Chapter 1 – Student Solutions Manual
3. Using the given conversion factors, we find
(a) the distance
d in rods
to be
(
)
(
)
4.0 furlongs
201.168 m furlong
4.0 furlongs =
160 rods,
5.0292 m rod
d
=
=
(b) and that distance
in chains
to be
(
)
(
)
4.0 furlongs
201.168 m furlong
40 chains.
20.117 m chain
d
=
=
5. Various geometric formulas are given in Appendix E.
(a) Expressing the radius of the Earth as
(
)
(
)
6
3
3
6.37
10 m
10
km m
6.37
10
km,
R
−
=
×
=
×
its circumference is
3
4
2
2
(6.37
10
km)
4.00
10
km.
s
R
π
π
=
=
×
=
×
(b) The surface area of Earth is
(
)
2
2
3
4
4
6.37
10
km
5.10
10
km .
A
R
=
π
=
π
×
=
×
8
2
(c) The volume of Earth is
(
)
3
3
3
4
4
6.37
10
km
1.08
10
km .
3
3
V
R
12
3
π
π
=
=
×
=
×
17. None of the clocks advance by exactly 24 h in a 24h period but this is not the most
important criterion for judging their quality for measuring time intervals. What is
important is that the clock advance by the same amount in each 24h period. The clock
reading can then easily be adjusted to give the correct interval. If the clock reading jumps
around from one 24h period to another, it cannot be corrected since it would impossible
to tell what the correction should be. The following gives the corrections (in seconds) that
must be applied to the reading on each clock for each 24h period. The entries were
determined by subtracting the clock reading at the end of the interval from the clock
reading at the beginning.
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 Spring '08
 drty
 mechanics, Orders of magnitude, Litre, Imperial units, Metre, 24h period

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