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respostas do volume 1 da 8ª edição do Halliday em inglês
Tipologia: Exercícios
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Lawrence Livermore National Laboratory
to accompany
EIGHTH EDITION
David Halliday University of Pittsburgh
Robert Resnick Rensselaer Polytechnic Institute
Jearl Walker Cleveland State University
Copyright 2007 John Wiley & Sons, Inc. All rights reserved.
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ISBN-10 0-470-11404-
Printed in the United States of America.
1 0 9 8 7 6 5 4 3 2 1
Printed and bound by Odyssey Press, Inc.
( ) ( )
The given measurement is 1.0 km (two significant figures), which implies our result
(b) We calculate the number of microns in 1 centimeter. Since 1 cm = 10−^2 m,
( ) ( )
(c) Since 1 yd = (3 ft)(0.3048 m/ft) = 0.9144 m,
1.0 yd = (^) ( 0.91m (^) ) (^) ( 10 6 μm m (^) )= 9.1 × 105 μm.
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(b) and that distance in chains to be
( 4.0 furlongs^ ) ( 201.168 m furlong^ ) 40 chains. 20.117 m chain
d = =
(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 = =
imply that
1 gry = (1/10)(1/12)(72 points) = 0.60 point.
Thus, 1 gry^2 = (0.60 point)^2 = 0.36 point^2 , which means that 0.50 gry = 0.18 point.^2
(a) In units of W, we have
( )
(b) In units of Z, we have
( )
2 2
V r z π =
where z is the ice thickness. Since there are 10^3 m in 1 km and 10^2 cm in 1 m, we have
( )
3 2 2000 km 10 m^ 10 cm 2000 105 cm. 1km 1m
r
In these units, the thickness becomes
( )
2 3000 m 3000 m 10 cm 3000 102 cm 1m
z
which yields (^) ( ) ( ) 5 2 2 22 3 2000 10 cm 3000 10 cm 1.9 10 cm. 2
π = × × = ×
1 acre ft = (43,560 ft ) ft = 43,560 ft⋅ 2 ⋅^3
Since 2 in. = (1/6) ft, the volume of water that fell during the storm is
V = (26 km )(1/6 ft)^2 = (26 km )(3281ft/km) (1/6 ft)^2 2 = 4.66 × 10 7 ft.^3
Thus,
V =
7 4
ft ft acre ft
acre ft.
3 3
3 7 10 14 86400
m m m day s day
m s
b gc h b gb g
μ = μ
(a) (^) ( 6 ) 100 y 365 day 24 h 60 min 1 century 10 century 52.6 min. 1 century 1 y 1 day 1 h
μ −^
(b) The percent difference is therefore
52.6 min 50 min 4.9%. 52.6 min
(b) In a regular day, there are 86400 seconds, but in the French system described in the problem, there would be 10^5 seconds. The ratio is therefore 0.864.
2 594 33 662 ,. C (^) 7 B (^) 7 B (^) 40 A 5 t = t + t = t −
These are used in obtaining the following results.
(a) We find
( )
495 s B B (^) 40 A A t ′^ − t = t ′ − t =
when t ' A − tA = 600 s.
(b) We obtain t (^) C ′ − t (^) C = t ′ − B tB = =
b g b^495 g 141 s.
(c) Clock B reads tB = (33/40)(400) − (662/5) ≈ 198 s when clock A reads tA = 400 s.
(d) From tC = 15 = (2/7) tB + (594/7), we get tB ≈ −245 s.
