10.12
(a) As the temperature drops by 20°C, the length of the pendulum changes by
()
i
LLT
α
∆=
∆
Thus, the final length of the rod is
()( )
( )
1
64
19
10
C
1.3 m
20 C
4.9
10
m
0.49 mm
−
−−
⎡⎤
=×
°
−
°=
−×
=
−
⎣⎦
1.3 m
0.49 mm
f
L
=
−
Assignment 10 solutions (Fall term Physics 102)
(b) From the expression for the period,
2
TL
π
=
g
, we see that as the length decreases
the period decreases. Thus, the pendulum will swing too rapidly and the clock will
run
fast .
10.15
From
(
fi
i
LL L
L T
)
, the final value of the linear dimension is
− =
∆
( )
i
LL LT
=+
∆
.
To remove the ring from the rod, the diameter of the ring must be at least as large as the
diameter of the rod. Thus, we require that
( ) ( )
ff
brass
al
LL
=
, or
( ) ( )
brass
brass
brass
T
L
L
T
+
∆= +
∆
This gives
T
αα
−
−
.
(a) If
10
L
=
,
.01 cm
()( ) ()( )
11
6
6
10.01
10.00
199 C
19
10
C
10.00
24
10
C
10.01
T
−
=
− °
×°
−
,
so
20.0 C
199 C=
TT T
°− °
179 C
which is attainable
−
°
.
(b) If
10
L
=
,
.02 cm
6
6
10.02
10.00
396 C
19
10
C
10.00
24
10
C
10.02
T
−
=
−
,
and
=
∆
376 C
which is below absolute zero and unattainable
−°
347
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 Fall '08
 VANDENBERG
 Physics, Energy, °C, al Lal, Lal − Lbrass

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