solucionario ejercicios zemansky 12

8/13/2019 SOLUCIONARIO EJERCICIOS ZEMANSKY 12

1/9

1.30:

eastofnorth38km,8.7

1.32:

a) 11.1 m @ 6.77 b) 28.5 m @ 202 c) 11.1 m @ 258 d) 28.5 m @ 22


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1.33:

west.ofsouth41m,144

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


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4/9

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

c) Thex-andy-components of the vector difference are26.4 m and for a

magnitude of m28.5 and a direction arctan .2024.268.10

Note that 180 must be added to

22arctanarctan4.268.10

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 given

vectors is:

a) 22 )cm20.5()cm6.8( = 10.0 cm, arctan 60.820.5

= 148.8o(which is 180

o

31.2

o

).

b) 22 )m45.2()m7.9( = 10.0 m, arctan 7.945.2

= 14

o+ 180

o= 194

o.

c) 22 )km70.2()km75.7( = 8.21 km, arctan 75.77.2 = 340.8

o(which is

360o19.2

o).

1.41:

The total northward displacement is km,75.1km50.1km3.25 , and the total

westward 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.

m,8.10


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b) Using Equations (1-8) and (1-9),22 )cm50.1()cm04.5( = 5.60 cm, arctan 40.5

50.1 = 344.5

occw.

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 80.2

00.6 = 295o(which is 360

o65

o).

1.43: a) The magnitude of BA

is

cm48.2

60.0sincm90.10.60sincm2.80

60.0coscm90.160.0coscm80.2

2

2

and the angle is

180.60coscm90.160.0coscm2.80

0.60sincm90.160.0sincm2.80arctan

b) The magnitudeof BA

is

cm10.460.0sincm90.10.60sincm2.80

60.0coscm90.160.0coscm80.2

2

2

and the angle is

840.60coscm90.160.0coscm2.80

0.60sincm90.160.0sincm2.80arctan

c) .26418084isangletheandcm4.10ismagnitudethe;

BAAB

1.44:

A = (12.0 m) i . More precisely,

.180sinm012180cosm012 jiA

..

jijiB m8.10m4.1437sinm01837cosm018

..


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1.45: jijiA m6.9m2.737.0cosm0.1237.0sinm0.12

jijiC

jijiB

m2.5m0.360.0sinm0.660.0cosm0.6

m6.9m5.1140.0sinm0.1540.0cosm0.15

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

jiji

94.14m01.12

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) 39.500.200.5,00.500.300.4 2222 BA

b) jijiBA 00.500.100.200.500.300.4

c)

3.1011.00-

5.00

arctan,10.500.51.00

22

d)

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

b) 222 zyx 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.


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c)

125

10.40.3

1

2

2222

a

aa

A

20.00.5

1a

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 the negative of

the other.

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


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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 angleas

.arccosarccos

AB

BABA

AB

yyxxBA

In the intermediate calculations given here, the significant figures in the dot

products and in the magnitudes of the vectors are suppressed.

a) ,13,40,22 BABA

and so

1651340

22arccos

.

b) ,136,34,60 BABA

2813634

60arccos

.

c) .90,0 BA


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solucionario ejercicios zemansky 12

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