Solucion Fisica Un_ 01

7/27/2019 Solucion Fisica Un_ 01

1/38

1.1: ( ) ( ) ( ) km61.1cm10km1.incm54.2ftin.12mift5280mi1 5 = Although rounded to three figures, this conversion is exact because the given conversionfrom inches to centimeters defines the inch.

1.2: .in9.28cm54.2

in1

L1

cm1000L473.0

3

33

=

1.3: The time required for light to travel any distance in a vacuum is the distancedivided by the speed of light;

ns.103.33s1033.3sm103.00

m10 368

3

==

1.4: .m

kg1013.1

m1

cm100

g1000

kg1

cm

g3.11

3

4

3

3

=

1.5: ( ) ( ) ( ) L.36.5cm1000L1incm54.2in327 333 =

1.6: ..oz16

bottle1

gal1

.oz128

L788.3

gal1

m1

L1000m1

3

3

bottles2112bottles9.2111 =

The daily consumption must then be

.da

bottles78.5

da24.365

yr1

yr

bottles1011.2 3 =

1.7: ( ) ( ) .hrkm2330mikm61.1hrmi1450 = ( ) ( ) s.m648s3600hr1kmm10hrkm2330 3 =

1.8: .h

mi67

h24

day1

day14

fortnight1

furlongs8

mile1

fortnight

furlongs000,180 =

1.9: .gal

mi3.35

gal1

L788.3

km1.609

mi1

L

km0.15 =


2/38

1.10: a)s

ft88

mi1

ft5280

s3600

h1

hr

mi60 =

b)22

s

m8.9

cm100

m1

ft1

cm48.30

s

ft32 =

c)3

3

3

3 m

kg10

g1000

kg1

m1

cm100

cm

g1.0 =

1.11: The density is mass per unit volume, so the volume is mass divided by density.333 cm3077cmg5.19g1060 ==V

Use the formula for the volume of a sphere, ,3

4 3rV =

to calculate ( ) cm0.943: 3/1 == Vrr

1.12: %58.0100)s10(3.16s)10s1016.3( 777 =

1.13:a) %.101.1m10890

m10 33

=

b) Since the distance was given as 890 km, the total distance should be 890,000meters.

To report the total distance as 890,010 meters, the distance should be given as890.01 km.

1.14: a) ( ) ( ) 2mm72mm98.5mm12 = (two significant figures).b)

mm12mm98.5 = 0.50 (also two significant figures).

c) 36 mm (to the nearest millimeter).d) 6 mm.e) 2.0.


3/38

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

%.13.0 b) If the chemical balance can measure to the nearest milligram, the error will be

about %.103.8 3 c) If a handheld stopwatch (as opposed to electric timing devices) canmeasure to the nearest tenth of a second, the error will be about %.108.2 2

1.16: The area is 9.69 0.07 cm2, where the extreme values in the pieces length andwidth are used to find the uncertainty in the area. The fractional uncertainty in the

area is 22

cm69.9

cm07.0 = 0.72%, and the fractional uncertainties in the length and width are

cm5.10cm01.0 = 0.20% and

cm1.9cm0.01 = 0.53%.

1.17: a) The average volume is

( )( ) 3

2

cm8.2cm050.04

cm50.8=

(two significant figures) and the uncertainty in the volume, found from the extreme

values of the diameter and thickness, is about 3cm3.0 , and so the volume of a

cookie is .cm3.08.2 3 (This method does not use the usual form for progation of errors,which is not addressed in the text. The fractional uncertainty in the thickness is so muchgreater than the fractional uncertainty in the diameter that the fractional uncertainty in thevolume is %10 , reflected in the above answer.)

b) .2017005.50.8 =

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 orderof 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 8000kernels.


4/38

1.21: Assuming the two-volume edition, there are approximately a thousand pages, andeach 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 610 .

1.22: Assuming about 10 breaths per minutes, 6024 minutes per day, 365 days peryear, and a lifespan of fourscore (80) years, the total volume of air breathed in a lifetime

is about 35 m102 . This is the volume of a room m20m100m100 , which is kind oftight for a major-league baseball game, but its the same order of magnitude as thevolume of the Astrodome.

1.23: This will vary from person to person, but should be of the order of 5101 .

1.24: With a pulse rate of a bit more than one beat per second, a heart will beat 105times per day. With 365 days in a year and the above lifespan of 80 years, the number of

beats in a lifetime is about 9103 . With20

1 L (50 cm3) per beat, and about4

1 gallon per

liter, this comes to about 7104 gallons.

1.25: The shape of the pile is not given, but gold coins stacked in a pile might well be inthe shape of a pyramid, say with a height of m2 and a base m3m3 . The volume ofsuch a pile is 3m6 , and the calculations of Example 1-4 indicate that the value of this

volume is.106$

8

1.26: The surface area of the earth is about 2142 m1054 =R , where R is the radius of

the earth, about m106 6 , so the surface area of all the oceans is about 214 m104 . An

average depth of about 10 km gives a volume of 324318 cm104m104 = . Characterizing

the size of a drop is a personal matter, but 25 3cmdrops is reasonable, giving a total o2610 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 studentsaveraging 40 pizzas a year each.


5/38

1.28: The moon is about mm104m104 118 = away. Depending on age, dollarbills can be stacked with about 2-3 per millimeter, so the number of bills in a stackto the moon would be about 1012. The value of these bills would be $1 trillion (1terabuck).

1.29: ( ) ( ) .billsofnumberbillAreaUSAofArea = ( ) ( )

.inhabitantmillion$3.6sinhabitant105.2bills109

bills109cm10m1cm6.7cm6.15kmm10km571,372,9

814

142422262

=

=

1.30:

1.31:

eastofnorth38km,8.7


6/38

1.32:

a) 11.1 m @ o6.77 b) 28.5 m @

o

202 c) 11.1 m @ o258 d) 28.5 m @ o22

1.33:

west.ofsouth41m,144


7/38

1.34:

1.35: ( ) ( ) m.6.937.0cosm0.12m,2.737.0sinm0.12; ====

yx AAA ( ) ( )

( ) ( ) m.2.560.0sinm0.6m,0.360.0cosm0.6;

m.6.940.0sinm0.15m,5.1140.0cosm0.15;

====

====

yx

yx

CC

BB

C

B

1.36: 500.0m2.00

m00.1tan(a) =

==

X

y

A

A

( )

( )

( )

( )

2076.26180500.0tan

500.0m2.00

m00.1tan(d)

1536.26180500.0tan

500.0m2.00

m00.1tan)(

6.26500.0tan

500.0m2.00

m00.1tan(b)

3336.26360500.0tan

1

1

1

1

=+==

=

==

===

=

==

==

===

===

A

A

A

Ac

A

A

x

y

x

y

x

y


8/38

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

second force has components N4334.32cos22 ==FFx and N.2754.32sin22 ==

FFy

The first force has components .0andN725 11 == yx FF

N115821 =+= xxx FFF and N27521 =+= yyy FFF

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 thenet eastward displacement is (4.0 km) + (3.1 km) cos 45o = 6.2 km. The

magnitude of the resultant displacement is 22 )km2.6()km8.4( + = 7.8 km, and

the direction is arctan ( )2.68.4 = 38o north of east.


9/38

1.39: Using components as a check for any graphical method, the components ofB arem4.14=

xB and m,8.10=yB A

has one component, m12=xA .

a) The -x and -y components of the sum are 2.4 m and 10.8 m, for a magnitude

of ( ) ( ) m,1.11m8.10m2.4 22 =+ , and an angle of .6.772.4

10.8=

b) The magnitude and direction ofA + B are the same as B + A.c) Thex- andy-components of the vector difference are 26.4 m and

m,8.10 for a magnitude of m28.5 and a direction arctan ( ) .2024.268.10 =

Note that180 must be added to ( ) ( ) 22arctanarctan

4.268.10

4.268.10 ==

in order to give an angle in the

third quadrant.

d) .m8.10m4.26m0.12m8.10m4.14 jiijiAB +=++=

( ) ( ) .2.2226.4

10.8arctanofangleandatm5.28m8.10m26.4Magnitude

22 =

=+=

1.40: Using Equations (1.8) and (1.9), the magnitude and direction of each of the givenvectors is:

a) 22 )cm20.5()cm6.8( + = 10.0 cm, arctan ( )60.8

20.5 = 148.8

o (which is

180o 31.2o).

b) 22 )m45.2()m7.9( + = 10.0 m, arctan ( )7.9

45.2 = 14o + 180o = 194o.

c) 22 )km70.2()km75.7( + = 8.21 km, arctan ( )75.7

7.2 = 340.8o (which is

360o 19.2o).


10/38

1.41:

The total northward displacement is km,75.1km50.1km3.25 = , and the totalwestward displacement is km4.75 . The magnitude of the net displacement is

( ) ( ) km.06.5km75.4km1.75 22 =+ The south and west displacements are the same, so

The direction of the net displacement is

69.80 West of North.

1.42: a) Thex- andy-components of the sum are 1.30 cm + 4.10 cm = 5.40 cm,2.25 cm + (3.75 cm) = 1.50 cm.

b) Using Equations (1-8) and (1-9),22 )cm50.1()cm04.5( = 5.60 cm, arctan ( )

40.550.1

+ = 344.5o ccw.

c) Similarly, 4.10 cm (1.30 cm) = 2.80 cm, 3.75 cm (2.25 cm) = 6.00 cm.d) 22 )cm0.6()cm80.2( + = 6.62 cm, arctan ( )

80.200.6 = 295o (which is 360o 65


11/38


12/38

1.46: a) ( ) ( ) ( ) ( )jijiA m38.3m23.10.70sinm60.30.70cosm60.3 +=+=

( ) ( ) ( ) ( )jijiB m20.1m08.230.0sinm40.230.0cosm40.2 +==

b)( ) ( )BAC

00.400.3 =

( )( ) ( )( ) ( )( ) ( )( )( ) ( )ji

jiji94.14m01.12

m20.100.4m08.200.4m38.300.3m23.100.3

+=+=

(Note that in adding components, the fourth figure becomes significant.)c) From Equations (1.8) and (1.9),

( ) ( ) 2.51m12.01

m14.94arctanm,17.19m94.14m01.12

22 =

=+=C

1.47: a) ( ) ( ) ( ) ( ) 39.500.200.5,00.500.300.42222

=+==+= BA b) ( ) ( )( ) ( ) ( )jijiBA 00.500.100.200.500.300.4 +=+=

c) ( ) ( ) 3.1011.00-

5.00arctan,10.500.51.00

22 =

=+

d)


13/38

1.48: a) 13111 222 =++=++ kji so it is not a unit vector

b) 222zyx AAA ++=A

If any component is greater than + 1 or less than 1, 1A , so it cannot be a unit

vector. A

can have negative components since the minus sign goes away when thecomponent is squared.

c)

( ) ( )

125

10.40.3

1

2

2222

=

=+

=

a

aa

A

20.00.5

1==a

1.49: a) Let , jiA yx AA +=

. jiB yx BB +=

( ) ( )

( ) ( )jiAB

jiBA

yyxx

yyxx

ABAB

BABA

+++=+

+++=+

Scalar addition is commutative, so ABBA

+=+

yyxx

yyxx

ABAB

BABA

+=

+=

AB

BA

Scalar multiplication is commutative, so ABBA

=

b) ( ) ( ) ( )kjiBA xyyxzxxzyzzy BABABABABABA ++=

( ) ( ) ( ) kjiAB xyyxzxxzyzzy ABABABABABAB ++=

Comparison of each component in each vector product shows that one is thenegative of the other.


14/38

1.50: Method 1: ( )cosmagnitudesofoductPr ( )

( )

( ) 2

2

2

m5.71187cosm6m12cosAC

m6.1580cosm6m15cosBC

m4.993cosm15m12cosAB

==

==

==

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

=+=

=+=

=+=

CA

CB

BA

1.51: a) From Eq.(1.21),

( )( ) ( )( ) .00.1400.200.300.500.4 =+=BA

b) ( ) ( )[ ] ( ) .7.58.5195arccos39.500.500.14arccosso,cosAB ==== BA

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

.arccosarccos

+=

=

AB

BABA

AB

yyxxBA

In the intermediate calculations given here, the significant figures in the dotproducts and in the magnitudes of the vectors are suppressed.

a) ,13,40,22 === BABA

and so

1651340

22arccos =

= .

b) ,136,34,60 === BABA

28

13634

60arccos =

= .

c) .90,0 == BA


15/38

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

( )( ) ( ) 2m13037180sinm0.18m0.12 =

The right-hand rule gives the direction as being into the page, or the z-direction. UsingEq. (1.27), the only non-vanishing component of the cross product is

( ) ( )( ) 2m13037sinm0.18m12 === yxz BAC

b) The same method used in part (a) can be used, but the relation given in Eq.(1.23) gives the result directly: same magnitude (130 m2), but the oppositedirection (+z-direction).

1.55: In Eq. (1.27), the only non-vanishing component of the cross product is

( )( ) ( )( ) ,00.2300.500.300.200.4 === xyyxz BABAC

so ( ) ,00.23 kBA =

and the magnitude of the vector product is 23.00.


16/38

1.56: a) From the right-hand rule, the direction of BA

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

( )( ) .cm61.4120sincm90.1cm80.2sin 2== AB Or, using Eq. (1.27) and noting that the only non-vanishing component is

( ) ( )

( ) ( )2cm61.4

0.60coscm90.10.60sincm80.2

60sincm90.10.60coscm80.2

=

=

=

xyyxz BABAC

gives the same result.

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

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

1.57: a) The area of one acre is ,mimimi 26401

801

81 = so there are 640 acres to a square

mile.

b) ( ) 222

ft560,43mi1

ft5280

acre640

mi1acre1 =

(all of the above conversions are exact).

c) (1 acre-foot) ( ) gal,1026.3ft1

gal477.7ft560,43 5

3

3 =

=

which is rounded to three significant figures.

1.58: a) .m12$mft7710)ft43560acre1(acres)102000,950,4($ 2222 = .

b) .in008$.)cm100m1(in)cm54.2()m12$( 2222 =

c) parcel.sizedstamppostagefor007$.)in87in1(in008$. 2 =


17/38

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

s.1004.7Hz10420.1

1 109

=

b)h

cycles1011.5h1

s3600s

cycles101.420 129 =

c) Using the conversion from years to seconds given in Appendix F,

( ) ( ) .1006.2y10600.4y1

s10156.3Hz101.42 269

79 =

d) ( )( ) s.104.60byoffbedclock woultheso,y1000.1104.60y104.600 4549 =

1.60: Assume a 70-kg person, and the human body is mostly water. Use Appendix D tofind the mass of one H2O molecule: 18.015 u 1.661 10

27 kg/u = 2.992 1026kg/molecule. (70 kg/2.992 1026 kg/molecule) = 2.34 1027 molecules. (Assumingcarbon 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:

343 m102.53

4 == rV

Mass is density times volume, and the density of water is 3mkg1000 , so

( )( )( ) kg5.0m102.5mkg100098.0343

==

m

b) Approximate as a sphere of radius m25.0 r= (probably an over estimate)

3203 m105.63

4 == rV

( )( )( ) g106kg106m105.6mkg100098.0 -14173203 === m

c) Estimate the volume as that of a cylinder of length 1 cm andradius 3 mm:

( )( )( ) g3.0kg103m108.2mkg100098.0m108.2

4373

372

===

==

m

lrV


18/38

1.62: a)

MV

V

M == so,

cm2.94m1094.2

m1054.2kg/m1086.7

kg200.0

2

35

33

3

==

=

=

x

x

b)

cm1.82m1082.1

m1054.23

4

2

353

==

=

R

R

1.63: Assume each person sees the dentist twice a year for checkups, for 2 hours. Assume2 more hours for restorative work. Assuming most dentists work less than 2000 hours per

year, this gives dentist.perpatients500patientperhours4hours2000 = Assuming onlyhalf 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 roomcould increase the number of patients seen in a single dental office.

1.64: a) atoms.106.21014

100.6)kg100.6( 50

mole

kg3

moleatoms23

24 =

b) The number of neutrons is the mass of the neutron star divided by themass of a neutron:

neutrons.104.2)neutronkg107.1(

)kg100.2()2( 5727

30

=

c) The average mass of a particle is essentially32 the mass of either the proton or

the neutron, 27107.1 kg. The total number of particles is the total mass divided by thisaverage, and the total mass is the volume times the average density. Denoting the density

by (the notation introduced in Chapter 14).

.102.1

)kg107.1(

)mkg10()m105.1()2(

3

23

4

79

27

318311

p

3

ave

=

=

=

m

R

m

M

Note the conversion from g/cm3 to kg/m3.


19/38

1.65: Let D

be the fourth force.

( )

N90.310.53sinN,07.240.53cos

N28.690.30cosN,00.400.30sin

N00.500.30cosN,6.860.30cos

so,0

===+=+=+===

+=+=+=+=

++==+++

CCCC

BBBB

AAAA

yx

yx

yx

CBADDCBA

Then N34.87N,53.22 == yx DD

N;2.9022 =+= yx DDD

53.22/34.87/tan ==xy

DD

54.75= axis-fromckwisecounterclo,256180 x +=+=

1.66:

( ) ( )

344.0km311.5

km2.107tan

km330km2.107km5.311

km2.10748sinkm)230(68coskm)170(

km5.31148coskm)230(68sinkm)170(

2222

===

=+=+=

==+=

=+=+=

x

y

R

yx

yyy

xxx

R

R

RRR

BAR

BAR

eastofsouth19=R


20/38

1.67: a)

b) Algebraically, areofcomponentsthesoand, ABCA

=

( ) ( )

( ) ( ) cm.10.863.0sincm40.622.0sincm40.6

cm03.363.0coscm40.622.0coscm40.6

=+==

===

yyy

xxx

BCA

BCA

( ) ( )

5.69cm3.03

cm8.10arctancm,65.8cm10.8cm03.3c)

22

=

=+=A


21/38


22/38

1.69:

Take the east direction to be the -x direction and the north direction to be the-y direction. Thex- andy-components of the resultant displacement of the

first three displacements are then

( ) ( ) ( )

( ) ( ) m,0.9430cosm28045cosm210

m,10830sinm28045sinm210m180

+=+

=++

keeping an extra significant figure. The magnitude and direction of this net displacementare

( ) ( ) .9.40m108

m94arctanm,144m0.94m108

22 =

=+

The fourth displacement must then be 144 m in a direction 9.40 south of west.


23/38

1.70:

The third leg must have taken the sailor east a distance

( ) ( ) ( ) km33.1km00.245coskm50.3km80.5 =

and a distance north

( ) ( )km47.245sinkm5.3 = The magnitude of the displacement is

km81.2)km47.2()km33.1( 22 =+

and the direction is arctan ( )33.147.2 = 62 north of east, which is = 286290 east

of north. A more precise answer will require retaining extra significant figures inthe intermediate calculations.

1.71: a)

b) The net east displacement is( ) ( ) ( ) km,37.122coskm30.330coskm40.745sinkm80.2 =+ and the net northdisplacement is ( ) ( ) ( ) km,48.022.0sinkm30.330sinkm40.745coskm80.2 =+

and so the distance traveled is ( ) ( ) km.45.1km48.0km1.37 22 =+


24/38

1.72: The eastward displacement of Manhattan from Lincoln is

( ) ( ) ( ) km3.34235sinkm166167sinkm10685sinkm147 =++ and the northward displacement is

( ) ( ) ( ) km7.185235coskm166167coskm10685coskm147 =++

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

a) km189)km7.185()km3.34( 22 =+

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

arctan

5.169km7.185

km3.34

=

and so the direction to fly in order to return to Lincoln is 5.3491805.169 ++ .


25/38

1.73: a) Angle of first line is ( ) .42tan10210202001 ==

Angle of

second line is .723042 =+ Therefore8772cos25010 =+= X 25872sin25020 =+= Y

for a final point of (87,258).b) The computer screen now looks something like this:

The length of the bottom line is ( ) ( ) 13625820087210 22 =+ and its direction is

( ) 25tan872102002581 =

below straight left.


26/38

1.74: a)

b) To use the method of components, let the east direction be thex-directionand the north direction be they-direction. Then, the explorers netx-displacement is, in units of his step size,

( ) ( ) 7.1160cos8045cos40 =

and they-displacement is

( ) ( ) .6.475060sin8045sin40 =+

The magnitude and direction of the displacement are

,49)6.47()7.11( 22 =+ arctan 1047.11

6.47 =

.

(More precision in the angle is not warranted, as the given measurements are tothe nearest degree.) To return to the hut, the explorer must take 49 steps in a

direction 1490104 = east of south.


27/38

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

be the displacement 285 km at 0.40 north

of west and let B

be the unknown displacement.

km.380

km.2.183km,333.3Then

0km,115

km2.18340.0sinkm,3.2180.40cos

eastkm,115where,

22

,

=+=

==

==

+=+===

== =

==+

yx

yx

yx

yx

yyyxxx

BBB

BB

RR

AAAA

ARBARB

ARB

RRBA

( ) ( )

eastofsouth,8.28

km3.333km2.183tan

=

==

BB xy

1.76:

N9600.35sin

N550

sin

sinc)(

cosb)(

sin(a)

0

par

par

perp

par

==

=

=

=

=


28/38


29/38

1.78: (a) Take the beginning of the journey as the origin, with north being they-direction, east thex-direction, and thez-axis vertical. The first displacement is then

,30k the second is j15 , the third is m)200km2.0(200 =i , and the fourth is j100 .Adding the four:

kjijijk 30852001002001530 +=++

(b) The total distance traveled is the sum of the distances of the individual segments: 30+ 15 + 200 + 100 = 345 m. The magnitude of the total displacement is:

( ) m2193085200 222222 =++=++= zyx DDDD

1.79: Let the displacement from your camp to the store be .A

062sin48sin32sinso,0

062cos48cos32cosso,0

northbeandeastbeLet

westofsouth62isandwestofsouth32is

eastofsouth32m,240

=+=++

==++

=++

++

=

CBACBA

CBACBA

yx

A

yyy

xxx

0CBA

CB

A is known so we have two equations in the two unknownsB andC. Solving givesB = 255 m and C = 70 m.

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

( ) ( ) .205.833.1923sin2123cos21 jiji = The displacement vector for Karl's tent is( ) ( ) jiji 26.1956.2537sin3237cos32 +=+ . The difference between the twodisplacements is:

( ) ( ) jiji 46.2723.625.19205.856.2533.19 =+ .

The magnitude of this vector is the distance between the two tents:

( ) ( ) m2.2846.2723.6 22 =+=D


30/38

1.81: a) becomes(1.22)Eq.,0With ==zz

BA

( )( ) ( )( )

( )

( )

cos

cos

sinsincoscos

sinsincosBcos

AB

AB

AB

BAABABA

BA

BABA

BABAyyxx

=

=

+=

+=+

where the expression for the cosine of the difference between two angles has been used(see Appendix B).

b) With zzzz CCCBA ==== and,0 kC

. From Eq. (1.27),

( )( ) ( )( )

( )sin

sin

cossinsincos

cossincoscos

AB

AB

AB

BABA

BABAC

AB

BABA

AABA

xyxx

==

=

=

=

: a) The angle between the vectors is ,14070210 = and so Eq. (1.18) gives

( )( ) 2m62.6140cosm40.2m60.3 ==

BA Or, Eq. (1.21) gives

( ) ( ) ( ) ( )2

m62.6

210sinm4.270sinm6.3210cosm4.270cosm60.3

=

+=

+=

yyxx BABABA

b) From Eq. (1.22), the magnitude of the cross product is( )( ) 2m55.5140sinm40.2m60.3 = ,

and the direction, from the right-hand rule, is out of the page (the

+z-direction). From Eq. (1-30), with thez-components of BA

and

vanishing, thez-component of the cross product is

( ) ( )

( ) ( )2m55.5

210cosm40.270sinm60.3

210sinm40.270cosm60.3

=

=

xyyx BABA


31/38

1.83: a) ABCtriangleofarea2arearamParallelog = ( )( ) ( )( )

BA

sinareamParellogra

sinAB21heightbase21areaTriangle

=

==

b) 90

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

of the page, ( ) ( ) ( ) .0,cm9.68,cm8.87 22 ===zyx

BABABA

1.85: a) ( ) ( ) ( ) .39.500.400.300.2 222 =++=A

( ) ( ) ( ) .36.400.300.100.3 222 =++=B

b)( ) ( ) ( )( ) ( ) ( )kji

kjiBA

00.700.200.5

++=

++= zzyyxx BABABA

( ) ( ) ( ) ,83.800.700.25.00c) 222 =++

and this will be the magnitude of AB

as well.

1.86: The direction vectors each have magnitude 3 , and their dot product is (1) (1) +

(1) (1) + (1) (1) = 1, so from Eq. (1-18) the angle between the bonds is arccos

=33

1 = arccos ( ) 10931 = .


32/38

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

,cos2222 ABBAC ++=

where is the angle between A

andB

.

a) If ,0cos,222 =+= BAC and the angle between A

andB

is 90 (the vectors are

perpendicular).

b) If ,0cos,222 +> BAC and the angle between A

andB

is less than 90 .


33/38

1.88: a) This is a statement of the law of cosines, and there are many ways toderive it. The most straightforward way, using vector algebra, is to assume the linearityof the dot product (a point used, but not explicitly mentioned in the text) to show that

the square of the magnitude of the sum BA

+ is

( ) ( )

cos2

2

2

22

22

ABBA

BA

++=

++=

++=+++=++

BA

BABBAA

BBABBAAABABA

Using components, if the vectors make angles A andB with thex-axis, the components

of the vector sum areA cos A +B cos B andA sin A +B sinB, and the square of themagnitude is

( ) ( )22 sinsincoscos BABA BABA +++( ) ( )

( )

( )

cos2

cos2

sinsincoscos2

sincossincos

22

22

222222

ABBA

ABBA

AB

BA

BA

BABA

BBAA

++=

++=

++

+++=

where = A B is the angle between the vectors.

b) A geometric consideration shows that the vectors BA , and the sum A + must be the sides of an equilateral triangle. The angle between BA

and, i

120o, since one vector must shift to add head-to-tail. Using the result of pa

(a), with ,BA = the condition is that cos2 2222 AAAA ++= , which solvfor 1 = 2 + 2 cos , cos = ,

21 and= 120o.

c) Either method of derivation will have the angle replaced by 180o, sothe cosine will change sign, and the result is .cos222 ABBA +

d) Similar to what is done in part (b), when the vectordifference has the samemagnitude, the angle between the vectors is 60o. Algebraically, is obtain

from 1 = 2 2 cos , so cos =21 and= 60o.


34/38

1.89: Take the length of a side of the cube to be L, and denote the vectors from a to b, a

to c anda to das DCB

and,, . In terms of unit vectors,

,kB L=

( ), kjC += L

( ). kjiD ++= L

Using Eq. (1.18),

( ) ( ),7.54

3arccosarccos

2

=

=

LL

L

BD

DB

( )( ).3.35

32

2arccosarccos

2

=

=

LL

L

CD

DC

1.90: From Eq. (1.27), the cross product is

.00.13

00.1100.13

00.6)00.1(13)00.11()00.6()00.13(

+=++ kjikji

The magnitude of the vector in square brackets is ,93.1 and so a unit vector in this

direction (which is necessarily perpendicular to both )andBA

is

+

93.1

)1300.11()00.1300.6()00.1( kji.

The negative of this vector,

+

93.1

)1300.11()00.1300.6()00.1( kji,

is also a unit vector perpendicular to BA

and .


35/38

1.91: A

andC

are perpendicular, so 0,.0 =+= yyxx CACACA

, which gives

.05.65.0 =yx

CC

0.150.75.3so,0.15 =+=yx

CCCB

We have two equations in two unknowns xC and yC . Solving gives

.1.6and0.8 == yx CC

1.92: AB sin=BA

( ) ( )( )( )

5984.000.300.3

00.200.5sin

22

=+

=

=AB

BA

( ) 8.365984.0sin 1 ==

1.93: a) Using Equations (1.21) and (1.27), and recognizing that the vectors

CBA

and,, do not have the same meanings as they do in those equations,

( ) ( ) ( ) ( )( ).

zxyzyxyzxyxzxyzxzy

xyyxzxxzyzzy

CBACBACBACBACBACBA

BABABABABABA

++=

++= CkjiCBA

A similar calculation shows that

( ) xyzyxzzxyxzyyzxzyx CBACBACBACBACBACBA ++= CBA

and a comparison of the expressions shows that they are the same.

b) Although the above expression could be used, the form given allows for ready

compuation of BA

the magnitude is ( ) 37.0sin20.00sin =AB and the directionis, from the right-hand rule, in the +z-direction, and so

( ) ( ) ( ) .2.7200.637.0sin20.00 +=+=

CBA


36/38

1.94: a) The maximum and minimum areas are

(L + l) (W+ w) =LW+ lW+Lw, (L l) (Ww) =LWlWLw,

where the common terms wl have been omitted. The area and its

uncertainty are then WL (lW+Lw), so the uncertainty in the area isa = lW + Lw.

b) The fractional uncertainty in the area is

,W

w

L

l

WL

WllW

A

a+=

+=

the sum of the fractional uncertainties in the length and width.

c) The similar calculation to find the uncertainty v in the volume will involveneglecting the terms lwH, lWh andLwh as well as lwh; the uncertainty in the volume is v

= lWH+LwH+LWh, and the fractional uncertainty in the volume is

,H

h

W

w

L

l

HLW

LWhLwHlWH

V

v++=

++=

the sum of the fractional uncertainties in the length, width and height.

1.95: The receiver's position is

( ) ( ) ( ) ( ) .0.280.160.180.40.110.50.120.60.90.1 jiji +=+++++++

The vector from the quarterback to the receiver is the receiver's position minus the

quarterback's position, or ( ) ( )ji 0.350.16 + , a vector with magnitude

( ) ( ) ,5.380.350.16 22 =+ given as being in yards. The angle is ( ) 6.24arctan0.350.16 = to the

right of downfield.


37/38

1.96: a)

b) i) In AU, 22 )9329.0()3182.0( + = 0.9857.ii) In AU, 222 )0414.()4423.()3087.1( ++ = 1.3820iii) In AU,

222

)0414.0())4423.(9329.0())3087.1(3182.0( ++ = 1.695.

c) The angle between the directions from the Earth to the Sun and toMars is obtained from the dot product. Combining Equations (1-18)and (1.21),

)695.1)(9857.0(

)9329.04423.0)(9329.0()3182.03087.1)(3182.0(arccos

+=

d) Mars could not have been visible at midnight, because the Sun-Mars angle is less

than 90

o

.


38/38

1.97: a)

The law of cosines (see Problem 1.88) gives the distance as

( ) ( ) ( )( ) ly,2.764.154cosly77ly1382ly77ly138 22 =++

where the supplement of 6.25 has been used for the angle between the directionvectors.

b) Although the law of cosines could be used again, it's far more convenient to use thelaw of sines (Appendix B), and the angle is given by

,1295.51180,5.51ly138ly76.2

25.6sinarcsin

==

where the appropriate angle in the second quadrant is used.

1.98: Define kjiS CBA ++=

CzByAxCBAzyx

++=++++= )()( kjikjiSr

If the points satisfyAx +By + Cz = 0, then 0= Sr

and all points r

are

perpendicular to S

.

Solucion Fisica Un_ 01

Documents

Transcript of Solucion Fisica Un_ 01