INSTRUCTOR'S SOLUTIONS MANUAL FOR SERWAY AND JEWETT'S PHYSICS FOR SCIENTISTS AND ENGINEERS SIXTH EDITION PDF

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INSTRUCTOR'S SOLUTIONS MANUAL FOR SERWAY AND JEWETT'S PHYSICS FOR SCIENTISTS AND ENGINEERS SIXTH EDITION Ralph V. McGrew Broome Community College James A. Currie Weston High School Australia • Canada • Mexico • Singapore • Spain • United Kingdom • United States 1 Physics and Measurement CHA...

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INSTRUCTOR'S SOLUTIONS MANUAL FOR

SERWAY AND JEWETT'S

PHYSICS FOR SCIENTISTS AND ENGINEERS SIXTH EDITION

Ralph V. McGrew Broome Community College James A. Currie Weston High School

Australia • Canada • Mexico • Singapore • Spain • United Kingdom • United States

1 Physics and Measurement CHAPTER OUTLINE 1.1 1.2 1.3 1.4 1.5 1.6 1.7

ANSWERS TO QUESTIONS

Standards of Length, Mass, and Time Matter and Model-Building Density and Atomic Mass Dimensional Analysis Conversion of Units Estimates and Order-ofMagnitude Calculations Significant Figures

Q1.1

Atomic clocks are based on electromagnetic waves which atoms emit. Also, pulsars are highly regular astronomical clocks.

Q1.2

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.

Q1.3

People have different size hands. Defining the unit precisely would be cumbersome.

Q1.4

(a) 0.3 millimeters (b) 50 microseconds (c) 7.2 kilograms

Q1.5

(b) and (d). You cannot add or subtract quantities of different dimension.

Q1.6

A dimensionally correct equation need not be true. Example: 1 chimpanzee = 2 chimpanzee is dimensionally correct. If an equation is not dimensionally correct, it cannot be correct.

Q1.7

If I were a runner, I might walk or run 10 1 miles per day. Since I am a college professor, I walk about 10 0 miles per day. I drive about 40 miles per day on workdays and up to 200 miles per day on vacation.

Q1.8

On February 7, 2001, I am 55 years and 39 days old. 55 yr

F 365.25 d I + 39 d = 20 128 dFG 86 400 s IJ = 1.74 × 10 GH 1 yr JK H 1d K

9

s ~ 10 9 s .

Many college students are just approaching 1 Gs. Q1.9

Zero digits. An order-of-magnitude calculation is accurate only within a factor of 10.

Q1.10

The mass of the forty-six chapter textbook is on the order of 10 0 kg .

Q1.11

With one datum known to one significant digit, we have 80 million yr + 24 yr = 80 million yr.

1

2

Physics and Measurement

SOLUTIONS TO PROBLEMS Section 1.1

Standards of Length, Mass, and Time

No problems in this section

Section 1.2 P1.1

Matter and Model-Building

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 = L2 + L2 . Thus, since the atoms are separated by a distance

L = 0.200 nm , the diagonal planes are separated by

Section 1.3 *P1.2

1 2 L + L2 = 0.141 nm . 2

Density and Atomic Mass

Modeling the Earth as a sphere, we find its volume as

4 3 4 π r = π 6.37 × 10 6 m 3 3

e

j

3

= 1.08 × 10 21 m 3 . Its

m 5.98 × 10 24 kg = = 5.52 × 10 3 kg m3 . This value is intermediate between the V 1.08 × 10 21 m 3 tabulated densities of aluminum and iron. Typical rocks have densities around 2 000 to 3 000 kg m3 . The average density of the Earth is significantly higher, so higher-density material must be down below the surface. density is then ρ =

P1.3

a

fb

g

e j

With V = base area height V = π r 2 h and ρ =

ρ=

a

fa

ρ = 2.15 × 10 kg m

3

3

9

3

.

m for both. Then ρ iron = 9.35 kg V and V 19.3 × 10 3 kg / m3 = 23.0 kg . = 9.35 kg 7.86 × 10 3 kg / m3

Let V represent the volume of the model, the same in ρ =

ρ gold =

P1.5

F 10 mm I f GH 1 m JK

1 kg m = 2 π r h π 19.5 mm 2 39.0 mm 4

*P1.4

m , we have V

m gold V

V = Vo − Vi =

ρ=

. Next,

ρ gold ρ iron

4 π r23 − r13 3

e

=

m gold 9.35 kg

and m gold

F GH

j

FG IJ e H K

e

4π ρ r23 − r13 m 4 , so m = ρV = ρ π r23 − r13 = V 3 3

j

j

.

I JK

Chapter 1

P1.6

3

4 4 3 π r and the mass is m = ρV = ρ π r 3 . We divide this equation 3 3 for the larger sphere by the same equation for the smaller:

For either sphere the volume is V =

m A ρ 4π rA3 3 rA3 = = = 5. m s ρ 4π rs3 3 rs3

a f

Then rA = rs 3 5 = 4.50 cm 1.71 = 7.69 cm . P1.7

*P1.8

Use 1 u = 1.66 × 10 −24 g .

F 1.66 × 10 GH 1 u F 1.66 × 10 = 55.9 uG H 1u F 1.66 × 10 = 207 uG H 1u

-24

I = 6.64 × 10 JK gI JK = 9.29 × 10 gI JK = 3.44 × 10 g

−24

g .

−23

g .

−22

g .

(a)

For He, m 0 = 4.00 u

(b)

For Fe, m 0

(c)

For Pb, m 0

(a)

The mass of any sample is the number of atoms in the sample times the mass m 0 of one atom: m = Nm 0 . The first assertion is that the mass of one aluminum atom is

-24

−24

m 0 = 27.0 u = 27.0 u × 1.66 × 10 −27 kg 1 u = 4.48 × 10 −26 kg . Then the mass of 6.02 × 10 23 atoms is m = Nm 0 = 6.02 × 10 23 × 4.48 × 10 −26 kg = 0.027 0 kg = 27.0 g . Thus the first assertion implies the second. Reasoning in reverse, the second assertion can be written m = Nm 0 . 0.027 0 kg = 6.02 × 10 23 m 0 , so m 0 =

0.027 kg 6.02 × 10 23

= 4.48 × 10 −26 kg ,

in agreement with the first assertion. (b)

The general equation m = Nm 0 applied to one mole of any substance gives M g = NM u , where M is the numerical value of the atomic mass. It divides out exactly for all substances, giving 1.000 000 0 × 10 −3 kg = N 1.660 540 2 × 10 −27 kg . With eight-digit data, we can be quite sure of the result to seven digits. For one mole the number of atoms is N=

F 1 I 10 GH 1.660 540 2 JK

−3 + 27

= 6.022 137 × 10 23 .

(c)

The atomic mass of hydrogen is 1.008 0 u and that of oxygen is 15.999 u. The mass of one molecule of H 2 O is 2 1.008 0 + 15.999 u = 18.0 u. Then the molar mass is 18.0 g .

(d)

For CO 2 we have 12.011 g + 2 15.999 g = 44.0 g as the mass of one mole.

b

g

b

g

4

Physics and Measurement

P1.9

b gFGH 101 kgg IJK = 4.5 × 10 kg I JK = 3.27 × 10 kg .

Mass of gold abraded: ∆m = 3.80 g − 3.35 g = 0.45 g = 0.45 g

F 1.66 × 10 GH 1 u

−27

Each atom has mass m 0 = 197 u = 197 u

3

−4

kg .

−25

Now, ∆m = ∆N m 0 , and the number of atoms missing is ∆N =

∆m

=

m0

4.5 × 10 −4 kg 3.27 × 10 −25 kg

= 1.38 × 10 21 atoms .

The rate of loss is ∆N ∆t ∆N ∆t P1.10

P1.11

=

FG H

1 yr 1.38 × 10 21 atoms 365.25 d 50 yr

IJ FG 1 d IJ FG 1 h IJ FG 1 min IJ K H 24 h K H 60 min K H 60 s K

= 8.72 × 10 11 atoms s .

e

je

j

(a)

m = ρ L3 = 7.86 g cm 3 5.00 × 10 −6 cm

(b)

N=

(a)

The cross-sectional area is

3

= 9.83 × 10 −16 g = 9.83 × 10 −19 kg

9.83 × 10 −19 kg m = = 1.06 × 10 7 atoms m 0 55.9 u 1.66 × 10 −27 kg 1 u

e

j

a

f a

fa

fa

A = 2 0.150 m 0.010 m + 0.340 m 0.010 m = 6.40 × 10

−3

2

m .

f.

The volume of the beam is

ja

e

f

V = AL = 6.40 × 10 −3 m 2 1.50 m = 9.60 × 10 −3 m3 . Thus, its mass is

e

FIG. P1.11

je9.60 × 10 m j = 72.6 kg . F 1.66 × 10 kg I = 9.28 × 10 The mass of one typical atom is m = a55.9 ufG H 1 u JK 3

m = ρV = 7.56 × 10 kg / m

(b)

−3

3

3

−27

0

m = Nm 0 and the number of atoms is N =

−26

kg . Now

72.6 kg m = = 7.82 × 10 26 atoms . −26 m 0 9.28 × 10 kg

Chapter 1

P1.12

(a)

F 1.66 × 10 GH 1 u

−27

The mass of one molecule is m 0 = 18.0 u

kg

I = 2.99 × 10 JK

−26

5

kg . The number of

molecules in the pail is N pail = (b)

1.20 kg m = = 4.02 × 10 25 molecules . m 0 2.99 × 10 −26 kg

Suppose that enough time has elapsed for thorough mixing of the hydrosphere. N both = N pail

F m I = (4.02 × 10 GH M JK pail

25

F 1.20 kg I , GH 1.32 × 10 kg JK

molecules)

total

21

or

N both = 3.65 × 10 4 molecules .

Section 1.4 P1.13

Dimensional Analysis

The term x has dimensions of L, a has dimensions of LT −2 , and t has dimensions of T. Therefore, the equation x = ka m t n has dimensions of

e

L = LT −2

j aTf m

n

or L1 T 0 = Lm T n − 2 m .

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 = 2 . The value of k, a dimensionless constant, cannot be obtained by dimensional analysis . *P1.14

(a)

Circumference has dimensions of L.

(b)

Volume has dimensions of L3 .

(c)

Area has dimensions of L2 .

e j

Expression (i) has dimension L L2

1/2

= L2 , so this must be area (c).

Expression (ii) has dimension L, so it is (a). Expression (iii) has dimension L L2 = L3 , so it is (b). Thus, (a) = ii; (b) = iii, (c) = i .

e j

6

Physics and Measurement

P1.15

*P1.16

(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 .

(a)

a f

a∝

∑F

or a = k

∑F

represents the proportionality of acceleration to resultant force and m m the inverse proportionality of acceleration to mass. If k has no dimensions, we have a = k

(b) P1.17

In units,

M ⋅L T

2

=

kg ⋅ m s2

F F L M ⋅L , F = , 2 =1 . m T M T2

, so 1 newton = 1 kg ⋅ m s 2 .

Inserting the proper units for everything except G,

LM kg m OP = G kg Ns Q m

2

2

Multiply both sides by m

Section 1.5 *P1.18

.

m3

2

and divide by kg ; the units of G are

kg ⋅ s 2

.

Conversion of Units

a

fa

f

Each of the four walls has area 8.00 ft 12.0 ft = 96.0 ft 2 . Together, they have area

e

4 96.0 ft 2 P1.19

2

2

jFGH 3.128mft IJK

2

= 35.7 m 2 .

Apply the following conversion factors: 1 in = 2.54 cm , 1 d = 86 400 s , 100 cm = 1 m , and 10 9 nm = 1 m

FG 1 H 32

IJ b2.54 cm inge10 m cmje10 K 86 400 s day −2

in day

9

j=

nm m

9.19 nm s .

This means the proteins are assembled at a rate of many layers of atoms each second! *P1.20

8.50 in 3 = 8.50 in 3

FG 0.025 4 m IJ H 1 in K

3

= 1.39 × 10 −4 m 3

Chapter 1

P1.21

Conceptualize: 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 m 2 .

a

fa

f

Categorize: We model the lot as a perfect rectangle to use Area = Length × Width. Use the conversion: 1 m = 3.281 ft .

a

Analyze: A = LW = 100 ft

1m I F 1 m IJ = 1 390 m 150 ft fG f FGH 3.281 a J K H 3.281 ft K ft

2

= 1.39 × 10 3 m 2 .

Finalize: Our calculated result agrees reasonably well with our initial estimate and has the proper units of m 2 . Unit conversion is a common technique that is applied to many problems. P1.22

(a)

a

fa

fa

f

V = 40.0 m 20.0 m 12.0 m = 9.60 × 10 3 m 3 3

3

b

g

V = 9.60 × 10 m 3.28 ft 1 m (b)

3

= 3.39 × 10 5 ft 3

The mass of the air is

e

je

j

m = ρ air V = 1.20 kg m3 9.60 × 10 3 m3 = 1.15 × 10 4 kg . The student must look up weight in the index to find

e

je

j

Fg = mg = 1.15 × 10 4 kg 9.80 m s 2 = 1.13 × 10 5 N . Converting to pounds,

jb

e

g

Fg = 1.13 × 10 5 N 1 lb 4.45 N = 2.54 × 10 4 lb . P1.23

(a)

Seven minutes is 420 seconds, so the rate is r=

(b)

30.0 gal = 7.14 × 10 −2 gal s . 420 s

Converting gallons first to liters, then to m3 ,

e

r = 7.14 × 10 −2 gal s

jFGH 3.1786galL IJK FGH 10 1 Lm IJK −3

3

r = 2.70 × 10 −4 m3 s . (c)

At that rate, to fill a 1-m3 tank would take t=

F 1m GH 2.70 × 10

3

−4

m

3

IF 1 h I = G J s JK H 3 600 K

1.03 h .

7

8

Physics and Measurement

*P1.24

(a)

(b)

(c)

(d) P1.25

FG 1.609 km IJ = 560 km = 5.60 × 10 m = 5.60 × 10 cm . H 1 mi K F 0.304 8 m IJ = 491 m = 0.491 km = 4.91 × 10 cm . Height of Ribbon Falls = 1 612 ftG H 1 ft K F 0.304 8 m IJ = 6.19 km = 6.19 × 10 m = 6.19 × 10 cm . Height of Denali = 20 320 ftG H 1 ft K F 0.304 8 m IJ = 2.50 km = 2.50 × 10 m = 2.50 × 10 cm . Depth of King’s Canyon = 8 200 ftG H 1 ft K 5

Length of Mammoth Cave = 348 mi

7

4

3

5

3

5

From Table 1.5, 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 =

ρ=

23.94 g 2.10 cm 3

F 1 kg I FG 100 cm IJ GH 1 000 g JK H 1 m K

3

m . We must convert to SI units in the calculation. V

= 1.14 × 10 4 kg m3

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. P1.26

It is often useful to remember that the 1 600-m race at track and field events is approximately 1 mile in length. To be precise, there are 1 609 meters in a mile. Thus, 1 acre is equal in area to mI a1 acrefFGH 6401 miacres IJK FGH 1 609 J mi K 2

*P1.27 P1.28

The weight flow rate is 1 200

FG H

ton 2 000 lb h ton

2

= 4.05 × 10 3 m 2 .

IJ FG 1 h IJ FG 1 min IJ = K H 60 min K H 60 s K

667 lb s .

1 mi = 1 609 m = 1.609 km ; thus, to go from mph to km h , multiply by 1.609. (a)

1 mi h = 1.609 km h

(b)

55 mi h = 88.5 km h

(c)

65 mi h = 104.6 km h . Thus, ∆v = 16.1 km h .

P1.29

Chapter 1

(a)

F 6 × 10 $ I F 1 h I FG 1 day IJ F 1 yr I = GH 1 000 $ s JK GH 3 600 s JK H 24 h K GH 365 days JK

(b)

The circumference of the Earth at the equator is 2π 6.378 × 10 3 m = 4.01 × 10 7 m . The length

12

190 years

e

j

of one dollar bill is 0.155 m so that the length of 6 trillion bills is 9.30 × 10 11 m. Thus, the 6 trillion dollars would encircle the Earth 9.30 × 10 11 m = 2.32 × 10 4 times . 4.01 × 0 7 m 1.99 × 10 30 kg mSun = = 1.19 × 10 57 atoms m atom 1.67 × 10 −27 kg

P1.30

N atoms =

P1.31

V = At so t =

V 3.78 × 10 −3 m 3 = = 1.51 × 10 −4 m or 151 µm 2 A 25.0 m

b

a

fe

13.0 acres 43 560 ft 2 acre 1 V = Bh = 3 3 7 3 = 9.08 × 10 ft ,

P1.32

g

j a481 ftf h

or

e

V = 9.08 × 10 7 ft 3

jFGH 2.83 ×110ft

−2

m3

3

I JK

B

FIG. P1.32

= 2.57 × 10 6 m3 P1.33 *P1.34

9

b

ge

jb

g

Fg = 2.50 tons block 2.00 × 10 6 blocks 2 000 lb ton = 1.00 × 10 10 lbs The area covered by water is

a fe

j a fa fe

j

2 = 0.70 4π 6.37 × 10 6 m A w = 0.70 AEarth = 0.70 4π REarth

2

= 3.6 × 10 14 m 2 .

The average depth of the water is

a

fb

g

d = 2.3 miles 1 609 m l mile = 3.7 × 10 3 m . The volume of the water is

e

je

j

V = A w d = 3.6 × 10 14 m 2 3.7 × 10 3 m = 1.3 × 10 18 m 3 and the mass is

e

je

j

m = ρ V = 1 000 kg m3 1.3 × 10 18 m3 = 1.3 × 10 21 kg .

10 P1.35

Physics and Measurement

(a)

d nucleus, scale = d nucleus, real

e

Fd I = 2.40 × 10 m FG 300 ft IJ = 6.79 × 10 jH 1.06 × 10 m K GH d JK e ft jb304.8 mm 1 ft g = 2.07 mm atom, scale

−15

−10

atom, real

d nucleus, scale = 6.79 × 10 −3

(b)

Vatom = Vnucleus

3 4π ratom 3 3 4π rnucleus 3

=

FG r Hr

IJ = FG d K Hd 3

atom

nucleus

IJ = F 1.06 × 10 K GH 2.40 × 10 3

atom

nucleus

−10 −15

m m

I JK

3

= 8.62 × 10 13 times as large *P1.36

scale distance between

P1.37

=

FG real IJ FG scale IJ ...