Chapter 1 Solutions

*1.1

With V = (base area) · (height) V = π r2 · h and ρ = m , we have V
9 3 m 1 kg  10 mm  = π r2 h π (19.5 mm)2 39.0 mm  1 m3 

ρ=

ρ = 2.15 × 104 kg/m3
1.2

ρ=

M M = V 4 πR3 3 3(5.64 × 1026 kg) = 623 kg/m3 4 π (6.00 × 107 m) 3 4 3 3 π (ro – ri ) 3
5.7cm 0.05 cm

ρ=

1.3

VCu = V0 − Vi =

VCu =

4 π [(5.75 cm)3 – (5.70 cm)3] = 20.6 cm3 3

ρ=m V
m = ρV = (8.92 g/cm3)(20.6 cm3) = 184 g 1.4 V = Vo – Vi = 4 3 3 π (r2 – r1 ) 3
3 3

ρ=
*1.5 (a)

4 πρ (r2 – r1 ) m 4 3 3 , so m = ρV = ρ  π  (r2 – r1) = V 3 3  The number of moles is n = m/M, and the density is ρ = m/V. Noting that we have 1 mole, V1 mol = mFe nFe MFe (1 mol)(55.8 g/mol) = = = 7.10 cm3 ρFe ρFe 7.86 g/cm3

2

Chapter 1 Solutions

(b)

In 1 mole of iron are NA atoms: V1 atom = V1 mol 7.10 cm3 = = 1.18 × 10–23 cm3 NA 6.02 × 1023 atoms/mol

= 1.18 × 10-29 m3
3

(c) (d)

datom =

1.18 × 10–29 m3 = 2.28 × 10–10 m = 0.228 nm (1 mol)(238 g/mol) = 12.7 cm3 18.7 g/cm3 V1 mol U 12.7 cm3 = = 2.11 × 10–23 cm3NA 6.02 × 1023 atoms/mol

V1 mol U =

V1 atom U =

= 2.11 × 10-29 m3
3 3

datom U =
3

V1 atom U =

2.11 × 10–29 m3 = 2.77 × 10–10 m = 0.277 nm

*1.6 1.7

r2 = r1

5 = (4.50 cm)(1.71) = 7.69 cm

Use m = molar mass/NA and 1 u = 1.66 × 10-24 g (a) For He, m = 4.00 g/mol = 6.64 × 10-24 g = 4.00 u 6.02 × 1023 mol-1

(b)

For Fe, m =

55.9 g/mol = 9.29 × 10-23 g = 55.9 u 6.02 × 1023 mol-1 207 g/mol -22 g = 207 u 23 -1 = 3.44 × 10 6.02 × 10 mol

(c)

For Pb, m =




Chapter 1 Solutions

3

Goal Solution Calculate the mass of an atom of (a) helium, (b) iron, and (c) lead. Give your answers in atomic mass units and in grams. The molar masses are 4.00, 55.9, and 207 g/mol, respectively, for the atoms given. Gather information: The mass of an atom of any element is essentially the mass of the protons and neutrons that make up its nucleus since the mass of the electrons is negligible (less than a 0.05% contribution). Since most atoms have about the same number of neutrons as protons, the atomic mass isapproximately double the atomic number (the number of protons). We should also expect that the mass of a single atom is a very small fraction of a gram (~10–23 g) since one mole (6.02 × 1023) of atoms has a mass on the order of several grams. Organize: An atomic mass unit is defined as 1/12 of the mass of a carbon-12 atom (which has a molar mass of 12.0 g/mol), so the mass of any atom in atomic mass units is simply the numerical value of the molar mass. The mass in grams can be found by multiplying the molar mass by the mass of one atomic mass unit (u): 1 u = 1.66 × 10–24 g. Analyze: For He, m = 4.00 u = (4.00 u)(1.66 × 10–24 g/u) = 6.64 × 10–24 g For Fe, m = 55.9 u = (55.9 u)(1.66 × 10–24g/u) = 9.28 × 10–23 g For Pb, m = 207 u = (207 u)(1.66 × 10–24 g/u) = 3.44 × 10–22 g Learn: As expected, the mass of the atoms is larger for bigger atomic numbers. If we did not know the conversion factor for atomic mass units, we could use the mass of a proton as a close approximation: 1u ≈ mp = 1.67 × 10–24 g. aˆ†m 3.80 g – 3.35 g = = 0.00228 mol M 197 g/mol

*1.8

aˆ†n =

aˆ†N = (aˆ†n)NA = (0.00228 mol)(6.02 × 1023atoms/mol) = 1.38 × 1021 atoms aˆ†t = (50.0 yr)(365 d/yr)(24.0 hr/d)(3600 s/hr) = 1.58 × 109 s aˆ†N 1.38 × 1021 atoms = = 8.72 × 1011 atoms/s aˆ†t 1.58 × 109 s 1.9 (a) m = ρL3 = (7.86 g/cm3)(5.00 × 10-6 cm)3 = 9.83 × 10-16 g NA   = (9.83 × 10-16 g)(6.02 × 1023 atoms/mol) Molar mass 55.9 g/mol

(b)

N=m

= 1.06 × 107 atoms




4

Chapter 1 Solutions

1.10

(a)

The cross-sectional area is A = 2(0.150 m)(0.010 m) + (0.340 m)(0.010 m) = 6.40 × 10-3 m2

15.0 cm

1.00 cm

36.0 cm

The volume of the beam is V = AL = (6.40 × 10-3 m2)(1.50 m) = 9.60 × 10-3 m3 Thus, its mass is m = ρV = (7.56 × 103 kg/m3)(9.60 × 10-3 m3) = 72.6 kg (b) Presuming that most of the atoms are of iron, we estimate the molar mass as M = 55.9 g/mol = 55.9 × 10-3 kg/mol. The number of moles is then n= m 72.6 kg = = 1.30 × 103 mol M 55.9 × 10-3 kg/mol
1.00 cm

The number of atoms is N = nNA = (1.30 × 103 mol)(6.02 × 1023 atoms/mol) = 7.82 × 1026 atoms *1.11 (a) n= m 1.20 × 103 g = = 66.7 mol, and M 18.0 g/mol Npail = nNA = (66.7 mol)(6.02 × 10 molecules/mol)= 4.01 × 1025 molecules (b) Suppose that enough time has elapsed for thorough mixing of the hydrosphere. Nboth = Npail 
23

 = (4.01 × 1025 molecules £« 1.20 kg  , or Mtotal 1.32 × 1021 kg

mpail

Nboth = 3.65 × 104 molecules 1.12 r, a, b, c and s all have units of L.

£® 

(s – a)(s – b)(s – c)  s =

L×L×L = L

L2 = L

Thus, the equation is dimensionally consistent.




Chapter 1 Solutions
1.13 The term s has dimensions of L, a has dimensions of LT -2, and t has dimensions of T. Therefore, the equation, s = kamtn has dimensions of L = (LT (T)
2 m n

5

or

L T =L T

1 0

m n-2m

The powers of L and T must be the same on each side of the equation. Therefore, L1 = Lm and m = 1 Likewise, equating terms in T, we see that n – 2m must equal 0. Thus, n = 2m = 2 The value of k, a dimensionless constant, cannot be obtained by dimensional analysis .

1.14

2π ï£
(a) (b)

g 

l =

L L/T 2

=

T2 = T

1.15

This is incorrect since the units of [ax] are m2/s2, while the units of [v] arem/s. This is correct since the units of [y] are m, and cos(kx) is dimensionless if [k] is in m-1.

1.16

Inserting the proper units for everything except G,

 kg m = G[kg]2  s2  [m]2
Multiply both sides by [m]2 and divide by [kg]2; the units of G are m3 kg · s2 1.17 One month is 1 mo = (30 day)(24 hr/day)(3600 s/hr) = 2.592 × 106 s Applying units to the equation, V = (1.50 Mft3/mo)t + (0.00800 Mft3/mo2)t2 Since 1 Mft3 = 106 ft3, V = (1.50 × 106 ft3/mo)t + (0.00800 × 106 ft3/mo2)t2




6

Chapter 1 Solutions

Converting months to seconds, V= 1.50 × 106 ft3/mo 0.00800 × 106 ft3/mo2 2 t+ t 2.592 × 106 s/mo (2.592 × 106 s/mo)2

Thus, V[ft3] = (0.579 ft3/s)t + (1.19 × 10 -9 ft3/s 2)t2 *1.18 Apply the following conversion factors: 1 in = 2.54 cm, 1 d = 86400 s, 100 cm = 1 m, and 109 nm = 1 m
-2 9  1 in/day (2.54 cm/in)(10 m/cm)(10 nm/m) = 9.19 nm/s 86400 s/day  32 

This means the proteins are assembled at a rate of many layers of atoms each second! 1.19 Area A = (100 ft)(150 ft) = 1.50 × 104 ft2, so A = (1.50 × 104ft2)(9.29 × 10-2 m2/ft2) = 1.39 × 103 m2

Goal Solution A rectangular building lot is 100 ft by 150 ft. Determine the area of this lot in m2. G: We must calculate the area and convert units. Since a meter is about 3 feet, we should expect the area to be about A ≈ (30 m 50 m) = 1 500 m2. O: Area = Length × Width. Use the conversion: 1 m = 3.281 ft. A: A = L × W = (100 ft)  1m  1m  (150 ft )  = 1 390 m2 3.281 ft 3.281 ft  

L: Our calculated result agrees reasonably well with our initial estimate and has the proper units of m2. Unit conversion is a common technique that is applied to many problems.

1.20

(a)

V = (40.0 m)(20.0 m)(12.0 m) = 9.60 × 103 m3 V = 9.60 × 103 m3 (3.28 ft/1 m)3 = 3.39 × 10 5 ft3




Chapter 1 Solutions
(b) The mass of the air is m = ρairV = (1.20 kg/m3)(9.60 × 103 m3) = 1.15 × 104 kg The student must look up weight in the index to find Fg = mg = (1.15 × 104 kg)(9.80 m/s2) = 1.13 × 105 N Converting to pounds, Fg = (1.13 × 105 N)(1 lb/4.45 N) = 2.54 × 104 lb *1.21 (a) Seven minutes is 420 seconds,so the rate is r= (b) 30.0 gal = 7.14 × 10-2 gal/s 420 s

7

Converting gallons first to liters, then to m3, gal  3.786 L  10-3 m3 r =  7.14 × 10 -2 s   1 gal   1 L   r = 2.70 × 10-4 m3/s

(c)

At that rate, to fill a 1-m3 tank would take t= 1 m3    1 hr  = 1.03 hr -4 m3/s 3600 s 2.70 × 10 

1.22

v =  5.00



furlongs   220 yd   0.9144 m  1 fortnight  1 day   1 hr  fortnight 1 furlong  1 yd   14 days  24 hrs 3600 s

= 8.32 × 10-4 m/s This speed is almost 1 mm/s; so we might guess the creature was a snail, or perhaps a sloth. 1.23 It is often useful to remember that the 1600-m race at track and field events is approximately 1 mile in length. To be precise, there are 1609 meters in a mile. Thus, 1 acre is equal in area to (1 acre £« 1 mi2   1609 m 2 = 4.05 × 103 m2 640 acres  mi 




8

Chapter 1 Solutions

1.24

Volume of cube = L3 = 1 quart (Where L = length of one side of the cube.) Thus, 1 gallon  3.786 liters  1000 cm3 L3 = (1 quart)  = 946 cm3, and 4quarts   1 gallon   1 liter   L = 9.82 cm

1.25

The mass and volume, in SI units, are m = (23.94 g) 1 kg  = 0.02394 kg 1000 g

V = (2.10 cm3)(10-2 m/cm)3 = 2.10 × 10-6 m3 Thus, the density is

ρ=

m 0.02394 kg = = 1.14 × 104 kg/m3 V 2.10 × 10-6 m 3

Goal Solution A solid piece of lead has a mass of 23.94 g and a volume of 2.10 cm3. From these data, calculate the density of lead in SI units (kg/m3). G: From Table 1.5, the density of lead is 1.13 × 104 kg/m3, 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. m O: Density is defined as mass per volume, in ρ = . We must convert to SI units in the calculation. V A: ρ = 23.94 g  1 kg   100 cm 3 = 1.14 × 104 kg/m3 2.10 cm3 1000 g  1 m 

L: 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 isprobably 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/cm3, and objects that float must be less dense than water.




Chapter 1 Solutions
1.26 (a) We take information from Table 1.1: 1 LY = (9.46 × 1015 m) (b) 1 AU  = 6.31 × 104 AU 1.50 × 1011 m

9

The distance to the Andromeda galaxy is 2 × 1022 m = (2 × 1022 m) 1 AU  = 1.33 × 1011 AU 1.50 × 1011 m 

1.27

Natoms =

mSun 1.99 × 1030 kg = = 1.19 × 1057 atoms matom 1.67 × 10-27 kg

1.28

1 mi = 1609 m = 1.609 km; thus, to go from mph to km/h, multiply by 1.609. (a) (b) (c) 1 mi/h = 1.609 km/h 55 mi/h = 88.5 km/h 65 mi/h = 104.6 km/h. Thus, aˆ†v = 16.1 km/h

1.29

(a) (b)

 6 × 10 12 $  1 hr   1 day  1 yr  = 190 years  1000 $/s  3600 s  24 hr   365 days
The circumference of the Earth at the equator is 2π (6378 × 103 m) = 4.01 × 107 m. The length of one dollar bill is 0.155 m so that the length of 6 trillion bills is9.30 × 1011 m. Thus, the 6 trillion dollars would encircle the Earth 9.30 × 1011 m = 2.32 × 104 times 4.01 × 107 m

Goal Solution At the time of this book’s printing, the U.S. national debt is about $6 trillion. (a) If payments were made at the rate of $1 000 per second, how many years would it take to pay off a $6-trillion debt, assuming no interest were charged? (b) A dollar bill is about 15.5 cm long. If six trillion dollar bills were laid end to end around the Earth’s equator, how many times would they encircle the Earth? Take the radius of the Earth at the equator to be 6 378 km. (Note: Before doing any of these calculations, try to guess at the answers. You may be very surprised.) (a) G: $6 trillion is certainly a large amount of money, so even at a rate of $1000/second, we might guess that it will take a lifetime (~ 100 years) to pay off the debt. O: Time to repay the debt will be calculated by dividing the total debt by the rate at which it is repaid.




10

Chapter 1 Solutions

A: T =

$6 trillion $6 × 10 12 = = 190 yr$1000/s ($1000/s 3.16 × 107 s/yr)

L: OK, so our estimate was a bit low. $6 trillion really is a lot of money! (b) G: We might guess that 6 trillion bills would encircle the Earth at least a few hundred times, maybe more since our first estimate was low. O: The number of bills can be found from the total length of the bills placed end to end divided by the circumference of the Earth. A: N = L (6 × 1012 15.5 cm)(1 m/100 cm) = = 2.32 × 104 times C 2π 6.37 × 106 m

L: OK, so again our estimate was low. Knowing that the bills could encircle the earth more than 20 000 times, it might be reasonable to think that 6 trillion bills could cover the entire surface of the earth, but the calculated result is a surprisingly small fraction of the earth’s surface area!

1.30

(a) (b)

(3600 s/hr)(24 hr/day)(365.25 days/yr) = 3.16 × 107 s/yr Vmm = 4 3 4 πr = π (5.00 × 10-7 m)3 = 5.24 × 10-19 m3 3 3 Vcube 1 m3 = = 1.91 × 1018 micrometeorites Vmm 5.24 × 10-19 m 3 This would take 1.91 × 1018 micrometeorites = 6.05 × 1010 yr 3.16 × 107 micrometeorites/yr

1.31

V = At, so

t=

V 3.78 × 10-3 m3 = =1.51 × 10-4 m (or 151 µm) A 25.0 m2

1.32

V=

1 [(13.0 acres)(43560 ft2/acre)] Bh = (481 ft) 3 3 = 9.08 × 107 ft3, or

 2.83 × 10-2 m3 V = (9.08 × 107 ft3)   1 ft 3  
= 2.57 × 10 m
6 3

h

B




Chapter 1 Solutions
6

11

1.33 1.34

Fg = (2.50 tons/block)(2.00 × 10 blocks)(2000 lb/ton) = 1.00 × 1010 lbs The area covered by water is Aw = 0.700 AEarth = (0.700)(4π REarth ) = (0.700)(4π)(6.37 × 106 m)2 = 3.57 × 1014 m2 The average depth of the water is d = (2.30 miles)(1609 m/l mile) = 3.70 × 103 m The volume of the water is V = Awd = (3.57 × 1014 m2)(3.70 × 103 m) = 1.32 × 1018 m3 and the mass is m = ρV = (1000 kg/m3)(1.32 × 1018 m3) = 1.32 × 1021 kg
2

*1.35

SI units of volume are in m3: V = (25.0 acre-ft) 43560 ft 2   0.3048 m 3 = 3.08 × 104 m3  1 acre   1 ft  datom, scale  datom, real

*1.36

(a)

dnucleus, scale = dnucleus, real  = (2.40 × 10-15 m)  = 6.79 × 10-3 ft, or

300 ft   1.06 × 10-10 m 

dnucleus, scale = (6.79 × 10-3 ft)(304.8 mm/1 ft) = 2.07 mm
3 V atom ratom  4 πratom/3 = = Vnucleus 4 π r3 rnucleus   nucleus/3 3

(b)

=

datom  dnucleus 

3

= 1.37

1.06 × 10-10 m  2.40 × 10-15 m

3

= 8.62 × 10 13 times as large

The scale factor used in the 'dinner plate' model is S= 0.25 m = 2.5 × 10-6 m/lightyears 1.0 × 10 5 lightyears

The distance to Andromeda in the scale model will be Dscale = DactualS = (2.0 × 106 lightyears)(2.5 × 10-6 m/lightyears) = 5.0 m




12

Chapter 1 Solutions
2 6 2 AEarth 4π rEarth  rEarth  =  (6.37 × 10 m)(100 cm/m) = 13.4 = 2 = r 8 A Moon 4πrMoon  Moon  1.74 × 10 cm  2

1.38

(a)

(b)

VEarth 4πrEarth /3  rEarth  3  (6.37 × 106 m)(100 cm/m) 3 = = = = 49.1 V Moon 4πrMoon3/3 rMoon  1.74 × 108 cm 
3

1.39

To balance, mFe = mAl or ρFeVFe = ρAlVAl 4 4 3 3 ρFe  π rFe = ρAl   π rAl 3 3 rAl = rFe 

ρFe 1/3 ρAl
7.86 1/3 = 2.86 cm 2.70

rAl = (2.00 cm)  1.40

The mass of each sphere is 4πρAlrAl mA1 = ρAlVAl = 3 Setting these masses equal, 4πρFerFe 4πρFerFe = and rAl = rFe 3 3
3 3 3 3

and

mFe =ρFeVFe

4π ρFerFe = 3

3

ρFe/ρAl

1.41

The volume of the room is 4 × 4 × 3 = 48 m3 , while the volume of one ball is 4π  0.038 m 3 = 2.87 × 10-5 m3. 106 ping-pong balls in the room.

Therefore, one can fit about

As an aside, the actual number is smaller than this because there will be a lot of space in the room that cannot be covered by balls. In fact, even in the best arrangement, the so-called 'best packing fraction' is

π 2 = 0.74 so that at least 26% of the space will be empty. Therefore, the 6 above estimate reduces to 1.67 × 106 × 0.740 106.




Chapter 1 Solutions

13

Goal Solution Estimate the number of Ping-Pong balls that would fit into an average-size room (without being crushed). In your solution state the quantities you measure or estimate and the values you take for them. G: Since the volume of a typical room is much larger than a Ping-Pong ball, we should expect that a very large number of balls (maybe a million) could fit in a room. O: Since we are only asked to findan estimate, we do not need to be too concerned about how the balls are arranged. Therefore, to find the number of balls we can simply divide the volume of an average-size room by the volume of an individual Ping-Pong ball. A: A typical room (like a living room) might have dimensions 15 ft × 20 ft × 8 ft. Using the approximate conversion 1 ft = 30 cm, we find Vroom ≈ 15 ft × 20 ft × 8 ft = 2400 ft3  30 cm 3 = 7 × 107 cm3  1 ft 

A Ping-Pong ball has a diameter of about 3 cm, so we can estimate its volume as a cube: Vball ≈ (3 × 3 × 3) cm3 = 30 cm3 The number of Ping-Pong balls that can fill the room is N≈ V room 7 × 107 cm3 = = 2 × 106 balls ~ 106 balls V ball 30 cm3

L: So a typical room can hold about a million Ping-Pong balls. This problem gives us a sense of how big a million really is.

*1.42

It might be reasonable to guess that, on average, McDonalds sells a 3 cm × 8 cm × 10 cm = 240 cm3 medium-sized box of fries, and that it is packed 3/4 full with fries that have a cross section of 1/2 cm × 1/2 cm. Thus, the typical box of fries would contain fries that stretched a total of L= 3  V   3   240 cm3  = = 720 cm = 7.2 m  4   A   4   (0.5 cm)2 

250 million boxes would stretch a total distance of (250 × 106 box)(7.2 m/box) = 1.8 × 109 m. But we require an order of magnitude, so our answer is 109 m = 1 million kilometers . *1.43 A reasonable guess for the diameter of a tire might be 2.5 ft, with a circumference of about 8 ft. Thus, the tire would make (50 000 mi)(5280 ft/mi)(1 rev/8 ft) = 3 × 107 rev 107 rev




14

Chapter 1 Solutions

1.44

A typical raindrop is spherical and might have a radius of about 0.1 inch. Its volume is then approximately 4 × 10-3 in3. Since 1 acre = 43,560 ft2, the volume of water required to cover it to a depth of 1 inch is (1 acre 1 inch) = (1 acre · in)  The number of raindrops required is n= volume of water required volume of a single drop ≈ 6.3 × 106 in3 = 1.6 × 109 109 - 3 in 3 4 × 10 43,560 ft2  144 in2  ≈ 6.3 × 106 in3. £­ 1 acre   1 ft 2 

*1.45

In order to reasonably carry on photosynthesis, we might expect a blade of grass to require atleast 1/16 in2 = 43 × 10-5 ft2. Since 1 acre = 43,560 ft2, the number of blades of grass to be expected on a quarter-acre plot of land is about n= total area (0.25 acre 43,560 ft2/acre) = area per blade 43 × 10-5 ft2/blade

= 2.5 × 107 blades 107 blades 1.46 Since you have only 16 hours (57,600 s) available per day, you can count only $57,600 per day. Thus, the time required to count $1 billion dollars is t= 10 9 dollars  1 year  = 47.6 years 4 5.76 × 10 dollars/day  365 days 

Since you are at least 18 years old, you would be beyond age 65 before you finished counting the money. It would provide a nice retirement, but a very boring life until then. We would not advise it. 1.47 Assume the tub measure 1.3 m by 0.5 m by 0.3 m. One-half of its volume is then V = (0.5 1.3 m)(0.5 m)(0.3 m) = 0.10 m3 The mass of this volume of water is mwater = ρwaterV= (1000 kg/m3)(0.10 m3) = 100 kg ~102 kg

Pennies are now mostly zinc, but consider copper pennies filling 50% of the volume of the tub. The mass of copper required is mcopper = ρcopperV = (8930 kg/m )(0.10 m ) = 893 kg
3 3

~10 kg

3

©2000 by Harcourt College Publishers. All rights reserved.


Chapter 1 Solutions
*1.48

15

The typical person probably drinks 2 to 3 soft drinks daily. Perhaps half of these were in aluminum cans. Thus, we will estimate 1 aluminum can disposal per person per day. In the U.S. there are 250 million people, and 365 days in a year, so (250 × 106 cans/day 365 days/year) ≈ 1010 cans are thrown away or recycled each year. Guessing that each can weighs around 1/10 of an ounce, we estimate this represents (1010 cans 0.1 oz/can)(1 lb/16 oz)(1 ton/2000 lb) ≈ 3.1 × 105 tons/year. 105 tons

1.49

Assume: Total population = 107; one out of every 100 people has a piano; one tuner can serve about 1,000 pianos (about 4 per day for 250 weekdays, assuming each piano is tuned once per year). Therefore, # tuners ~  1 tuner   1 piano  (107 people) = 100 1000 pianos 100 people  (c) 3 (d) 2

1.50 1.51

(a) (a)

2

(b) 4

πr2 = π (10.5 m ± 0.2 m)2
= π [ (10.5 m)2 ± 2(10.5 m)(0.2 m) + (0.2 m)2] = 346 m2 ± 13 m2

(b) 1.52

2πr = 2π (10.5 m ± 0.2 m) = 66.0 m ± 1.3 m

( a ) 756.?? 37.2?0.83 + 2.5? 796./3 = 797 5/ (b) (c) 0.0032 (2 s.f.) × 356.3 (4 s.f.) = 1.14016 = (2 s.f.) 5.620 (4 s.f.) × π (> 4 s.f.) = 17.656 = (4 s.f.) 1.1 17.66




16

Chapter 1 Solutions

1.53

r = (6.50 ± 0.20) cm = (6.50 ± 0.20) × 10-2 m m = (1.85 ± 0.02) kg

ρ=

m

 4  π r3 3

also,

δρ δm 3δr = + ρ m r
In other words, the percentages of uncertainty are cumulative. Therefore,

δρ 0.02 3(0.20) = + = 0.103 ρ 1.85 6.50
1.85 = 1.61 × 103 kg/m3

ρ=

 4 π (6.5 × 10 -2 m) 3  3

and ρ ± δρ = (1.61 ± 0.17) × 103 kg/m3 1.54 1.55 (a) 3 (b) 4 (c) 3 (d) 2

The distance around is 38.44 m + 19.5 m + 38.44 m + 19.5 m = 115.88 m, but this answer must be rounded to 115.9 m because the distance 19.5 m carries information to only one place past the decimal. 115.9 m
19.0 m

1.56

V = 2V1 + 2V2 = 2(V1 + V2) V1 = (17.0 m + 1.0 m + 1.0 m)(1.0 m)(0.09 m) = 1.70 m3
36.0 cm 10.0 m

V2 = (10.0 m)(1.0 m)(0.090 m) = 0.900 m V = 2(1.70 m3) + 2(0.900 m3) = 5.2 m3

3

  δV δw 0.01 m = = 0.010  = 0.006 + 0.010 + 0.011 = 0.027 =2.7% w 1.0 m V  δt 0.1 cm = = 0.011  t 9.0 cm
δ l1 0.12 m l1 = 19.0 m = 0.0063
1 1 1 1




Chapter 1 Solutions
x = 100 m

17

*1.57

It is desired to find the distance x such that

1000 m (i.e., such that x is the same multiple of 100 m as the multiple that 1000 m is of x) . x Thus, it is seen that x = (100 m)(1000 m) = 1.00 × 10 m , and therefore x = 316 m . 9.00 × 10-7 kg = 9.80 × 10–10 m3. If the diameter of a molecule is d, 918 kg/m3 then that same volume must equal d(πr2) = (thickness of slick)(area of oil slick) where r = 0.418 m. Thus, 9.80 × 10-10 m3 -9 2 = 1.79 × 10 m π (0.418 m) Vtotal  Vtotal (A ) =  3  (4πr2) Vdrop drop  4πr /3
2 5 2

1.00 × 10 m

5

2

=

1.58

The volume of oil equals V =

d=

1.59

Atotal = (N)(Adrop) =  =

3Vtotal  30.0 × 10-6 m 3 = 3 = 4.50 m2 -5 m   r   2.00 × 10 

1.60

α' (deg) 15.0 20.0 25.0 24.0 24.4 24.5 24.6 24.7
2πr = 15.0 m h = tan55.0° r

α (rad) 0.262 0.349 0.436 0.419 0.426 0.428 0.429 0.431
r = 2.39 m

tan( α ) 0.268 0.364 0.4660.445 0.454 0.456 0.458 0.460

sin( α ) 0.259 0.342 0.423 0.407 0.413 0.415 0.416 0.418

difference 3.47% 6.43% 10.2% 9.34% 9.81% 9.87% 9.98% 10.1%

24.6°

1.61

h = (2.39 m)tan(55.0°) = 3.41 m




18

Chapter 1 Solutions

h

55° r




Chapter 1 Solutions
*1.62 (a) [V] = L3, [A] = L2, [h] = L [V] = [A][h] L3 = L3L = L3. Thus, the equation is dimensionally correct. (b) V cylinder = πR 2h = (πR 2)h = Ah, where A = π R 2 Vrectangular object = l wh = ( l w)h = Ah, where A = l w 1.63 The actual number of seconds in a year is (86,400 s/day)(365.25 day/yr) = 31,557,600 s/yr The percentage error in the approximation is thus (π × 107 s/yr) – (31,557,600 s/yr) 31,557,600 s/yr *1.64

19

× 100% = 0.449%

From the figure, 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, Ldiag = distance L2 + L2 . Thus, since the atoms areseparated by a

L = 0.200 nm, the diagonal planes are separated *1.65 (a) The speed of flow may be found from v= (b)

1 2

L2 + L2 = 0.141 nm

(Vol rate of flow) 16.5 cm3/s = = 0.529 cm/s 2 (Area: π D /4) π (6.30 cm)2/4

Likewise, at a 1.35 cm diameter, v= 16.5 cm3/s = 11.5 cm/s π (1.35 cm)2/4

*1.66

t=

V V 4(12.0 cm3 ) 1 m   106 µm  = = = 0.0289 cm  = 289 µm A π D2/4 π (23.0 cm)2  100 cm   1 m  (108 cars)(104 mi/yr) = 5.0 × 1010 gal/yr 20 mi/gal (108 cars)(104 mi/yr) 10 = 4.0 × 10 gal/yr 25 mi/gal

1.67

V20 mpg =

V25 mpg =

Fuel saved = V25 mpg – V20 mpg = 1.0 × 1010 gal/yr




20

Chapter 1 Solutions




Chapter 1 Solutions
1.68 (a) 1 cubic meter of water has a mass m = ρV = (1.00 × 10-3 kg/cm3)(1.00 m3)(102 cm/m)3 = 1000 kg (b) As a rough calculation, we treat each item as if it were 100% water. cell: m = ρV = ρ Error! πR3 ) = ρ Error! D3 ) 1 = ( 1000 kg/m3)  π  (1.0 × 10-6 m)3 = 5.2 × 10-16 kg 6  kidney: m = ρV = ρ Error! R3 ) =(1.00 × 10-3 kg/cm3 )Error! 3 = Error! fly:

21

π m = ρ  D 2h  4  π = (1 × 10-3 kg/cm3)  (2.0 mm) 2(4.0 mm)(10-1 cm/mm)3 4 
= 1.3 × 10-5 kg

1.69

The volume of the galaxy is

πr2t = π (1021 m)2 1019 m ~ 1061 m3
If the distance between stars is 4 × 1016 m, then there is one star in a volume on the order of (4 × 1016 m)3 ~ 1050 m3. The number of stars is about 1061 m3 ~ 10 11 stars 3 10 m /star
50




22

Chapter 1 Solutions

1.70

The density of each material is ρ =

m m 4m = = V π r2h π D 2h g  The tabulated value  2.70 is 2% cm3  g  The tabulated value  8.92 is 5% cm3  smaller. smaller.

Al:

ρ=

4(51.5 g) g = 2.75 π (2.52 cm)2(3.75 cm) cm3 4(56.3 g) π (1.23 cm)2(5.06 cm) = 9.36 g cm3

Cu: ρ =

Brass: ρ = Sn: Fe:

4(94.4 g) g = 8.91 2 π (1.54 cm) (5.69 cm) cm3

ρ= ρ=

4(69.1 g) g = 7.68 2 π (1.75 cm) (3.74 cm) cm3 4(216.1 g) g = 7.88 3 π (1.89 cm)2(9.77 cm) cm g  The tabulated value  7.86 is 0.3% smaller. cm3 