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Physics and Measurement
CHAPTER OUTLINE 1.1 Standards of Length, Mass, and Time 1.2 Matter and Model-Building 1.3 Dimensional Analysis 1.4 Conversion of Units 1.5 Estimates and Order-of- Magnitude Calculations 1.6 Significant Figures
ANSWERS TO QUESTIONS
* An asterisk indicates an item new to this edition.
Q1.1 Density varies with temperature and pressure. It would be necessary to measure both mass and volume very accurately in order to use the density of water as a standard.
Q1.2 (a) 0.3 millimeters (b) 50 microseconds (c) 7.2 kilograms
*Q1.3 In the base unit we have (a) 0.032 kg (b) 0.015 kg (c) 0.270 kg (d) 0.041 kg (e) 0.27 kg. Then the ranking is c = e > d > a > b
Q1.4 No: A dimensionally correct equation need not be true. Example: 1 chimpanzee = 2 chimpanzee is dimension- ally correct. Yes: If an equation is not dimensionally correct, it cannot be correct.
*Q1.5 The answer is yes for (a), (c), and (f). You cannot add or subtract a number of apples and a number of jokes. The answer is no for (b), (d), and (e). Consider the gauge of a sausage, 4 kg2 m, or the volume of a cube, (2 m)^3. Thus we have (a) yes (b) no (c) yes (d) no (e) no (f) yes
*Q1.6 41 € ≈ 41 € (1 L1.3 €)(1 qt1 L)(1 gal4 qt) ≈ (101.3) gal ≈ 8 gallons, answer (c)
*Q1.7 The meterstick measurement, (a), and (b) can all be 4.31 cm. The meterstick measurement and (c) can both be 4.24 cm. Only (d) does not overlap. Thus (a) (b) and (c) all agree with the meterstick measurement.
*Q1.8 0.02(1.365) = 0.03. The result is (1.37 ± 0.03) × 10^7 kg. So (d) 3 digits are significant.
1
SOLUTIONS TO PROBLEMS
Section 1.1 Standards of Length, Mass, and Time
P1.1 Modeling the Earth as a sphere, we find its volume as 4 3
π r^3 = π( 6 37. × 10 6 m) = 3 1 08. × 1021 m 3. Its density is then ρ = = × ×
= ×
m V
24 21 3
kg m kg m. This value is intermediate between the tabulated densities of aluminum and iron. Typical rocks have densities around 2 000 to 3 000 kgm^3. The average density of the Earth is significantly higher, so higher-density material must be down below the surface.
P1.2 With V = ( base area )( height ) V = ( π r^2 ) h and^ ρ =^
m V , we have
ρ π π
m r h^2
kg mm mm
mm m
3
..^3
⎝⎜^
ρ = 2 15. × 10 4 kg m^3.
P1.3 Let V represent the volume of the model, the same in ρ = m V for both. Then^ ρiron =^ 9 35.^ kg V
and ρgold = gold
m V
. Next,^ ρ ρ
gold iron
gold kg
m 9 35.
and m gold
3 3 kg (^3) 19.3 10 kg/m kg/m
= ×
×
⎝⎝⎜^.
= 23 0. kg (^).
*P1.4 ρ = m / V and V = ( 4 / 3 ) π r^3 = ( 4 / 3 ) π ( d / 2 ) 3 =π d^3 / 6 where d is the diameter.
Then ρ π π
×
×
= ×
− (^6) −
27 m / d 15 3
kg m 11017 kg/m^3
2.3 × 10 17 kg/m /(11.3^3 × 10 3 kg/m ) =^3 it is 20 × 1012 times the density of lead.
P1.5 For either sphere the volume is V = 4 r 3
π 3 and the mass is m = ρ V =ρ 4 π r 3
(^3). We divide
this equation for the larger sphere by the same equation for the smaller: m m
r r
r s s rs
(^) = ρ π (^) = = ρ π
3 3
3
Then r = rs^3 5 = 4 50. cm ( 1 71. ) = 7 69. cm.
Section 1.2 Matter and Model-Building
P1.6 From the fi gure, we may see that the spacing between diagonal planes is half the distance between diagonally adjacent atoms on a flat plane. This diagonal distance may be obtained from the Pythagorean theorem, (^) L (^) diag = L^2^ + L^2. Thus, since the atoms are separated by a distance L = 0 200. nm, the diagonal planes are separated by
L^2^ + L^2 = 0 141. nm.
Section 1.3 Dimensional Analysis
P1.7 (a) This is incorrect since the units of [ ax ] are m 2 s^2 , while the units of [ v ] are m s.
(b) This is correct since the units of [ ] y are m, and cos^ ( kx^ )^ is dimensionless if [ k ] is in m−^1.
P1.8 (a) Circumference has dimensions of L.
(b) Volume has dimensions of L^3.
(c) Area has dimensions of L^2. Expression (i) has dimension L L^2 L 1 2 (^2) ( ) =
/ , so this must be area (c). Expression (ii) has dimension L, so it is (a). Expression (iii) has dimension L L ( 2 ) = L^3 , so it is (b). Thus, (^) (a) = ii; (b) = iii;(c) = i.
2 Chapter 1
4 Chapter 1
P1.14 (a) Seven minutes is 420 seconds, so the rate is
r = = × −
gal s gal s.
(b) Converting gallons first to liters, then to m 3 ,
r = (^) ( × )
⎝⎜^
7 14. 10 − 2 gal s 3 786.^ L^10 −^3 1 gal
m 3 1 1 L m 3 s
⎝⎜^
r = 2 70. × 10 −^4. (c) At that rate, to fill a 1-m 3 tank would take
t = ×
⎝⎜^
⎝⎜^
m m s
h 3 600
3
. 3.^ h^.
P1.15 From Table 14.1, the density of lead is 1 13. × 10 4 kg m 3 , so we should expect our calculated value to be close to this number. This density value tells us that lead is about 11 times denser than water, which agrees with our experience that lead sinks.
Density is defined as mass per volume, in ρ = m V
. We must convert to SI units in the calculation.
ρ =
g cm
kg g
cm ⎝⎝⎜ m
⎠⎟^
3 3
g cm
kg g
000 cm 1 14 10 4 m
. kg m
3 3
= ×
At one step in the calculation, we note that one million cubic centimeters make one cubic meter. Our result is indeed close to the expected value. Since the last reported significant digit is not certain, the difference in the two values is probably due to measurement uncertainty and should not be a concern. One important common-sense check on density values is that objects which sink in water must have a density greater than 1 g cm 3 , and objects that float must be less dense than water.
P1.16 The weight flow rate is 1 200 ton 2 000 1 1 h
lb ton
h 60 min
min 60
⎠ (^) s lb s
⎠ =^667.
P1.17 (a) 8 10 1 1
⎛ ×^12
⎝⎜^
⎝⎜^
1 000 $ s
h 3 600 s
day 24 h
yr 365 days
⎛⎝ ⎞⎠⎛ years ⎝⎜^
(b) The circumference of the Earth at the equator is 2 π (^) (6 378. × 10 3 m) = 4 01. × 107 m. The length of one dollar bill is 0.155 m so that the length of 8 trillion bills is 1 24. × 1012 m. Thus, the 8 trillion dollars would encircle the Earth 1 24 10 1
12
. ×. 4. ×
m= × 4.01 0 m 7 times
P1.18 (^) V = Bh = 1 ⎡⎣( )(^ )⎤⎦ 3
. acres ft 2 acre fft
ft 3
= 9 08. × 10 7 ,
or
V = (^) ( × )
⎛ ×
⎝⎜^
− 9 08 10
. 7.^2
ft m ft
3 3 3
77 × 10 6 m 3
FIG. P1.
B
h
B
h
Physics and Measurement 5
P1.19 Fg = ( 2 50. tons block ) (^) ( 2 00. × 10 6 blocks (^) )( 2 000lb tton ) = 1 00. × 1010 lbs
P1.20 (a) d d d nucleus, scale nucleus, real = atom, scale dd atom, real
m ft
= (^) ( 2 40. × 10 − (^15) )^300 ××
− ⎠ =^ ×
− 10 m 10 6 79.^103 ft, or
d nucleus, scale = (^) ( 6 79. × 10 −^3 ft (^) )(304 8. mm 1 ft)) = 2 07. mm
(b) V V
r r
atom r nucleus
atom nucleus
= 4 3 = atom 4 3
3 3
π π
/ rr
d nucleus d
atom nucleus
⎠⎟^
⎠⎟^
×
3 3 1 06. 110 2 40 10 8 62 10
10 15
3
13
− × −
⎝⎜^
= ×
m m times
. as large
P1.21 V = At so t V A
= = ×^ = ×
.^34
m. m
m or
3 2 (^ μm)
P1.22 (a) A A
r r
r r
Earth Moon
Earth Moon 2 Earth Moon
π^2 π ⎜⎜
⎠⎟^
( × )(^ ) ×
8
m cm m cm (^) ⎠⎠⎟^
2 13 4.
(b) V V
r r
r r
Earth Moon
Earth Moon
Earth Mo
3 3
π π
/ (^) oon
(^3) m cm m cm
⎠⎟^
( × )(^ ) ×
6 8
3 49 1.
P1.23 To balance, m (^) Fe = m Alor ρ (^) Fe V Fe (^) =ρAl V Al
ρ π ρ π
ρ ρ
Fe Fe Al Al
Al Fe Fe A
⎝⎜^
⎠⎟^
r r
r r ll
cm c
⎠⎟^
1 3 (^) 1 3 2 00
/ (^) / .
. mm.
P1.24 The mass of each sphere is
m (^) Al = ρAl V (^) Al=^4 π ρAl^ r Al 3
3
and
m V r Fe =^ ρFe Fe= Fe^ Fe
4 π ρ 3
3 .
Setting these masses equal, 4 3
π ρ 3 π ρ^3 Al r Al^ (^) = Fe r Fe and r Al (^) r Fe Fe Al
ρ ρ
The resulting expression shows that the radius of the aluminum sphere is directly proportional to the radius of the balancing iron sphere. The sphere of lower density has larger radius. The fraction ρ ρ
Fe Al
is the factor of change between the densities, a number greater than 1. Its cube root is a number much closer to 1. The relatively small change in radius implies a change in volume suffi cient to compensate for the change in density.
Physics and Measurement 7
P1.31 (a) 3 (b) 4 (c) 3 (d) 2
P1.32 (^) r m
= ( ± ) = ( ± ) ×
.. cm.. 2 m . .. 02 kg
43 3
( )
ρ π
m r
also,^ δ ρ ρ
δ δ = + m m
r r
In other words, the percentages of uncertainty are cumulative. Therefore, δ ρ ρ
ρ π
( ) (^) ( × )
− =^ ×
4 1 61^10
3 2 3
m
kg m
and
ρ ± δ ρ= ( 1 61. ±0 17. ) × 10 3 kg m 3 = ( 1 6. ±0 2.) × 103 kg m^33.
P1.33 (a) 756.?? 37.2?
- / 5 / 3 = 797
(b) 0 003 2. ( 2 s.f. ) × 356 3. (4 s.f. ) = 1 140 16. = ((2 s.f. ) 1 1.
(c) 5.620 4 s.f.( ) × π ( >4 s.f. ) = 17.656= 4 s.f.( ) 17 ..
P1.34 We work to nine significant digits:
yr yr d 1 yr
h 1 d
⎝⎜^
min 1 h
s 1 min
⎛⎝ ⎞⎠⎛⎝ ⎞⎠ =. s.
*P1.35 The tax amount is $1.36 − $1.25 = $0.11. The tax rate is $0.11$1.25 = 0.088 0 = 8.80%
*P1.36 (a) We read from the graph a vertical separation of 0.3 spaces = 0.015 g.
(b) Horizontally, 0.6 spaces = 30 cm^2.
(c) Because the graph line goes through the origin, the same percentage describes the vertical and the horizontal scatter: 30 cm^2 380 cm^2 = 8%.
(d) Choose a grid point on the line far from the origin: slope = 0.31 g600 cm^2 = 0.000 52 gcm^2 = (0.000 52 gcm 2 )(10 000 cm^2 1 m 2 ) = 5.2 g/m 2.
(e) For any and all shapes cut from this copy paper, the mass of the cutout is proportional to its area. The proportionality constant is 5.2 g/m^2 ± 8%, where the uncertainty is estimated.
(f ) This result should be expected if the paper has thickness and density that are uniform within the experimental uncertainty. The slope is the areal density of the paper, its mass per unit area.
ISMV1_5103_01.inddISMV1_5103_01.indd 77 10/28/0610/28/06 2:41:45 AM2:41:45 AM
8 Chapter 1
*P1.37 15 players = 15 players (1 shift 1.667 player) = 9 shifts
*P1.38 Let o represent the number of ordinary cars and s the number of trucks. We have o = s + 0.947 s = 1.947 s , and o = s + 18. We eliminate o by substitution: s + 18 = 1.947 s 0.947 s = 18 and s = 18 0.947 = 19.
*P1.39 Let s represent the number of sparrows and m the number of more interesting birds. We have s m = 2.25 and s + m = 91. We eliminate m by substitution: m = s 2. s + s 2.25 = 91 1.444 s = 91 s = 91 1.444 = 63.
*P1.40 For those who are not familiar with solving equations numerically, we provide a detailed solution. It goes beyond proving that the suggested answer works.
The equation 2^ x^4^^ −^3 x^^3 +^5 x −^70 =^0 is quartic, so we do not attempt to solve it with algebra. To fi nd how many real solutions the equation has and to estimate them, we graph the expression:
x − 3 − 2 − 1 0 1 2 3 4
y = 2 x^4 − 3 x^3 + 5 x − 70 158 − 24 − 70 − 70 − 66 − 52 26 270
We see that the equation y = 0 has two roots, one around x = −2 2. and the other near x = +2 7.. To home in on the first of these solutions we compute in sequence: When x = −2 2. , y = −2 20.. The root must be between x = −2 2. and x = − 3. When x = −2 3. , y = 11 0.. The root is between x = −2 2. and x = −2 3.. When x = −2 23. , y = 1 58.. The root is between x = −2 20. and x = −2 23.. When x = −2 22. , y = 0 301.. The root is between x = −2 20. and −2.22. When x = −2 215. , y = −0 331.. The root is between x = −2 215. and −2.22. We could next try x = −2 218. , but we already know to three-digit precision that the root is x = −2 22..
*P1.41 We require sin θ = − 3 cosθ, or sin cos
θ θ
= − 3 , or^ tan^ θ = −^3.
For tan −^1 ( − 3 ) = a rc tan( − 3 ), your calculator may return −71.6°,
but this angle is not between 0° and 360° as the problem requires. The tangent function is negative in the second quad- rant (between 90° and 180°) and in the fourth quadrant (from 270° to 360°). The solutions to the equation are then 360 ° − 71 6. ° = 288 ° and 180 ° − 71 6. ° = 108 °.
y
x
FIG. P1.
tan θ
θ
FIG. P1.
10 Chapter 1
Additional Problems
P1.46 It is desired to find the distance x such that x 100 x
m
= m
(i.e., such that x is the same multiple of 100 m as the multiple that 1 000 m is of x ). Thus, it is seen that
x^2 = ( 100 m )(1 000 m) = 1 00. × 105 m 2
and therefore x = 1 00. × 10 5 m 2 = 316 m.
*P1.47 (a) The mass is equal to the mass of a sphere of radius 2.6 cm and density 4.7 gcm^3 , minus the mass of a sphere of radius a and density 4.7 gcm 3 plus the mass of a sphere of radius a and density 1.23 gcm 3.
m = ρ 14 π r^3 3 − ρ 1 4 π a^3 3 + ρ 24 π a^3 3
= (4.7 gcm 3 )4π(2.6 cm)^3 3 − (4.7 gcm 3 )4 π( a ) 3 3 + (1.23 gcm^3 )4π( a ) 3 3
m = 346 g − (14.5 g/cm 3 ) a^3
(b) For a = 0 the mass is a maximum, (c) 346 g. (d) Yes. This is the mass of the uniform sphere we considered in the first term of the calculation.
(e) For a = 2.60 cm the mass is a minimum, (f ) 346 − 14.5(2.6)^3 = 90.6 g. (g) Yes. This is the mass of a uniform sphere of density 1.23 gcm 3.
(h) (346 g + 90.6 g) 2 = 218 g (i) No. The result of part (a) gives 346 g − (14.5 gcm^3 ) (1.3 cm)^3 = 314 g, not the same as 218 g.
(j) We should expect agreement in parts b-c-d, because those parts are about a uniform sphere of density 4.7 g/cm^3. We should expect agreement in parts e-f-g, because those parts are about a uniform liquid drop of density 1.23 g/cm^3. The function m ( a ) is not a linear function, so a halfway between 0 and 2.6 cm does not give a value for m halfway between the minimum and maximum values. The graph of m versus a starts at a = 0 with a horizontal tangent. Then it curves down more and more steeply as a increases. The liquid drop of radius 1.30 cm has only one eighth the volume of the whole sphere, so its presence brings down the mass by only a small amount, from 346 g to 314 g.
(k) No change, so long as the wall of the shell is unbroken.
*P1.48 (a) We have B + C (0) = 2.70 gcm^3 and B + C (14 cm) = 19.3 gcm^3. We know B = 2.70 g/cm^3 and we solve for C by subtracting: C (14 cm) = 16.6 gcm^3 so C = 1.19 g/cm^4.
(b) m = (^) ∫ (2.70 g/cm 3 + g/cm 4 x cm^2 dx
cm 0 1 19^9
14
. )( )
= 24.3 g/cm 0 g/cm 2
14 cm c ∫^ dx^ +10 7^0 xdx
14 .
mm
= (24.3 g/cm)(14 cm – 0) + (10.7 g/c
∫ mm cm) = 340 g + 1046 g = 1.
2 )[( 14 2 − 0 ] / 2
3 39 kg
Physics and Measurement 11
P1.49 The scale factor used in the “dinner plate” model is
S = ×
0 25 = × −
m lightyears
.5 m lightyeears.
The distance to Andromeda in the scale model will be
D (^) scale = D (^) actual S = (2.0 × 10 6 lightyears ) ( 2.5 × 10 −^66 m lightyears ) = 5 0. m.
*P1.50 The rate of volume increase is
dV dt
d dt
r r dr dt
r dr dt
π 3 π 3 2 4 π 2.
(a) dV dt = 4 π(6.5 cm)^2 (0.9 cms) = 478 cm^3 /s
(b) dr dt
dV dt r
π π
cm s cm)
cm
3 2
(^33) /s
(c) When the balloon radius is twice as large, its surface area is four times larger. The new volume added in one second in the inflation process is equal to this larger area times an extra radial thickness that is one-fourth as large as it was when the balloon was smaller.
P1.51 One month is
1 mo = ( 30 day )( 24 h day )(3 600 s h ) = 2 592. × 10 6 s.
Applying units to the equation,
V = ( 1 50. Mft 3 mo) + ( t 0 008 00. Mft 3 mo^2 ) t^2.
Since 1 Mft 3 = 10 6 ft^3 ,
V = (^) ( 1 50. × 10 6 ft 3 mo (^) ) + t (^) ( 0 008 00. × 106 ft 3 mo^2 ) t^2.
Converting months to seconds,
V = t
×
×
1 50. 10 6 ft mo 0 008 00. × 106 2.592 10 s mo
3 6
fft mo 2.592 10 s mo
3 2 ( ×^6 )^2
t^2.
Thus, V [ft 3 ] = ( 0 579. ft 3 s ) + t ( 1 19. × 10 − 9 ft 3 s^2 ) t^2.
*P1.
α ′(deg) α(rad) tan( α ) sin( α ) difference between α and tan α
We see that in radians, tan() and sin() start out together from zero and diverge only slightly in value for small angles. Thus 31 0. º is the largest angle for which tan tan
α α α
Physics and Measurement 13
P1.58 The density of each material is ρ π π
m V
m r h
m (^2) D h 2
Al: g cm cm
g cm 3
ρ π
= (^ )
. TThe tabulated value g cm
⎛⎝2 70. 3 ⎞⎠ is 2 % smalller.
Cu: g .23 cm .06 cm
g cm 3 ρ π
( )
. TThe tabulated value g cm
⎛⎝8 92. 3 ⎞⎠ is 5 % smalller.
Brass: 4.4 g .54 cm .69 cm
ρ g π
= (^ )
2 8 91.^
ccm 3
Sn: g .75 cm .74 cm
g cm 3 ρ π
( )
Fe: 16.1 g .89 cm .77 cm
g cm
ρ π
= (^ )
2 7 88.^33 The tabulated value^ g 3 cm
⎛⎝7 86. ⎞⎠ is 0 3. % ssmaller.
P1.59 (^) V 20 mpg cars^ mi yr mi gal
= = 5.0 × 10
(^8 ) gal yr
V 25 mpg cars^ mi yr mi gal
= (^10 )(^10 )= 4.0 × 10
8 4 (^110) gal yr
Fuel saved = V 25 mpg (^) − V 20 mpg = 1 0. × 1010 gal yr
P1.60 The volume of the galaxy is
π r t^2 = π( 10 21 m) ( 2 10 19 m (^) )~ 1061 m 3.
If the distance between stars is 4 × 1016 m, then there is one star in a volume on the order of
( 4 ×^10 16 m)^3 ~^1050 m^3.
The number of stars is about 10 10
61 50
m 11 m star
stars
3 3 ~^.
ANSWERS TO EVEN-NUMBERED PROBLEMS
P1.2 2.15 × 104 kgm^3
P1.4 2.3 × 1017 kgm^3 is twenty trillion times larger than the density of lead.
P1.6 0.141 nm
P1.8 (a) ii (b) iii (c) i
P1.10 9.19 nm s
P1.12 (a) 3.39 × 105 ft^3 (b) 2.54 × 104 lb
P1.14 (a) 0.071 4 gal s (b) 2.70 × 10 −^4 m^3 s (c) 1.03 h
14 Chapter 1
P1.16 667 lbs
P1.18 2 57.^ ×^10 6 m^3
P1.20 (a) 2.07 mm (b) 8.57 × 1013 times as large
P1.22 (a) 13.4; (b) 49.
P1.24 r Al (^) r Fe Fe Al
ρ ρ
1 3
P1.26 ~10^7 rev
P1.28 No. There is a strong possibility that you would die before finishing the task, and you have much more productive things to do.
P1.30 ( 209 ± 4 ) cm^2
P1.32 (1.61 ± 0.17) × 103 kgm 3
P1.34 31 556 926.0 s
P1.36 (a) 0.015 g (b) 30 cm^2 (c) 8% (d) 5.2 gm^2 (e) For any and all shapes cut from this copy paper, the mass of the cutout is proportional to its area. The proportionality constant is 5.2 gm^2 ± 8%, where the uncertainty is estimated. (f) This result is to be expected if the paper has thickness and density that are uniform within the experimental uncertainty. The slope is the areal density of the paper, its mass per unit area.
P1.38 19
P1.40 see the solution
P1.42 1.38 km
P1.44 either 3.46 or −3.
P1.46 316 m
P1.48 (a) ρ = 2.70 gcm^3 + 1.19 gcm 4 x (b) 1.39 kg
P1.50 (a) 478 cm^3 s (b) 0.225 cm s (c) When the balloon radius is twice as large, its surface area is four times larger. The new volume added in one increment of time in the inflation process is equal to this larger area times an extra radial thickness that is one-fourth as large as it was when the bal- loon was smaller.
P1.52 0.542 rad
P1.54 3 64. cents; no
P1.56 8 32.^ ×^10 −^4 m s; a snail
P1.58 see the solution
P1.60 ~10^11 stars
