




























































































Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Community
Ask the community for help and clear up your study doubts
Discover the best universities in your country according to Docsity users
Free resources
Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors
Instructor’s Solutions Manual for University Physics with Modern Physics 12 Solution Manual - Young
Typology: Exercises
1 / 1179
This page cannot be seen from the preview
Don't miss anything!
On special offer
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
3
3 1 L = 1000 cm. 1 in. =2.54 cm
3 3 1000 cm 1 in. 3 0.473 L 28.9 in.. 1 L 2.54 cm
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.
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
3 4 3 3
3
3
3
3
EXECUTE: (^) ( ) ( ) (^) ( )
3 3 3
3
3
3
3
3
2
2
4 2
2 2
4 2
2 2 2
9
EXECUTE: (^) ( )
1-2 Chapter 1
7
EXECUTE: ( )
EXECUTE: (a)
2
EXECUTE: (a)
(b) 2 2
(c)
3 3 3 3
3 3 3
2 2
3
critical
4 3 3
3 critical 3
( )
r
7
7 7
EXECUTE: (a)
3 3
−
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.
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
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
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.
9 beats
60 min 24 h 365 days 80 yr 75 beats/min 3 10 beats/lifespan 1 h 1 day yr lifespan
( )
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
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
3 cm :
3 3 3 5 3
The density of gold is
3 19.3 g/cm , so the mass of this volume of gold is
3 5 3 6
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.
4 3 1 3 3 6
3 3 10 cm = 1 L.
1 3 3 3 6
− = = ×. The number of drops in 1.0 L is
3 5 3 3
1000 cm 2 10 4 10 cm
−
EVALUATE: Since
3
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.
8 3 12 12 bills
3.8 10 m 10 mm 3.8 10 bills 4 10 bills 0.1 mm/bill 1 m
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
2
( ) ( ) 16 2 2 14
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-
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.
EVALUATE: The signs of the components depend on the quadrant in which the vector lies.
Figure 1.
, use that cos x
y
axis, measured counterclockwise from the axis.
x
y
x
y
x
y
x
6.02 m y
EVALUATE: The signs of the components correspond to the quadrant in which the vector lies.
1.36. IDENTIFY: tan
y
x
y
lies.
(a)
1.00 m tan 0. 2.00 m
y
X
= = = −. (^) ( )
1
−
(b)
1.00 m tan 0. 2.00 m
y
x
= = =. (^) ( )
1
−
(c)
1.00 m tan 0. 2.00 m
y
x
. (^) ( )
1
−
(d)
1.00 m tan 0. 2.00 m
y
x
. (^) ( )
1
−
EVALUATE: The angles 26.6° and 207° have the same tangent. Our sketch tells us which is the correct value
1.37. IDENTIFY: Find the vector sum of the two forces.
upward.
1-8 Chapter 1
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.
EXECUTE: Rx = Ax + Bx + Cx = 0 + 4.0 km + (^ 3.1 km cos 45)^ (^ )=6.2 km
2.6 km + 0 + (3.1 km) sin45^ (^ )=4.8 km
;
2 2
1
− = ⎡^ ⎤ ⎣ ⎦ =^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
cos30.0 13.0 m y
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
(b) B + A = A + B
, so B + A
(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
and
is in the 3
rd
(d) B − A = − ( A − B )
, so B − A
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
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° ).
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.
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
2.425 cm 1.645 cm 0.78 cm. y y y
2 2 2 2 (2.35 cm) (0.78 cm) x y
R = 2.48 cm
0.78 cm tan 0. 2.35 cm
y
x
Figure 1.43b
is
is in the 1st quadrant, with
Ry < Rx ,in agreement with
our calculation.
Units, Physical Quantities and Vectors 1-
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 = 4.09 cm
4.070 cm tan 9. 0.45 cm
y
x
EVALUATE: The vector addition diagram for R = A + ( − B )
is
is in the 1st quadrant,
with , x y R < R in
agreement with our
calculation.
B − A = − (^) ( A − B )
are
equal in magnitude and
opposite in direction.
R = 4.09 cmand
Units, Physical Quantities and Vectors 1-
components.
are shown in Figure 1.46. Let + y be in the direction of
the resultant. A = B.
only a component of each force is upward.
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
à à
à à 5.0 B = 5.0(4 i − 6 ) j = 20 i − 30 j
sum. Eq.(1.14) expresses a vector in terms of its components.
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
( ) ( )
2 2 14.94 m 12.01 m 14.94 m 19.17 m, arctan 51. 12.01 m
x
y
x
y
2 2 2 2 (4.00) (3.00) 5. x y
1-14 Chapter 1
à à B = 5.00 i −2.00 ; j
2 2 2 2
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
à à R = A − B = −1.00 i +5.00. j
2 2 x y
2 2
tan 5.
y
x
θ = = = − −
θ = − 78.7 ° + 180 ° = 101.3 .°
Figure 1.
x
y
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
. 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
2
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,
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
and à^ à x y
. (^) ( ) à^ ( )à x x y y
( ) ( ) à à x x y y
. Scalar addition is commutative, so A + B = B + A
x x y y
and x x y y
. 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
( ) ( ) ( ) à à à y z z y z x x z x y y x
. 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.
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
à à à à à à à à
But à à à à
à à à
à à à j × i = − k ,so (^) ( ) à à à
is 23.0.
Figure 1.
above calculation that used unit vectors.
1.59. IDENTIFY: The right-hand rule gives the direction and Eq.(1.22) gives the magnitude.
is (^) ( )( )
2
.
has magnitude 4.61 cm
2 and is in
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
6
4 3 3
3 3
EXECUTE: (a) The planet has mass
25
25 22 3 3
3.28 10 kg 1.86 10 m 1760 kg/m
1/ 3 1/ 3 22 3 3 3[1.86 10 m ] (^7 ) 1.64 10 m 1.64 10 km 4 4
1/ 3
1/ 3
the planet and earth had the same density its radius would be
1/ 3
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
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.
4 3 3
2
3 1000 kg m.
EXECUTE: (a) Estimate the volume as that of a sphere of diameter 10 cm:
4 3
− = ×.
( ) ( )( )
3 4 3
− = × =.
20 3
− = ×.
( ) ( )( )
3 20 3 17 14
− − − = × = × = ×.
(c) Estimate the volume as that of a cylinder of length 1 cm and radius 3 mm:
2 7 3
− = = ×.
( ) ( )( )
3 7 3 4
− − = × = × =.
EVALUATE: The mass is directly proportional to the volume.
3
4 3 3
EXECUTE: (a)
3 5 3 3 3
0.200 kg 2.54 10 m 7.86 10 kg/m
− = = × ×
2
− = × =.
(b)
2.54 10 m 3
− = ×.
2
− = × =.
4 3
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.
1 3 6
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
− − = × = ×. 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-
Figure 1.
x x x
y y y
( ) ( )
2 2 2 2
x y
y R x
R
Figure 1.
EXECUTE: (a)
Figure 1.71a
1-20 Chapter 1
(c)
2 2 x y
2 2
y
x
Figure 1.71b
y
x
Figure 1.
so D = −( A + B + C )
x x x x
y y y y
x
y
x
y
x
y
Figure 1.73a
x x x x