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
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentCHAPTER 11
10.55
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 VANDENBERG
 Physics, Energy, °C, al Lal, Lal − Lbrass

Click to edit the document details