1-1 ???????1-2 ??

1
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• 1-1 ??????? 1-2 ??
• 1-3 ?????? 1-4 ????
• 1-5 ?? 1-6 ??
• 1-7??

2
1-1 ???????

• ?????? ??????Physics?
• ????????,???????? ?-???
• ?????????
• ???????????

3
1-2 ? ?
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• ???????

??????,??????
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1.????? 2.?????
4
1-3 ??????
????(SI??) 1971??14???????????,?????? ?
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• ????????

??? ???? ??
? ? ?? m
? ? ?? kg
? ? ? s
? ? ?? A
? ? ?? K
??? ?? mol
???? ?? cd
?????????????????
5
?1.2 SI ?????
?? ?? ?? ?? ?? ??
1024 yotta- Y 10-1 deci- d
1021 zetta- Z 10-2 centi- c
1018 exa- E 10-3 milli- m
1015 peta- P 10-6 micro- m
1012 tera- T 10-9 nano- n
109 giga- G 10-12 pico- p
106 maga- M 10-15 femto- f
103 kilo- k 10-18 atto- a
102 hecto- h 10-21 zepto- z
101 deka- da 10-24 yocto- y
6
1-4 ????
????????????????,???????????
Exp?2??????????
7
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1 ? ? ????
8
1-5 ? ?

• 1792????????????????
• 1959?????0.9144?
• 1960????85??????????? ??1,650,763.73 ?
• 1983??????????299,992,458? ????????

9
???????????

• ???????????(????)????????(????)?
• ??????50nm?

10
1-5 ? ?

• 1.??????????315576000???
• 2.????133????????????????? 9,192,631,770?

11
?????????
12
????????????
????38min???????,????????,????h?????

• ?(?/360o) (t/24h)
• t 38min??,??9.5o
• ????
• cos? r/(rh)
• ? r 6.37106m,? 9.5o
• ?????
• h 8.86104m89km

13
1-7 ? ?
1. ???(kg)???????????????????
2.???????(u)??12?????12???

• 1 u 1.6605402 10-27 kg

14
?7, 16, 23
15
7. The volume of ice is given by the product of
the semicircular surface area and the thickness.
The are of the semicircle is A pr2/2, where r
is the radius. Therefore, the volume is
where z is the ice thickness. Since there are 103
m in 1 km and 102 cm in 1 m, we have
In these units, the thickness becomes
which yields,
16
16. We denote the pulsar rotation rate f (for
frequency).
(a) Multiplying f by the time-interval t 7.00
days (which is equivalent to 604800 s, if we
ignore significant figure considerations for a
moment), we obtain the number of rotations
which should now be rounded to 3.88108 rotations
since the time-interval was specified in the
problem to three significant figures.
17
(b) We note that the problem specifies the exact
number of pulsar revolutions (one million). In
this case, our unknown is t, and an equation
similar to the one we set up in part (a) takes
the form N ft, or
which yields the result t 1557.80644887275 s
(though students who do this calculation on their
calculator might not obtain those last several
digits).
(c) Careful reading of the problem shows that the
time-uncertainty per revolution is . We therefore
expect that as a result of one million
revolutions, the uncertainty should be .
18
23. We introduce the notion of density, rm/V ,
and convert to SI units 1000 g 1 kg, and 100
cm 1 m. (a) The density r of a sample of iron
is therefore
which yields r 7870 kg/m3. If we ignore the
empty spaces between the close-packed spheres,
then the density of an individual iron atom will
be the same as the density of any iron sample.
That is, if M is the mass and V is the volume of
an atom, then
19
(b) We set V 4pR3/3, where R is the radius of
an atom . Solving for R, we find
The center-to-center distance between atoms is
twice the radius, or 2.82 10-10 m.