fisica1-youn-e-freedman-exercicios-resolvidos

1.59: a) To three significant figures, the time for one cycle is
1.420×
10
1 −10
9
= 7.04×
10
Hz
s.
b)
⎛
⎜1.420×
10
⎝
9
cycles⎞
s
⎟×
⎠
⎛ 3600 s ⎞
⎜ ⎟ = 5.11×
10
⎝ 1h ⎠
12
cycles
h
c) Using the conversion from years to seconds given in Appendix F,
7
9
⎛ 3.156×
10 s ⎞
9
26
( 1.42×
10 Hz) ×
⎜
( 4.600×
10 y) = 2.06×
10 .
1 y
⎟ ×
⎝
⎠
9
4
5
4
d) 4.600×
10 y = ( 4.60×
10 )( 1.00×
10 y ),
so the clock would be off by 4.60×
10 s.
1.60: Assume a 70-kg person, and the human body is mostly water. Use Appendix D to
find the mass of one H 2 O molecule: 18.015 u × 1.661 × 10 –27 kg/u = 2.992 × 10 –26
kg/molecule. (70 kg/2.992 × 10 –26 kg/molecule) = 2.34 × 10 27 molecules. (Assuming
carbon to be the most common atom gives 3 × 10 27 molecules.
1.61: a) Estimate the volume as that of a sphere of diameter 10 cm:
4 3
− 4
3
V = π r = 5.2 × 10 m
3
3
Mass is density times volume, and the density of water is 1000 kg m , so
m =
3
−4
3
( 0.98)( 1000 kg m )( 5.2×
10 m ) = 0.5 kg
b) Approximate as a sphere of radius r = 0 .25µ
m (probably an over estimate)
4 3
− 20
V = π r
3
m = 0.98
= 6.5×
10
m
3
3
−20
3
−17
-14
( )( 1000 kg m )( 6.5×
10 m ) = 6×
10 kg = 6 × 10 g
c) Estimate the volume as that of a cylinder of length 1 cm and
radius 3 mm:
2
−7
3
V = πr l = 2.8 × 10 m
m =
3
−7
3
−4
( 0.98)( 1000 kg m )( 2.8×
10 m ) = 3×
10 kg = 0.3g