60 m)(2 40 m)sin140 ° = 5 55 m2 and the direction, from the right-hand rule, is out of the page (the +z-direction)

E VALUATE : We could also use Eqs (1.21) and (1.27), with the components of AGand BG

1.52 I DENTIFY : Use Eq (1.27) for the components of the vector product

S ET U P : Use coordinates with the +x-axis to the right, +y-axis toward the top of the page, and +z-axisout of the page A x=0, A y=0 and A z= − 3 50 cm The page is 20 cm by 35 cm, so B x= −20 cmand

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

B A A B 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

1.54 I DENTIFY : Area is length times width Do unit conversions

S ET U P : 1 mi 5280 ft.= 1 ft3= 7 477 gal

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

= ×⎜⎝ ⎟⎠= × which is rounded to three significant figures

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

1.55 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 38 10 m.6 The volume of a sphere is 3

1.56 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

(c) Calculate the number of seconds in 4600 million years 4 6 10 y= × 9 and divide by the time for 1 cycle:

26 10

(d) The clock is off by 1 s in 100,000 y 1 10 y,= × 5 so in 4 60 10 y × 9 it is off by

9

4 5

1.57 I DENTIFY : 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

S ET U P : 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 V=l3

E XECUTE : (a) (12 breaths/min) 60 min 24 h 17,280 breaths/day

(b) V = 8.64 m3 and V = l3, so l V= 1/3=2.1 m

E VALUATE : 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

1.58 I DENTIFY : Use the extreme values in the piece’s length and width to find the uncertainty in the area

S ET U P : The length could be as large as 7.61 cm and the width could be as large as 1.91 cm

E XECUTE : The area is 14.44 ± 0.095 cm2 The fractional uncertainty in the area is 0.095 cm

2

14.44 cm2 =0.66%, and the fractional uncertainties in the length and width are 0.01 cm

7.61 cm=0.13% and 0.01 cm

1.9 cm =0.53% The sum of these fractional uncertainties is 0.13% 0.53% 0.66%,+ = in agreement with the fractional uncertainty in the area

E VALUATE : The fractional uncertainty in a product of numbers is greater than the fractional uncertainty in

any of the individual numbers

1.59 I DENTIFY : Calculate the average volume and diameter and the uncertainty in these quantities

S ET U P : 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

E XECUTE : (a) The volume of a disk of diameter d and thickness t is V=π( /2)d 2t

The average volume is V=π(8 50 cm/2) (0 50 cm) 2 837 cm 2 = 3. But t is given to only two significant

figures so the answer should be expressed to two significant figures: V= 2 8 cm3

We can find the uncertainty in the volume as follows The volume could be as large as

(8 52 cm/2) (0 055 cm) 3 1 cm ,

V=π = which is 0 3 cm 3 larger than the average value The volume could be as small as V=π(8 48 cm/2) (0 045 cm) 2 5 cm , 2 = 3 which is 0 3 cm 3 smaller than the average value The uncertainty is ± 0 3 cm ,3 and we express the volume as V= ± 2 8 0 3 cm3

(b) The ratio of the average diameter to the average thickness is 8 50 cm/0 050 cm 170 = By taking the largest possible value of the diameter and the smallest possible thickness we get the largest possible value for this ratio: 8 52 cm/0 045 cm 190 = The smallest possible value of the ratio is 8 48/0 055 150 = Thus the uncertainty is 20± and we write the ratio as 170 20±

E VALUATE : The thickness is uncertain by 10% and the percentage uncertainty in the diameter is much

less, so the percentage uncertainty in the volume and in the ratio should be about 10%