Environmental Engineering Reference
In-Depth Information
Appendix B
Solutions to Problems
PROBLEMS 1
=
−
=−
+
−
1.1
sfasfd
(a)
c
4
i
k
;
d
2
i
2
j
3
k
.
√
1
2
√
6;
√
11;
√
17.
|
| =
+
1
2
+
2
2
=
|
| =
|
| =
(b)
a
b
c
3
k
.
a
1
1
(c)
a
=
|
a
|
=
√
6
i
+
√
6
j
−
1.2 Choose
x
−
axis East and
y
−
axis North.
100
(
sin 45
◦
,
cos 45
◦
)
m
For the first leg:
l
1
=
(
70
.
7
,
70
.
7
)
m and similar
calculations for the second and third legs
l
2
and
l
3
.
The sum of all three legs is
l
≈
=
l
1
+
l
2
+
l
3
≈
(
95
.
7
,
−
59
.
2
)
m. Then you
122
◦
.
should obtain
|
l
| =
112 m and the direction to North
θ
=
1.3 Construct displacement vectors
r
AB
=
r
B
−
r
A
etc. The sum of these gives
the null vector.
1.4 Consider an observer in galaxy G. Using Hubble's Law for galaxy G and
galaxy
i
and subtracting we get
v
i
−
v
G
=
H
0
(
r
i
−
r
G
).
This is the velocity of galaxy
i
as seen from galaxy G. Since the right-
hand-side is just the displacement of
i
from G we see that the observer in G
also “discovers” Hubble's Law.
√
3;
2
√
3;
| =
√
26;
| =
√
3.
1.5
|
A
| =
|
B
| =
|
C
|
D
A
·
B
=−
2;
A
·
C
=
0;
A
·
D
=
1;
B
·
D
=−
6.
B
|
A
||
B
|
=−
A
·
1
3
π
2
cos
θ
AB
=
so
θ
AB
=
1
.
91 rad;
θ
AC
=
rad;
θ
AD
=
1
.
23 rad;
θ
BD
=
π
rad;
A
×
B
=
4
j
−
4
k
;
A
×
D
=−
2
j
+
2
k
;
B
×
D
=
0
.
10
13
ms
−
2
.
1.6 2
×
