Download - SOLUCIONARIO EJERCICIOS ZEMANSKY 12
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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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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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