Exercises 3.5
(e)
Using the equation
v
e
=
√
2
gR
for the escape velocity and converting meters to kilometers
we have
v
e
=
p
2
=
±
2
·
9
.
81 m
/
sec
2
·
(1 km
/
1000 m)(6370 km)
≈
11
.
18 km
/
sec
.
(f)
Similarly to (e), we Fnd
v
e
=
p
2(
g/
6)
R
=
p
2(9
.
81
/
6)(1
/
1000)(1738) = 2
.
38 (km
/
sec)
.
EXERCISES 3.5:
Electrical Circuits, page 122
1.
In this problem,
R
=5Ω
,
L
=0
.
05 H, and the voltage function is given by
E
(
t
) = 5 cos 120
t
V.
Substituting these data into a general solution (3) to the Kirchho±’s equation (2) yields
I
(
t
)=
e
−
Rt/L
²Z
e
Rt/L
E
(
t
)
L
dt
+
K
³
=
e
−
5
t/
0
.
05
e
5
t/
0
.
05
5 cos 120
t
0
.
05
dt
+
K
³
=
e
−
100
t
²
100
Z
e
100
t
cos 120
tdt
+
K
³
.
Using the integral tables, we evaluate the integral in the righthand side and obtain
I
(
t
e
−
100
t
´
100
e
100
t
(100 cos 120
t
+ 120 sin 120
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Velocity, Cos, Orders of magnitude, Metre, Kilometre, .km

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