Solutions
Assignment
1,
Chapter
2:
4,
6, 11,
20, 23, 24,
27
3.
Using
the
vector
triangle
shown,
the
speed
of
the
light
coming
toward
the
mirror
is
Qc
2
—v
and
the
same
onthe
oc
return
trip.
Therefore
the
total
time
is
distance
2(2
=
speed
=
t/c2
—
Notice
that
smO=
soO=
siir
1
(!‘\
c
c)
032km/hl_
199
and
4.
As
in
Problem
3,
sinS
=
vj/v
2
,
so
S
=
sin(vi/v
2
)
=
sin
[
Ij
=
—
=
J(0.94
km/h)
2
—
(0.32
km/h)
2
=
0.88
km/h.
Ad
*
6.
Let
ii
=
the
number
of
fringes
shifted,
then
i.=
—i.
Smce
Ad
=
c(At’
—
At),
we
have
(At’
At)
—
2
(i
+
£2)
1?
A
—
c
2
A
Solving
for
v
and
noting
that
(i
+
2
=
22
m,
=
/__nA
=
(3.00
x
108
m/s)
J
(0.005)
(589
10
m)
=
3.47
km/s
22
in
V
i+C2
*1Li,3flienv
cwefindl
6
2
—1,so
i—3d
=
x
—
3d
=
x
—
vt
—
=
t
—
Ox/c
t
—
Ox/c
t
_32
x’
+
3d’
_sx’+/Id’=x’+vt’
—
52
=
I’
+
Ox’/c
t’
+
flx’/c
—
52
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12.
a)
Conversion
100
km/h
=
27.77
rn/s
so
6
=
v/c
=
(27
77
m/s)
(3
00
108
mfs)
=
9.3
x
10
b)
fi
=
v/c
=
(290
m/s)/
(3.00
x
108
m/s)
=
9.7
x
1O
c)u=2.3vj=(2.3330m/s)
and
6
=
v/c
=
(2.3
.330
m/s)
/
(3.00 x
108
rn/a)
2.5 x
10
d)
Conversion
27,000
km/h
=
7500
in/s
so
=
t,/c
=
(7500m/s)/
(3.00
x
10
8
m/s)
=
2.5
x
10
e)
(25cm)
/
(2ns)
=
1.25
x
108
rn/s
so
fl
u/c
=
(1.28 x
108
m/s)/
(3.00
x
io
m/s)
=
042
f)
(1
x
1O
14
m
)
/
(0.3.5
x
1022s)
=
2.857
x
108
rn/s
6
=
v
c
=
(2.857
x
108
m/s)
(3.00
x
108
m/s)
=
0.95
20.
The
roundtrip
distance
is
d
=
40
ly.
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 Spring '08
 Adair
 Physics, Light, m/s, Frame of reference, Miles per hour

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