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UNITS , PHYSICAL QUANTITIES AND VECTORS
1.1. IDENTIFY: Convert units from mi to km and from km to ft.
S ET UP: 1 in. = 2.54 cm, 1 km = 1000 m , 12 in. = 1 ft, 1 mi = 5280 ft.
EXECUTE: (a) 2 3
5280 ft 12 in. 2.54 cm 1 m 1 km 1.00 mi (1.00 mi) 1.61 km 1 mi 1 ft 1 in. 10 cm 10 m
(b)
3 2
10 m 10 cm 1 in. 1 ft 3
1.00 km (1.00 km) 3.28 10 ft
1 km 1 m 2.54 cm 12 in.
= ⎜ ⎟⎜ ⎟⎜ = ×
⎝ ⎠⎝ ⎠⎝^ ⎠⎝^ ⎠
EVALUATE: A mile is a greater distance than a kilometer. There are 5280 ft in a mile but only 3280 ft in a km.
1.2. IDENTIFY: Convert volume units from L to
3
in..
S ET UP:
3 1 L = 1000 cm. 1 in. =2.54 cm
EXECUTE:
3 3 1000 cm 1 in. 3 0.473 L 28.9 in.. 1 L 2.54 cm
× ⎜ ⎟ × =
⎝ ⎠ ⎝^ ⎠
EVALUATE:
3 1 in. is greater than
3 1 cm , so the volume in
3 in. is a smaller number than the volume in
3 cm ,
which is
3 473 cm.
1.3. IDENTIFY: 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 UP: The speed of light is
8 v = 3.00 × 10 m/s. 1 ft = 0.3048 m.
9 1 s = 10 ns.
EXECUTE:
9 8
0.3048 m 1.02 10 s 1.02 ns 3.00 10 m/s
t
− = = × = ×
EVALUATE: In 1.00 s light travels
8 5 5 3.00 × 10 m = 3.00 × 10 km = 1.86 × 10 mi.
1.4. IDENTIFY: Convert the units from g to kg and from
3 cm to
3
m.
S ET UP: 1 kg = 1000 g. 1 m = 1000 cm.
EXECUTE:
3 4 3 3
g 1 kg 100 cm kg
cm 1000 g 1 m m
× ⎜ ⎟ × ⎜ ⎟= ×
EVALUATE: The ratio that converts cm to m is cubed, because we need to convert
3
cm to
3
m.
1.5. IDENTIFY: Convert volume units from
3
in. to L.
S ET UP:
3
1 L = 1000 cm. 1 in. = 2.54 cm.
EXECUTE: (^) ( ) ( ) (^) ( )
3 3 3
327 in. × 2.54 cm in. × 1 L 1000 cm =5.36 L
EVALUATE: The volume is
3
5360 cm.
3
1 cm is less than
3
1 in. , so the volume in
3
cm is a larger number than the
volume in
3
in..
1.6. IDENTIFY: Convert
2
ft to
2
m and then to hectares.
S ET UP:
4 2
1.00 hectare = 1.00 × 10 m. 1 ft = 0.3048 m.
EXECUTE: The area is
2 2
4 2
43,600 ft 0.3048 m 1.00 hectare
(12.0 acres) 4.86 hectares
1 acre 1.00 ft 1.00 10 m
⎝ ⎠⎝^ ⎠ ⎝^ × ⎠
EVALUATE: Since 1 ft = 0.3048 m,
2 2 2
1 ft = (0.3048) m.
1.7. IDENTIFY: Convert seconds to years.
S ET UP:
9
1 billion seconds = 1 × 10 s. 1 day = 24 h. 1 h = 3600 s.
EXECUTE: (^) ( )
9 1 h^ 1 day^ 1 y
1.00 billion seconds 1.00 10 s 31.7 y
3600 s 24 h 365 days
⎛ ⎞⎛ ⎞^ ⎛^ ⎞
= × ⎜ ⎟=
1-2 Chapter 1
EVALUATE: The conversion
7
1 y = 3.156 × 10 sassumes 1 y = 365.24 d, which is the average for one extra day
every four years, in leap years. The problem says instead to assume a 365-day year.
1.8. IDENTIFY: Apply the given conversion factors.
S ET UP: 1 furlong = 0.1250 mi and 1 fortnight = 14 days.1 day =24 h.
EXECUTE: ( )
0.125 mi 1 fortnight 1 day
180,000 furlongs fortnight 67 mi/h
1 furlong 14 days 24 h
⎝ ⎠⎝ ⎠⎝^ ⎠
EVALUATE: 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. IDENTIFY: Convert miles/gallon to km/L.
S ET UP: 1 mi = 1.609 km. 1 gallon =3.788 L.
EXECUTE: (a)
1.609 km 1 gallon
55.0 miles/gallon (55.0 miles/gallon) 23.4 km/L
1 mi 3.788 L
(b) The volume of gas required is
1500 km
64.1 L
23.4 km/L
64.1 L
1.4 tanks
45 L/tank
EVALUATE: 1 mi/gal = 0.425 km/L. A km is very roughly half a mile and there are roughly 4 liters in a gallon,
so
2
1 mi/gal ∼ 4 km/L, which is roughly our result.
1.10. IDENTIFY: Convert units.
S ET UP: Use the unit conversions given in the problem. Also, 100 cm = 1 mand 1000 g = 1 kg.
EXECUTE: (a)
mi 1h 5280 ft ft
h 3600s 1mi s
⎛ ⎞ ⎛^ ⎞ ⎛^ ⎞
(b) 2 2
ft 30.48cm 1 m m
s 1ft 100 cm s
⎛ ⎞ ⎛^ ⎞⎛ ⎞
(c)
3 3 3 3
g 100 cm 1 kg kg
cm 1 m 1000 g m
⎛ ⎞ ⎛ ⎞ ⎛^ ⎞
EVALUATE: The relations 60 mi/h = 88 ft/sand
3 3 3
1 g/cm = 10 kg/mare exact. The relation
2 2
32 ft/s = 9.8 m/sis
accurate to only two significant figures.
1.11. IDENTIFY: 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 UP:
3
Density = 19.5 g/cm and
critical
m = 60.0 kg.For a sphere
4 3 3
V = π r.
EXECUTE:
3 critical 3
60.0 kg 1000 g
/ density 3080 cm
19.5 g/cm 1.0 kg
V m
( )
3080 cm 9.0 cm
V
r
EVALUATE: The density is very large, so the 130 pound sphere is small in size.
1.12. IDENTIFY: Use your calculator to display
7
π × 10. Compare that number to the number of seconds in a year.
S ET UP: 1 yr = 365.24 days, 1 day = 24 h,and 1 h =3600 s.
EXECUTE:
24 h 3600 s 7
(365.24 days/1 yr) 3.15567... 10 s
1 day 1 h
⎜ ⎟⎜ =^ ×
⎝ ⎠⎝^ ⎠
7 7
π × 10 s = 3.14159... × 10 s
The approximate expression is accurate to two significant figures.
EVALUATE: The close agreement is a numerical accident.
1.13. IDENTIFY: The percent error is the error divided by the quantity.
S ET UP: The distance from Berlin to Paris is given to the nearest 10 km.
EXECUTE: (a)
3 3
10 m
890 10 m
−
= ×
×
(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.
EVALUATE: In this case a very small percentage error has disastrous consequences.
1.14. IDENTIFY: 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.
1-4 Chapter 1
(d) 200 mm = 0.200 m = 7.9 inches. This is much too short.
(e)
1 y 200 months (200 mon) 17 y 12 mon
. This is the age of a teenager; a middle-aged man is much older than this.
EVALUATE: None are plausible. When specifying the value of a measured quantity it is essential to give the units
in which it is being expressed.
1.20. IDENTIFY: The number of kernels can be calculated as N = V bottle (^) / V kernel.
S ET UP: Based on an Internet search, Iowan 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
3 1 L = 1000 cm and 1 cm =10 mm.
EXECUTE: The volume of the kernel is: (^ )(^ )(^ )^
3 kernel V = 10 mm 6 mm 3 mm = 180 mm. The bottleís volume is:
( ) ( ) ( ) ( ) ( ) 3 3 3 6 3 bottle V = 2.0 L ⎡^ 1000 cm 1.0 L ⎤⎡^ 10 mm 1.0 cm ⎤= 2.0 × 10 mm ⎣ ⎦ ⎣ ⎦
. The number of kernels is then
( ) ( ) 6 3 3 kernels bottle kernels N = V / V ≈ 2.0 × 10 mm 180 mm = 11,000 kernels.
EVALUATE: 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.
1.21. IDENTIFY: Estimate the number of pages and the number of words per page.
S ET UP: 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).
EXECUTE: An estimate for the number of words is about
6
EVALUATE: We can expect that this estimate is accurate to within a factor of 10.
1.22. IDENTIFY: Approximate the number of breaths per minute. Convert minutes to years and
3 cm to
3 m to find the
volume in
3 m breathed in a year.
S ET UP: Assume 10 breaths/min.
24 h 60 min 5 1 y (365 d) 5.3 10 min 1 d 1 h
= ⎜ ⎟⎜ ⎟= ×
2 10 cm = 1 mso
6 3 3 10 cm = 1 m. The volume of a sphere is
4 3 1 3 3 6
V = π r = π d , where r is the radius and d is the diameter. Donít
forget to account for four astronauts.
EXECUTE: (a) The volume is
5 6 3 5.3^10 min 4 3 (4)(10 breaths/min)(500 10 m ) 1 10 m / yr 1 y
− ⎛^ × ⎞
× ⎜ ⎟= ×
(b)
1/ 3 4 3 1/ 3 6 6[1 10 m ] 27 m
V
d
⎛ ⎞ ⎛^ × ⎞
EVALUATE: Our estimate assumes that each
3 cm of air is breathed in only once, where in reality not all the
oxygen is absorbed from the air in each breath. Therefore, a somewhat smaller volume would actually be
required.
1.23. IDENTIFY: Estimate the number of blinks per minute. Convert minutes to years. Estimate the typical lifetime in
years.
S ET UP: Estimate that we blink 10 times per minute. 1 y = 365 days. 1 day = 24 h, 1 h = 60 min. Use 80 years
for the lifetime.
EXECUTE: The number of blinks is
60 min 24 h 365 days 8 (10 per min) (80 y/lifetime) 4 10 1 h 1 day 1 y
⎛ ⎞ ⎛^ ⎞⎛^ ⎞
⎜ ⎟⎜ ⎟ =^ ×
EVALUATE: 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.
1.24. IDENTIFY: 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 UP: 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.
EXECUTE: (^ )^
9 beats
60 min 24 h 365 days 80 yr 75 beats/min 3 10 beats/lifespan 1 h 1 day yr lifespan
N
= ⎜ ⎟⎜ = ×
( )
9 3 7 blood (^3)
1 L 1 gal 3 10 beats 50 cm /beat 4 10 gal/lifespan 1000 cm 3.788 L lifespan
V
⎛ ⎞⎛ ⎞⎛^ × ⎞
= ⎜ ⎟⎜ ⎟⎜ ⎟= ×
EVALUATE: This is a very large volume.
Units, Physical Quantities and Vectors 1-
1.25. IDENTIFY: Estimation problem
S ET UP: 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.
EXECUTE: The volume of gold in the pile is
3
V = 18 in. × 18 in. × 68 in. = 22,000 in. .Convert to
3 cm :
3 3 3 5 3
V = 22,000 in. (1000 cm / 61.02 in. ) = 3.6 × 10 cm.
The density of gold is
3 19.3 g/cm , so the mass of this volume of gold is
3 5 3 6
m = (19.3 g/cm )(3.6 × 10 cm ) = 7 × 10 g.
The monetary value of one gram is $10, so the gold has a value of
6 7 ($10 / gram)(7 × 10 grams) = $7 ×10 , or about
6 $100 × 10 (one hundred million dollars).
EVALUATE: This is quite a large pile of gold, so such a large monetary value is reasonable.
1.26. IDENTIFY: Estimate the diameter of a drop and from that calculate the volume of a drop, in
3 m. Convert
3 m to L.
S ET UP: Estimate the diameter of a drop to be d = 2 mm. The volume of a spherical drop is
4 3 1 3 3 6
V = π r = π d.
3 3 10 cm = 1 L.
EXECUTE:
1 3 3 3 6
V π (0.2 cm) 4 10 cm
− = = ×. The number of drops in 1.0 L is
3 5 3 3
1000 cm 2 10 4 10 cm
−
= ×
×
EVALUATE: Since
3
V ∼ d , if our estimate of the diameter of a drop is off by a factor of 2 then our estimate of the
number of drops is off by a factor of 8.
1.27. IDENTIFY: Estimate the number of students and the average number of pizzas eaten by each student in a school year.
SET UP: Assume a school of thousand students, each of whom averages ten pizzas a year (perhaps an underestimate)
EXECUTE: They eat a total of 10
4 pizzas.
EVALUATE: The same answer applies to a school of 250 students averaging 40 pizzas a year each.
1.28. IDENTIFY: The number of bills is the distance to the moon divided by the thickness of one bill.
S ET UP: Estimate the thickness of a dollar bills by measuring a short stack, say ten, and dividing the
measurement by the total number of bills. I obtain a thickness of roughly 1 mm. From Appendix F, the distance
from the earth to the moon is
8 3.8 × 10 m.
EXECUTE:
8 3 12 12 bills
3.8 10 m 10 mm 3.8 10 bills 4 10 bills 0.1 mm/bill 1 m
N
⎛ × ⎞⎛ ⎞
= ⎜ ⎟⎜ ⎟= × ≈ ×
EVALUATE: This answer represents 4 trillion dollars! The cost of a single space shuttle mission in 2005 is
significantly less ñ roughly 1 billion dollars.
1.29. IDENTIFY: The cost would equal the number of dollar bills required; the surface area of the U.S. divided by the
surface area of a single dollar bill.
S ET UP: By drawing a rectangle on a map of the U.S., the approximate area is 2600 mi by 1300 mi or
2 mi. This estimate is within 10 percent of the actual area, 3,794,
2 mi. The population is roughly 8 3.0 × 10 while the area of a dollar bill, as measured with a ruler, is approximately
1 8 6 in. by
5 8 2 in.
EXECUTE: (^ )^ [(^ )^ (^ )]^ (^ ) (^ )^
2 2 2 16 2
A U.S. = 3,380,000 mi 5280 ft / 1 mi ⎡⎣ 12 in. 1 ft ⎤⎦ = 1.4 × 10 in.
2
A bill = 6.125 in. 2.625 in. =16.1 in.
( ) ( ) 16 2 2 14
Total cost = N bills = A U.S. A bill= 1.4 × 10 in. 16.1 in. / bill = 9 × 10 bills
14 8 6 Cost per person = (9 × 10 dollars) /(3.0 × 10 persons) = 3 ×10 dollars/person
EVALUATE: The actual cost would be somewhat larger, because the land isnít flat.
1.30. IDENTIFY: The displacements must be added as vectors and the magnitude of the sum depends on the relative
orientation of the two displacements.
S ET UP: 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.
EXECUTE: The orientations of the displacements that give the desired sum are shown in Figure 1.30.
EVALUATE: 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.
Units, Physical Quantities and Vectors 1-
EVALUATE: D
is equal in magnitude and opposite in direction to the sum A + B + C
Figure 1.
1.34. IDENTIFY and S ET UP: Use a ruler and protractor to draw the vectors described. Then draw the corresponding
horizontal and vertical components.
EXECUTE: (a) Figure 1.34 gives components 4.7 m, 8.1 m.
(b) Figure 1.34 gives components −15.6 km, 15.6 km.
(c) Figure 1.34 gives components 3.82 cm, − 5.07 cm.
EVALUATE: The signs of the components depend on the quadrant in which the vector lies.
Figure 1.
1.35. IDENTIFY: For each vector V
, use that cos x
V = V θand sin
y
V = V θ, when θ is the angle V
makes with the + x
axis, measured counterclockwise from the axis.
S ET UP: For A
, θ = 270.0°. For B
, θ = 60.0°. For C
, θ = 205.0°. For D
EXECUTE: 0
x
A = , 8.00 m
y
A = −. 7.50 m
x
B = , 13.0 m
y
B =. 10.9 m
x
C = − , 5.07 m
y
C = −. 7.99 m
x
D = − ,
6.02 m y
D =.
EVALUATE: The signs of the components correspond to the quadrant in which the vector lies.
1.36. IDENTIFY: tan
y
x
A
A
θ = , for θ measured counterclockwise from the + x -axis.
S ET UP: A sketch of Ax ,
y
A and A
tells us the quadrant in which A
lies.
EXECUTE:
(a)
1.00 m tan 0. 2.00 m
y
X
A
A
= = = −. (^) ( )
1
θ tan 0.500 360 26.6 333
−
(b)
1.00 m tan 0. 2.00 m
y
x
A
A
= = =. (^) ( )
1
θ tan 0.500 26.
−
(c)
1.00 m tan 0. 2.00 m
y
x
A
A
. (^) ( )
1
θ tan 0.500 180 26.6 153
−
(d)
1.00 m tan 0. 2.00 m
y
x
A
A
. (^) ( )
1
θ tan 0.500 180 26.6 207
−
EVALUATE: The angles 26.6° and 207° have the same tangent. Our sketch tells us which is the correct value
of θ.
1.37. IDENTIFY: Find the vector sum of the two forces.
S ET UP: Use components to add the two forces. Take the + x -directionto be forward and the + y -directionto be
upward.
1-8 Chapter 1
EXECUTE: The second force has components F 2 x = F 2 cos32.4 ° = 433 Nand F 2 y = F 2 sin 32.4 ° = 275 N.The
first force has components F 1 x = 725 Nand F 1 y =0.
Fx = F 1 x + F 2 x = 1158 N and Fy = F 1 y + F 2 y =275 N
The resultant force is 1190 N in the direction 13.4° above the forward direction.
EVALUATE: Since the two forces are not in the same direction the magnitude of their vector sum is less than the
sum of their magnitudes.
1.38. IDENTIFY: Find the vector sum of the three given displacements.
S ET UP: Use coordinates for which + x is east and + y is north. The driverís vector displacements are:
A = 2.6 km, 0 ° of north; B = 4.0 km, 0 ° of east; C = 3.1 km, 45 °north of east
EXECUTE: Rx = Ax + Bx + Cx = 0 + 4.0 km + (^ 3.1 km cos 45)^ (^ )=6.2 km
; R y = Ay + B y + Cy =
2.6 km + 0 + (3.1 km) sin45^ (^ )=4.8 km
;
2 2
R = Rx + Ry = 7.8 km; (^ ) (^ )
1
θ tan 4.8 km 6.2 km
− = ⎡^ ⎤ ⎣ ⎦ =^38
;
R =7.8 km, 38 north of east.
This result is confirmed by the sketch in Figure 1.38.
EVALUATE: Both Rx and Ry are positive and R
is in the first quadrant.
Figure 1.
1.39. IDENTIFY: If C = A + B
, then x x x C = A + B and y y y C = A + B. Use x C and y C to find the magnitude and
direction of C
S ET UP: From Figure 1.34 in the textbook, 0 x A = , 8.00 m y A = − and sin 30.0 7.50 m x
B = + B ° = ,
cos30.0 13.0 m y
B = + B ° =.
EXECUTE: (a) C = A + B
so 7.50 m x x x C = A + B = and 5.00 m y y y C = A + B = +. C = 9.01 m.
5.00 m tan 7.50 m
y
x
C
C
θ = = and θ = 33.7°.
(b) B + A = A + B
, so B + A
has magnitude 9.01 m and direction specified by 33.7°.
(c) D = A − B
so 7.50 m x x x D = A − B = − and 21.0 m y y y D = A − B = −. D = 22.3 m.
21.0 m tan 7.50 m
y
x
D
D
and
φ = 70.3°. D
is in the 3
rd
quadrant and the angle θ counterclockwise from the + x axis is 180 ° + 70.3 ° =250.3°.
(d) B − A = − ( A − B )
, so B − A
has magnitude 22.3 m and direction specified by θ = 70.3°.
EVALUATE: These results agree with those calculated from a scale drawing in Problem 1.32.
1.40. IDENTIFY: Use Equations (1.7) and (1.8) to calculate the magnitude and direction of each of the given vectors.
S ET UP: A sketch of x
A ,
y A and A
tells us the quadrant in which A
lies.
EXECUTE: (a)
2 2 ( 8.60 cm)− + (5.20 cm) = 10.0 cm,
arctan 148.
(which is 180 ° − 31.2° ).
(b)
2 2 ( 9.7 m)− + ( 2.45 m)− = 10.0 m,
arctan 14 180 194.
(c)
2 2 (7.75 km) + ( 2.70 km)− = 8.21 km,
arctan 340.
(which is 360 ° − 19.2° ).
EVALUATE: In each case the angle is measured counterclockwise from the + x axis. Our results for θ agree with
our sketches.
1-10 Chapter 1
(c) Similarly, 4 10 cm. − (^) ( 1 30 cm. (^) )= 2 80 cm,. 2 3 75 cm. − (^) ( 2 25 cm. (^) )= 2 6 00 cm..
(d)
2 2 (2.80cm) + ( 6.00cm)− = 6.62cm,
arctan 295
(which is 360 ° − 65 ° ).
EVALUATE: We can draw the vector addition diagram in each case and verify that our results are qualitatively
correct.
1.43. IDENTIFY: Vector addition problem. A − B = A + ( − B ).
S ET UP: Find the x - and y -components of A
and B.
Then the x - and y -components of the vector sum are
calculated from the x - and y -components of A
and B.
EXECUTE:
Ax = A cos(60.0 )°
Ax = (2.80 cm)cos(60.0 )° = + 1.40 cm
Ay = A sin(60.0 )°
Ay = (2.80 cm)sin(60.0 )° = + 2.425 cm
Bx = B cos( 60.0 )− °
Bx = (1.90 cm)cos( 60.0 )− ° = + 0.95 cm
B (^) y = B sin( 60.0 )− °
B (^) y = (1.90 cm)sin( 60.0 )− ° = − 1.645 cm
Note that the signs of the components
correspond to the directions of the component
vectors.
Figure 1.43a
(a) Now let R = A + B.
1.40 cm 0.95 cm 2.35 cm. x x x
R = A + B = + + = +
2.425 cm 1.645 cm 0.78 cm. y y y
R = A + B = + − = +
2 2 2 2 (2.35 cm) (0.78 cm) x y
R = R + R = +
R = 2.48 cm
0.78 cm tan 0. 2.35 cm
y
x
R
R
Figure 1.43b
EVALUATE: The vector addition diagram for R = A + B
is
R
is in the 1st quadrant, with
Ry < Rx ,in agreement with
our calculation.
Figure 1.43c
Units, Physical Quantities and Vectors 1-
(b) EXECUTE: Now let R = A − B.
Rx = Ax − Bx = + 1.40 cm − 0.95 cm = +0.45 cm.
R (^) y = Ay − By = + 2.425 cm + 1.645 cm = +4.070 cm.
2 2 2 2 (0.45 cm) (4.070 cm) x y
R = R + R = +
R = 4.09 cm
4.070 cm tan 9. 0.45 cm
y
x
R
R
Figure 1.43d
EVALUATE: The vector addition diagram for R = A + ( − B )
is
R
is in the 1st quadrant,
with , x y R < R in
agreement with our
calculation.
Figure 1.43e
(c) EXECUTE:
B − A = − (^) ( A − B )
B − A
and A − B
are
equal in magnitude and
opposite in direction.
R = 4.09 cmand
Figure 1.43f
Units, Physical Quantities and Vectors 1-
1.46. IDENTIFY: We know the vector sum and want to find the magnitude of the vectors. Use the method of
components.
SET UP: The two vectors A
and B
and their resultant C
are shown in Figure 1.46. Let + y be in the direction of
the resultant. A = B.
EXECUTE: C y = Ay + By. 372 N = 2 A cos 43.0° and A = 254 N.
EVALUATE: 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.
1.47. IDENTIFY: Find the components of each vector and then use Eq.(1.14).
SET UP: Ax = 0 , Ay = −8.00 m. Bx = 7.50 m , By = 13.0 m. Cx = −10.9 m , Cy = −5.07 m. Dx = −7.99 m ,
D y = 6.02 m.
EXECUTE:
à A = ( 8.00 m)− j
à à B = (7.50 m) i + (13.0 m) j
à à C = ( 10.9 m)− i + ( 5.07 m)− j
à à D = ( 7.99 m)− i + (6.02 m) j
EVALUATE: All these vectors lie in the xy -plane and have no z -component.
1.48. IDENTIFY: The general expression for a vector written in terms of components and unit vectors is
à à
A = A x i + Ay j
SET UP:
à à 5.0 B = 5.0(4 i − 6 ) j = 20 i − 30 j
EXECUTE: (a) Ax = 5.0, Ay = −6.3 (b) Ax = 11.2, Ay = −9.91 (c) Ax = −15.0 , Ay =22.
(d) Ax = 20 , Ay = − 30
EVALUATE: The components are signed scalars.
1.49. IDENTIFY: 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.
SET UP: Use the coordinates in the figure that accompanies the problem.
EXECUTE: (a) A = (^) ( 3.60 m cos70.0) ° i + à^ (^) ( 3.60 m sin 70.0) °à j =^ (^) ( 1.23 m) i + à^ (^) ( 3.38 m)à j
( ) ( ) ( ) ( ) B = − 2.40 m cos 30.0 ° − i à^ 2.40 m sin 30.0 ° à j =^ −2.08 m i + à^ −1.20 m à j
(b) C = (^) ( 3.00) A −( 4.00) B
( )( ) ( )( ) ( )( ) ( )( ) = 3.00 1.23 m i + à^ 3.00 3.38 m à j^ − 4.00 −2.08 m i à^ − 4.00 −1.20 m à j
à à = (12.01 m) i^ +(14.94) j
(c) From Equations (1.7) and (1.8),
( ) ( )
2 2 14.94 m 12.01 m 14.94 m 19.17 m, arctan 51. 12.01 m
C
EVALUATE:
x
C and
y
C are both positive, so θ is in the first quadrant.
1.50. IDENTIFY: Find A and B. Find the vector difference using components.
SET UP: Deduce the x - and y -components and use Eq.(1.8).
EXECUTE: (a) A = 4.00 i à^ +3.00 ;à j
x
A = + 3.
y
A = +
2 2 2 2 (4.00) (3.00) 5. x y
A = A + A = + =
1-14 Chapter 1
à à B = 5.00 i −2.00 ; j
Bx = +5.00; By = −2.
2 2 2 2
B = Bx + By = (5.00) + −( 2.00) =5.
EVALUATE: Note that the magnitudes of A
and B
are each larger than either of their components.
EXECUTE: (b) (^) ( ) à à à à à à A − B = 4.00 i + 3.00 j − 5.00 i − 2.00 j = (4.00 − 5.00) i + (3.00 +2.00) j
à à A − B = −1.00 i +5.00 j
(c) Let
à à R = A − B = −1.00 i +5.00. j
Then Rx = −1.00, Ry =5.00.
2 2 x y
R = R + R
2 2
R = ( 1.00)− + (5.00) =5.10.
tan 5.
y
x
R
R
θ = = = − −
θ = − 78.7 ° + 180 ° = 101.3 .°
Figure 1.
EVALUATE: 0
x
R < and 0,
y
R > so R
is in the 2nd quadrant.
1.51. IDENTIFY: A unit vector has magnitude equal to 1.
S ET UP: The magnitude of a vector is given in terms of its components by Eq.(1.12).
EXECUTE: (a) à à à 2 2 2 i + j + k = 1 + 1 + 1 = 3 ≠ 1 so it is not a unit vector.
(b)
2 2 2 x y z
A = A + A + A
. If any component is greater than + 1 or less than −1, A > 1
, so it cannot be a unit
vector. A
can have negative components since the minus sign goes away when the component is squared.
(c) A = 1
gives (^) ( ) ( )
2 2 2 2
a 3.0 + a 4.0 = 1 and
2
a 25 = 1.
a = ± = ±.
EVALUATE: The magnitude of a vector is greater than the magnitude of any of its components.
1.52. IDENTIFY: If vectors A
and B
commute for addition, A + B = B + A
. If they commute for the scalar product,
A B ⋅ = B ⋅ A
S ET UP: Express the sum and scalar product in terms of the components of A
and B
EXECUTE: (a) Let à^ à x y
A = A i + A j
and à^ à x y
B = B i + B j
. (^) ( ) à^ ( )à x x y y
A + B = A + B i + A + B j
( ) ( ) à à x x y y
B + A = B + A i + B + A j
. Scalar addition is commutative, so A + B = B + A
x x y y
A B ⋅ = A B + A B
and x x y y
B ⋅ A = B A + B A
. Scalar multiplication is commutative, so A B ⋅ = B ⋅ A
(b) (^) ( ) à^ ( ) à^ ( )à y z z y z x x z x y y x
A ◊ B = A B − A B i + A B − A B j + A B − A B k
( ) ( ) ( ) à à à y z z y z x x z x y y x
B ◊ A = B A − B A i + B A − B A j + B A − B A k
. Comparison of each component in each vector
product shows that one is the negative of the other.
EVALUATE: The result in part (b) means that A ◊ B
and B ◊ A
have the same magnitude and opposite direction.
1.53. IDENTIFY: A B ⋅ = AB cos φ
S ET UP: For A
and B
, φ = 150.0°. For B
and C
, φ = 145.0°. For A
and C
, φ = 65.0°.
EXECUTE: (a)
2 A B ⋅ = (8.00 m)(15.0 m)cos150.0 = −104 m
(b)
2 B C ⋅ = (15.0 m)(12.0 m)cos145.0 = −148 m
(c)
2 A C ⋅ = (8.00 m)(12.0 m)cos65.0 =40.6 m
EVALUATE: When φ < 90 ° the scalar product is positive and when φ > 90 ° the scalar product is negative.
1.54. IDENTIFY: Target variables are A B ⋅
and the angle φ between the two vectors.
S ET UP: We are given A
and B
in unit vector form and can take the scalar product using Eq.(1.19). The angle
φ can then be found from Eq.(1.18).
1-16 Chapter 1
( ) ( )
à à à à
A × B = 4.00 i + 3.00 j × 5.00 i − 2.00 j =
à à à à à à à à
20.0 i × i − 8.00 i × j + 15.0 j × i − 6.00 j × j
But à à à à
i × i = j × j = 0 and
à à à
i × j = k ,
à à à j × i = − k ,so (^) ( ) à à à
A × B = −8.00 k + 15.0 − k = −23.0. k
The magnitude of A × B
is 23.0.
EVALUATE: Sketch the vectors A
and B
in a coordinate system where the xy -plane is in the plane of the paper
and the z -axis is directed out toward you.
Figure 1.
By the right-hand rule A × B
is directed into the plane of the paper, in the − z -direction.This agrees with the
above calculation that used unit vectors.
1.59. IDENTIFY: The right-hand rule gives the direction and Eq.(1.22) gives the magnitude.
S ET UP: φ = 120.0°.
EXECUTE: (a) The direction of A◊ B
is into the page (the − z -direction). The magnitude of the vector product
is (^) ( )( )
2
AB sin φ = 2.80 cm 1.90 cm sin120 =4.61 cm
.
(b) Rather than repeat the calculations, Eq. (1.23) may be used to see that B ◊ A
has magnitude 4.61 cm
2 and is in
the + z -direction(out of the page).
EVALUATE: For part (a) we could use Eq. (1.27) and note that the only non-vanishing component is
C (^) z = A Bx y − A By x = (^) ( 2.80 cm cos 60.0) ° −( 1.90 cm sin 60 (^) ) °
( ) ( )
2 − 2.80 cm sin 60.0 ° 1.90 cm cos60.0 ° = −4.61 cm.
This gives the same result.
1.60. IDENTIFY: Area is length times width. Do unit conversions.
S ET UP: 1 mi = 5280 ft.
3 1 ft = 7.477 gal.
EXECUTE: (a) The area of one acre is 1 1 1 2 8 80 640 mi × mi = mi ,so there are 640 acres to a square mile.
(b) (^) ( )
2 2 1 mi 5280 ft 2 1 acre 43,560 ft 640 acre 1 mi
× ⎜ ⎟ × =
⎝ ⎠ ⎝^ ⎠
(all of the above conversions are exact).
(c) (1 acre-foot) (^) ( )
3 5 3
7.477 gal 43,560 ft 3.26 10 gal, 1 ft
= × = ×
which is rounded to three significant figures.
EVALUATE: 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.
1.61. IDENTIFY: The density relates mass and volume. Use the given mass and density to find the volume and from
this the radius.
S ET UP: The earth has mass
24
m E = 5.97 × 10 kgand radius
6
r E = 6.38 × 10 m. The volume of a sphere is
4 3 3
V = π r.
3 3
ρ = 1.76 g/cm = 1760 km/m.
EXECUTE: (a) The planet has mass
25
m = 5.5 m E= 3.28 × 10 kg.
25 22 3 3
3.28 10 kg 1.86 10 m 1760 kg/m
m
V
×
= = = ×.
1/ 3 1/ 3 22 3 3 3[1.86 10 m ] (^7 ) 1.64 10 m 1.64 10 km 4 4
V
r
⎛ ⎞ ⎛^ × ⎞
= ⎜ ⎟ = ⎜ ⎟ = × = ×
(b) r =2.57 r E
EVALUATE: Volume V is proportional to mass and radius r is proportional to
1/ 3
V , so r is proportional to
1/ 3
m. If
the planet and earth had the same density its radius would be
1/ 3
(5.5) r E = 1.8 r E. The radius of the planet is greater
than this, so its density must be less than that of the earth.
Units, Physical Quantities and Vectors 1-
1.62. IDENTIFY and S ET UP: Unit conversion.
EXECUTE: (a)
9
f = 1.420 × 10 cycles/s,so
10 9
s 7.04 10 s 1.420 10
− = × ×
for one cycle.
(b)
12 10
3600 s/h 5.11 10 cycles/h 7.04 10 s/cycle
−
= ×
×
(c) Calculate the number of seconds in 4600 million
9 years = 4.6 × 10 yand divide by the time for 1 cycle:
9 7 6 10
(4.6 10 y)(3.156 10 s/y) 2.1 10 cycles 7.04 10 s/cycle
2 −
× ×
= ×
×
(d) The clock is off by 1 s in
5 100,000 y = 1 × 10 y,so in
9 4.60 × 10 yit is off by
9 4 5
(1 s) 4.6 10 s 1 10
⎛ × ⎞
⎜ ⎟=^ ×
×
(about 13 h).
EVALUATE: In each case the units in the calculation combine algebraically to give the correct units for the answer.
1.63. IDENTIFY: The number of atoms is your mass divided by the mass of one atom.
S ET UP: Assume a 70-kg person and that the human body is mostly water. Use Appendix D to find the mass of
one H2O molecule:
27 26 18 015 u. × .1 661 × 10 kg/u = 2 992. × 10 kg/molecule.
2 2
EXECUTE: ( ) (^) ( )
26 27 70 kg / 2 992. × 10 kg/molecule = 2 34. × 10
2 molecules. Each H O molecule has 3 atoms, so 2
there are about
27 6 × 10 atoms.
EVALUATE: Assuming carbon to be the most common atom gives
27 3 × 10 molecules, which is a result of the
same order of magnitude.
1.64. IDENTIFY: Estimate the volume of each object. The mass m is the density times the volume.
S ET UP: The volume of a sphere of radius r is
4 3 3
V = π r. The volume of a cylinder of radius r and length l is
2
V = π r l. The density of water is
3 1000 kg m.
EXECUTE: (a) Estimate the volume as that of a sphere of diameter 10 cm:
4 3
V 5.2 10 m
− = ×.
( ) ( )( )
3 4 3
m 0.98 1000 kg m 5.2 10 m 0.5 kg
− = × =.
(b) Approximate as a sphere of radius r = 0.25 m μ (probably an over estimate):
20 3
V 6.5 10 m
− = ×.
( ) ( )( )
3 20 3 17 14
m 0.98 1000 kg m 6.5 10 m 6 10 kg 6 10 g
− − − = × = × = ×.
(c) Estimate the volume as that of a cylinder of length 1 cm and radius 3 mm:
2 7 3
V π r l 2.8 10 m
− = = ×.
( ) ( )( )
3 7 3 4
m 0.98 1000 kg m 2.8 10 m 3 10 kg 0.3 g
− − = × = × =.
EVALUATE: The mass is directly proportional to the volume.
1.65. IDENTIFY: Use the volume V and density ρ to calculate the mass M : ,so
M M
ρ V
V ρ
S ET UP: The volume of a cube with sides of length x is
3
x. The volume of a sphere with radius R is
4 3 3
π R.
EXECUTE: (a)
3 5 3 3 3
0.200 kg 2.54 10 m 7.86 10 kg/m
x
− = = × ×
2
x 2.94 10 m 2.94 cm
− = × =.
(b)
2.54 10 m 3
π R
− = ×.
2
R 1.82 10 m 1.82 cm
− = × =.
EVALUATE:
4 3
π = 4.2, so a sphere with radius R has a greater volume than a cube whose sides have length R.
1.66. IDENTIFY: Estimate the volume of sand in all the beaches on the earth. The diameter of a grain of sand determines
its volume. From the volume of one grain and the total volume of sand we can calculate the number of grains.
S ET UP: The volume of a sphere of diameter d is
1 3 6
V = π d. Consulting an atlas, we estimate that the continents
have about
5 1.45 × 10 kmof coastline. Add another 25% of this for rivers and lakes, giving
5 1.82 × 10 kmof
coastline. Assume that a beach extends 50 m beyond the water and that the sand is 2 m deep.
9 1 billion = 1 × 10.
EXECUTE: (a) The volume of sand is
8 10 3 (1.82 × 10 m)(50 m)(2 m) = 2 × 10 m. The volume of a grain is
1 3 3 12 3 6
V π (0.2 10 m) 4 10 m
− − = × = ×. The number of grains is
10 3 21 12 3
2 10 m 5 10 4 10 m
−
×
= ×
×
. The number of grains of sand
is about
22
(b) The number of stars is
9 9 22 (100 × 10 )(100 ×10 ) = 10. The two estimates result in comparable numbers for these
two quantities.
Units, Physical Quantities and Vectors 1-
EVALUATE: For the first solution, with A
east of north, each worker has to exert less force to produce the given
resultant force and this is the sensible direction for the worker to pull.
Figure 1.
1.70. IDENTIFY: Find the vector sum of the two displacements.
S ET UP: Call the two displacements A
and B
, where A = 170 kmand B = 230 km. A + B = R
. A
and B
are
as shown in Figure 1.70.
EXECUTE: (170 km) sin 68 (230 km) cos 48 311.5 km
x x x
R = A + B = ° + ° =.
(170 km) cos 68 (230 km) sin 48 107.2 km
y y y
R = A + B = ° − ° = −.
( ) ( )
2 2 2 2
311.5 km 107.2 km 330 km
x y
R = R + R = + − =.
107.2 km
tan 0.
311.5 km
y R x
R
R
19 south of east
R
EVALUATE: Our calculation using components agrees with R
shown in the vector addition diagram, Figure 1.70.
Figure 1.
1.71. IDENTIFY: A + B = C
(or B + A = C
). The target variable is vector A.
S ET UP: Use components and Eq.(1.10) to solve for the components of A.
Find the magnitude and direction of
A
from its components.
EXECUTE: (a)
Cx = Ax + Bx ,so Ax = Cx − Bx
Cy = Ay + By ,so Ay = C y − By
Cx = C cos 22.0 ° = (6.40 cm)cos 22.0°
Cx = + 5.934 cm
Cy = C sin 22.0 ° = (6.40 cm)sin 22.0°
Cy = + 2.397 cm
Bx = B cos(360 ° − 63.0 )° = (6.40 cm)cos 297.0°
Bx = + 2.906 cm
By = B sin 297.0 ° = (6.40 cm)sin 297.0°
By = − 5.702 cm
Figure 1.71a
(b) Ax = C x − Bx = +5.934 cm − 2.906 cm = +3.03 cm
Ay = Cy − By = + 2.397 cm − −( 5.702) cm = +8.10 cm
1-20 Chapter 1
(c)
2 2 x y
A = A + A
2 2
A = (3.03 cm) + (8.10 cm) =8.65 cm
8.10 cm
tan 2.
3.03 cm
y
x
A
A
Figure 1.71b
EVALUATE: The A
we calculated agrees qualitatively with vector A
in the vector addition diagram in part (a).
1.72. IDENTIFY: Add the vectors using the method of components.
S ET UP: Ax = 0 , Ay = −8.00 m. Bx = 7.50 m , By = 13.0 m. Cx = −10.9 m , C y = −5.07 m.
EXECUTE: (a) Rx = Ax + Bx + Cx = −3.4 m. Ry = Ay + By + Cy = −0.07 m. R = 3.4 m.
0.07 m
tan
3.4 m
θ = 1.2° below the − x -axis.
(b) S x = Cx − Ax − Bx = −18.4 m. S y = Cy − Ay − By = −10.1 m. S = 21.0 m.
10.1 m
tan
18.4 m
y
x
S
S
below the − x -axis.
EVALUATE: The magnitude and direction we calculated for R
and S
agree with our vector diagrams.
Figure 1.
1.73. IDENTIFY: Vector addition. Target variable is the 4th displacement.
S ET UP: Use a coordinate system where east is in the + x -directionand north is in the + y -direction.
Let A ,
B ,
and C
be the three displacements that are given and let D
be the fourth unmeasured displacement.
Then the resultant displacement is R = A + B + C + D.
And since she ends up back where she started, R =0.
0 = A + B + C + D ,
so D = −( A + B + C )
x x x x
D = − A + B + C and ( )
y y y y
D = − A + B + C
EXECUTE:
180 m,
x
A = − 0
y
A =
cos315 (210 m)cos315 148.5 m
x
B = B ° = ° = +
sin 315 (210 m)sin 315 148.5 m
y
B = B ° = ° = −
cos 60 (280 m)cos60 140 m
x
C = C ° = ° = +
sin 60 (280 m)sin 60 242.5 m
y
C = C ° = ° = +
Figure 1.73a
( ) ( 180 m 148.5 m 140 m) 108.5 m
x x x x
D = − A + B + C = − − + + = −
