Tài liệu Physics exercises_solution: Chapter 01 ppt

1.21: 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 e

1

cm1000L

473

3 3

1

cm100g

1000

kg1cm

g3

bottle1gal

1

.oz128L

788.3

gal1m

1

L1000m

bottles9



The daily consumption must then be

.da

bottles78

.5da24.365

yr1yr

bottles10

11

h

mi67h

24

day1day

14

fortnight1

furlongs8

mile1fortnight

furlongs000

1

L788.3km1.609

mi1L

1.10: a)

s

ft88mi

1

ft5280s

3600

h1hr

m1ft

1

cm48.30s

kg1m

1

cm100cm

1.14: a) 12mm  5.98mm72mm2 (two significant figures)

b)512 98mmmm = 0.50 (also two significant figures)

c) 36 mm (to the nearest millimeter)

d) 6 mm

e) 2.0

1.15: a) If a meter stick can measure to the nearest millimeter, the error will be about

1.16: The area is 9.69  0.07 cm2, where the extreme values in the piece’s length and

width are used to find the uncertainty in the area The fractional uncertainty in the

2

cm 69 9 cm 07

0 = 0.72%, and the fractional uncertainties in the length and width are

b) 8.05 50 17020

1.18: (Number of cars  miles/car.day)/mi/gal = gallons/day

(2  108 cars  10000 mi/yr/car  1 yr/365 days)/(20 mi/gal) = 2.75  108 gal/day

1.19: Ten thousand; if it were to contain ten million, each sheet would be on the order

of a millionth of an inch thick

1.20: If it takes about four kernels to fill 1 cm3, a 2-L bottle will hold about 8000

kernels

1.21: 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 exercise and problems), so an estimate for the number of words is about 106

1.22: Assuming about 10 breaths per minutes, 2460 minutes per day, 365 days per year, and a lifespan of fourscore (80) years, the total volume of air breathed in a lifetime

is about 2105m3 This is the volume of a room 100m100m20m, which is kind of tight for a major-league baseball game, but it’s the same order of magnitude as the

volume of the Astrodome

1.23: This will vary from person to person, but should be of the order of 1105

1.24: With a pulse rate of a bit more than one beat per second, a heart will beat 105

times per day With 365 days in a year and the above lifespan of 80 years, the number of beats in a lifetime is about 3109 With

20

1 L (50 cm3) per beat, and about

4

1 gallon per liter, this comes to about 4107 gallons

1.25: The shape of the pile is not given, but gold coins stacked in a pile might well be in

the shape of a pyramid, say with a height of 2m and a base 3m3m The volume of such a pile is 6m3, and the calculations of Example 1-4 indicate that the value of this volume is 6108

1.26: The surface area of the earth is about 4R2 51014m2, where R is the radius of

the earth, about 6106m, so the surface area of all the oceans is about 41014m2 An average depth of about 10 km gives a volume of 41018m3 41024cm3 Characterizing the size of a “drop” is a personal matter, but 25 drops cm3 is reasonable, giving a total of

26

10 drops of water in the oceans

1.27: This will of course depend on the size of the school and who is considered a

"student'' A school of thousand students, each of whom averages ten pizzas a year (perhaps an underestimate) will total 104 pizzas, as will a school of 250 students

averaging 40 pizzas a year each

1.28: The moon is about 4108m41011mm away Depending on age, dollar bills can be stacked with about 2-3 per millimeter, so the number of bills in a stack

to the moon would be about 1012 The value of these bills would be $1 trillion (1 terabuck)

1.29: Areaof USA Area billnumber of bills

.inhabitantmillion

$3.6sinhabitant10

2

bills

10

9

bills109cm10m1cm6.7cm6.15kmm10km

14 2

4 2 2

2 6 2

m,

y x

C C

B B

C

B

m2.00

m00.1 tan

.26180500.0tan

500.0m2.00

m00.1 tan

(d)

1536

.26180500.0tan

500.0m2.00

m00.1 tan

)(

6.26500.0tan

500.0m2.00

m00.1 tan

(b)

3336

.26360500

.0tan

1 1

1 1

A

A θ c

θ A

A θ θ

x y

x y x y

1.37: Take the +x-direction to be forward and the +y-direction to be upward Then the

second force has components F2x  F2cos32.4433Nand F2y F2sin32.4275N The first force has components F1x725NandF1y 0

N1158

The resultant force is 1190 N in the direction 13.4 above the forward direction.

1.38: (The figure is given with the solution to Exercise 1.31).

The net northward displacement is (2.6 km) + (3.1 km) sin 45o = 4.8 km, and the net eastward displacement is (4.0 km) + (3.1 km) cos 45o = 6.2 km The

magnitude of the resultant displacement is (4.8km)2(6.2km)2 = 7.8 km, and the direction is arctan  64..28 = 38o north of east

1.39: Using components as a check for any graphical method, the components of B

are m

b) The magnitude and direction of A + B are the same as B + A.

c) The x- and y-components of the vector difference are – 26.4 m and

at m5.28m

8.10m

cm04.5(  = 5.60 cm, arctan  51 .4050



 = 344.5o ccw

c) Similarly, 4.10 cm – (1.30 cm) = 2.80 cm, –3.75 cm – (2.25 cm) = –6.00 cm.d) (2.80cm)2 (6.0cm)2 = 6.62 cm, arctan  26.80 00 = 295o (which is 360o– 65

0.60sincm90.160.0sin cm2.80

.60coscm90.160.0coscm2.80

0.60sincm90.160.0sin cm2.80

j i

B

ˆm2.5ˆm0.3ˆ60.0sin m0.6ˆ60.0cosm0.6

ˆm6.9ˆm5.11ˆ40.0sin m0.15ˆ40.0cosm0.15

j i

ˆ94.14ˆm01.12

ˆm20.100.4ˆm08.200.4ˆm38.300.3ˆm23.100.3

m14.94arctan m,

17.19m94.14m

01

5.00arctan ,10.500.5

1.48: a) iˆ ˆjkˆ  12 12 12  3 1 so it is not a unit vector

z y

10.40

.3

1

2

2 2 2 2

A

20.00.5

B

A

ˆˆ

ˆˆ

y y x

x

y y x

x

A B A

B

B A B

y y x x

A B A B

B A B A

1.50: Method 1: Product of magnitudescosθ

2 2

m5.71187

cosm6m12cos

AC

m6.1580cosm6m15cos

BC

m4.993cosm15m12cos

Method 2: (Sum of products of components)

2 2 2

m5.71)20.5(9.58)(

)0.3(22.7(

m6.15)20.59.64)(

()0.311.49)(

(

m4.99.64)(9.58)(

(11.49)22

.7

1.52: For all of these pairs of vectors, the angle is found from combining Equations

(1.18) and (1.21), to give the angle as

.arccos

y y x x

1.53: Use of the right-hand rule to find cross products gives (a) out of the page and b)

into the page

1.54: a) From Eq (1.22), the magnitude of the cross product is

12.0m18.0msin18037130m2

The right-hand rule gives the direction as being into the page, or the – z-direction Using

Eq (1.27), the only non-vanishing component of the cross product is

12m  18.0msin37 130m2



y x

1.56: a) From the right-hand rule, the direction of A B

 is into the page (the –

z-direction) The magnitude of the vector product is, from Eq (1.22),

0.60coscm90.10.60sincm80.2

60sincm90.10.60coscm80.2

C

gives the same result

b) Rather than repeat the calculations, Eq (1-23) may be used to see that

A

B 

 has magnitude 4.61 cm2 and is in the +z-direction (out of the page).

1.57: a) The area of one acre is mi mi mi2,

6401

8018

mile

2 2

ft560,43mi

1

ft5280acre

640

mi1acre

ft1

gal477.7ft560,

1.58: a) ($4,950,000 102acres)(1acre 43560ft2)10.77ft2 m2$12 m2

b) ($12 m2)(2.54cm in)2(1m 100cm)2 $.008 in2

c) $.008 in2(1in7 8in)$.007for postagestampsizedparcel

1.59: a) To three significant figures, the time for one cycle is

s

1004.7Hz10420.1

11.5h1

s3600s

cycles10

s10156.3Hz10

d) 4.600109 y4.601041.00105 y,so theclock wouldbeoff by 4.60104 s

1.60: Assume a 70-kg person, and the human body is mostly water Use Appendix D to

find the mass of one H2O molecule: 18.015 u  1.661  10–27 kg/u = 2.992  10–26kg/molecule (70 kg/2.992  10–26 kg/molecule) = 2.34  1027 molecules (Assuming carbon to be the most common atom gives 3  1027 molecules

1.61: a) Estimate the volume as that of a sphere of diameter 10 cm:

3 4

3 5.2 10 m3

3 6.5 10 m3

4 3

7 3

3 7 2

V 

1.62: a)

ρ

M V V

M

ρ ,so 

cm2.94m

1094.2

m1054.2kg/m1086.7

kg200.0

2

3 5 3

3 3

b)

cm1.82m1082.1

m1054.23

4

2

3 5 3

1.63: Assume each person sees the dentist twice a year for checkups, for 2 hours Assume

2 more hours for restorative work Assuming most dentists work less than 2000 hours per year, this gives 2000hours hoursper patient 500patientsper dentist.Assuming only half of the people who should go to a dentist do, there should be about 1 dentist per 1000 inhabitants Note: A dental assistant in an office with more than one treatment room could increase the number of patients seen in a single dental office

1014

100.6)kg100.6

mole kg 3 mole atoms 23

107.1(

)kg100.2(2

by  (the notation introduced in Chapter 14)

.102.1)

kg107.1(

)mkg10()m105.1(2(3

23

4

79 27

3 18

3 11

M

Note the conversion from g/cm3 to kg/m3

1.65: Let D

be the fourth force

N90.310

.53sin

N,07.240

.53

cos

N28.690

.30cos

N,00.400

.30

sin

N00.500

.30cos

N,6.860

.30

cos

so,0

C

C

B B

B

B

A A

A

A

y x

y x

y x

C B A D D

tanα D y D x 



54.75



α

axis-fromckwisecounterclo,

km2.107tan

km330km

2.107km

5.311

km2.10748

sin km)230(68coskm)170(

km5.31148

coskm)230(68sin km)170(

2 2

2 2

y x

y y y

x x x

R

R θ

R R R

B A R

B A R

19



R

θ

x x x

B C

A

B C

A

cm3.03

cm8.10arctan cm,65.8cm10.8cm03

and m,7.15

60180cosm0.640cosm00.153790cosm0.12

R

The magnitude of the resultant is R R x2  R2y 16.6m, and the direction from

the positive x-axis is arctan  15 5..73 18.6 Keeping extra significant figures in the intermediate calculations gives an angle of 18.49, which when considered as a

positive counterclockwise angle from the positive x-axis and rounded to the

nearest degree is 342 

m;

71.21m49.11m22.7m00.3

)14.5(



S

Take the east direction to be the x direction and the north direction to be the -

-y direction The x- and y-components of the resultant displacement of the

first three displacements are then

210mcos45 280mcos30 94.0m,

m,10830sin m28045

sin m210m

m94arctan m,144m

0.94m

The third leg must have taken the sailor east a distance

5.80km  3.50kmcos452.00km1.33kmand a distance north

3.5kmsin45 2.47km

The magnitude of the displacement is

km81.2)km47.2()km33.1

and the direction is arctan  12..3347 = 62 north of east, which is  906228east

of north A more precise answer will require retaining extra significant figures in the intermediate calculations

1.71: a)

b) The net east displacement is

2.80kmsin 45 7.40kmcos30 3.30kmcos22 1.37km,

displacement is 2.80kmcos457.40kmsin 303.30kmsin 22.00.48km,and so the distance traveled is 1.37km 2 0.48km2 1.45km

1.72: The eastward displacement of Manhattan from Lincoln is

147kmsin85106kmsin 167166kmsin23534.3km and the northward displacement is

147kmcos85106kmcos167166kmcos235185.7km

(A negative northward displacement is a southward displacement, as indicated in Fig (1.33) Extra figures have been kept in the intermediate calculations.)

b) The direction from Lincoln to Manhattan, relative to the north, is

km7.185

km3

1.73: a) Angle of first line is θtan 1200210102042 Angle of

second line is 423072 Therefore

8772cos250

X

25872

sin 250

Y

for a final point of (87,258)

b) The computer screen now looks something like this:

The length of the bottom line is 21087 2  2002582 136 and its direction is

  25

tan 1 25821020087  below straight left

1.74: a)

b) To use the method of components, let the east direction be the x-direction

and the north direction be the y-direction Then, the explorer’s net

x-displacement is, in units of his step size,

 40 cos45 80 cos6011.7

and the y-displacement is

 40 sin45 80 sin605047.6.The magnitude and direction of the displacement are

,49)6.47()7.11

7.11

6.47

1.75: Let +x be east and +y be north Let A

be the displacement 285 km at 40.0 north

of west and let B

be the unknown displacement

km

380

km

2.183km,

333.3

Then

0km,

115

km2.18340.0

sin km,

3.2180

.40

cos

east km,115 where

y x

y x

y y y x

R R

A A A

A

A R B A

B

A

eastofsouth ,8.28

km3.333km2.183tan

1.76:

N9600

.35sin

N550sin

sin

c)

(

cos

b)

(

sin

(a)

0 par

par perp par

ω ω

ω ω

ω ω

exerts.biceps

theforce theis

B

N;

160

N2.37N,

2.158

Then

N5.132,

0

N7.16943

cosN,

2.15843

sin

,

upward

isandN5.132 where

,

exerts

elbow the

y

x

y x

y y y x

R

R

B B B

B

B R E B

,13

2.1582.37tan

1.78: (a) Take the beginning of the journey as the origin, with north being the

y-direction, east the x-y-direction, and the z-axis vertical The first displacement is then

i j

2 2

sin 32

sin so

,0

062cos48

cos32

cosso ,0

northbeandeast

be

Let

westofsouth 62

isand west ofsouth

32

is

eastofsouth 32

A C

B

A

C B

A C

B

A

y x

A is known so we have two equations in the two unknowns B and C Solving gives

B = 255 m and C = 70 m

1.80: Take your tent's position as the origin The displacement vector for Joe's tent is

21cos23 iˆ 21sin 23ˆj19.33iˆ8.205ˆj The displacement vector for Karl's tent is

32cos37 iˆ 32sin 37jˆ25.56iˆ19.26jˆ The difference between the two

displacements is:

19.3325.56 iˆ 8.20519.25jˆ6.23iˆ27.46ˆj.The magnitude of this vector is the distance between the two tents:

6.23 2  27.462 28.2m



D

1.81: a) With A z B z 0,Eq.(1.22)becomes

coscos

sinsincos

cos

sinsin

cosBcos

AB

θ θ AB

θ θ θ

θ AB

θ B θ A θ θ

A B A B A

B A

B A B

A

B A

B A

y y x x

sin

cossinsin

cos

cossin

coscos

AB

θ θ AB

θ θ θ

θ AB

θ B θ A θ B θ A

B A B

A

C

A B

B A B

A

A A

B A

x y x

210sin m4.270sinm6.3210cosm4.270cosm60.3

A

B A

b) From Eq (1.22), the magnitude of the cross product is

3.60m2.40msin 1405.55m2,and the direction, from the right-hand rule, is out of the page (the

+z-direction) From Eq (1-30), with the z-components of A B

210cosm40.270sinm60.3

210sinm40.270cosm60.3

y

A

1.83: a) Parallelogramarea2areaof triangleABC

θ BA

θ

sin area

mParellogra

sin

A B21heightbase

21areaTriangle





b) 90

1.84: With the +x-axis to the right, +y-axis toward the top of the page, and +z-axis out

i B

A

ˆ00.7ˆ00.2ˆ00.5

ˆˆ

1.87: The best way to show these results is to use the result of part (a) of Problem 1-65,

a restatement of the law of cosines We know that

,cos2

2 2

where  is the angle between A and B

.a) If C2  A2B2,cos 0,and the angle between A

and B

is less than 90 

1.88: a) This is a statement of the law of cosines, and there are many ways to

derive it The most straightforward way, using vector algebra, is to assume the linearity

of the dot product (a point used, but not explicitly mentioned in the text) to show that the square of the magnitude of the sum A B

cos2

22

2 2

2 2

AB B

A

B A

B A B

B A A

B B A B B A A A B A B A

Using components, if the vectors make angles A and B with the x-axis, the components

of the vector sum are A cos  A + B cos  B and A sin  A + B sinB, and the square of the magnitude is

Acosθ ABcosθ B 2 Asin θ ABsinθ B2

cos2

sinsincos

cos2

sincos

sincos

2 2

2 2

2 2

2 2

2 2

AB B

A

θ AB B

A

θ θ θ

θ AB

θ θ

B θ θ

A

B A

B A B

A

B B

A A

where  = A–B is the angle between the vectors

b) A geometric consideration shows that the vectors A B

, and the sum A B

c) Either method of derivation will have the angle  replaced by 180o –, so the cosine will change sign, and the result is A2  B2 2AB cos.d) Similar to what is done in part (b), when the vector difference has the same

magnitude, the angle between the vectors is 60o Algebraically, is obtained from 1 = 2 – 2 cos , so cos  = 21 and  = 60o