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Instructor Manual Fundamentals of Physics 10th Edition Halliday, Res Instant download and all chapters Instructor Manual Fundamentals of Edition Halliday, Resnick, Walker
https://testbankdata.com/download/instructor-m https://testbankdata.com/downl oad/instructor-manual-fund anual-fundamentals-ph amentals-ph edition-halliday-resnick-walker/
Chapter 1
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CHAPTER 1
2
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(b) and in chains to be d = 4.0 furlongs =(4.0 furlongs)
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1 0
chains
1
40 chains.
furlong
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CHAPTER 1
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the smallest range of correction, B has the next smallest range, and E has the greatest range. From best to worst, the ranking of the clocks is C, D, A, B, E. CLOCK
Sun. -Mon.
A
-16
-16
-15
-17
Thurs . -15
B
-3
+5
-10
+5
+6
C
i n -
8
Mon. -Tues.
-
n
i
Tues. -Wed.
8
Wed. -Thurs.
-58
-58 -
Fri. -Sat. -15 -7
i n
8
-58
D
+67
+67
+67
+67
+67
+67
E
+70
+55
+2
+2 0
+1 0
+1 0
LEARN Of the five clocks, the readings in clocks A, B and E jump around from one 24- h period to another, making it difficult to correct them. 7.
The last day of the 20 centuries is longer than the first day by (20 century) ( 0.001 s/century)
= 0.02 s.
The average day during the 20 centuries is (0 + 0.02)/2a =Preview 0.01 s longer than the first day. Since the You're Reading increase occurs uniformly, the cumulative effect T is Unlock full access with a free trial.
T = ( average increase in length of a day)( number of days)
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='
V 265^]( 2000 y )
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8
CHAPTER 1
or d = 2 rh + h , where r is the radius of the Earth. Since r » h, the second term can be
dropped, leading to d 2 « 2rh. Now the angle between the two radii to the two tangent points A and
B is 0, which is also the angle through which the Sun moves about Earth during the time interval = 11.1 s. The value of 0 can be obtained by using 0
_ t 360°
= 24 h . This yields 0
= ---------------------
(360 )(11 1 s)
.
-------------------------------------------- =
0.04625°.
(24 h)(60 min/h)(60 s/min) 2
2
Using d = r tan0, we have d = r tan 2 0 = 2rh, or
2h r = ------ ; — tan 0
Using the above value for 0and h = 1.7 m, we have r = 5.2 x 10 6 m.
You're Reading a Preview 9.
(a) We find the volume in cubic centimeters Unlock full access with a free trial. 193 gal = (193 gal)
2.54 cm
231
Download With Free Trial= 7.31 x 105 cm 3 in 1
v
n
J 3
and subtract this from 1 x 10 6 cm3 to obtain 2.69 x 10 5 cm . The conversion gal ^ in 3 is given in
Master your semester Scribd Appendix D (immediatelywith below the table of Volume conversions). Read Free Foron 30this Days Sign up to vote title & The New Times Useful by Not) useful (b) York The volume found in part (a) is converted (by dividing (100 cm/m) to 0.731 m , which 3
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N = M. = __________ ^ ________________ m(40 u) (1.661 x 10 ~ 27 kg/u)
=
90
x ,0.9.
11. The density of gold is . m 19.32g p = - = - ---------- r = 19.32 g/cm . V 1 cm
3
(a) We take the volume of the l eaf to be its area A multiplied by its thickness z. With density p = 19.32 g/cm3 and mass m = 27.63 g, the volume of the leaf is found to be
m -> 3 V = — = 1.430 cm . P We convert the volume to SI units: 3
v
10 0 cm
/
1m
V = (1.430 cm ) |
X
3
J = 1.430 x 10 6 m .
Since V = A z with z = 1 x 10 -6 m (metric prefixes can be found in Table 1-2), we obtain
You're Reading a Preview 1.430 x 106 m 3 full access with a free trial. m . A = Unlock --- ----------------= 1.430 1 x 10 6 m
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(b) The volume of a cylinder of length t is V = At where the cross-section area is that of a circle: 3
2 = wr . Therefore, with r = 2.500 x 10 _6 m and V = 1.430 x 10 _6 m , we obtain
V Master your semester with Scribd „ Read £ = — - =7.284 x 10 m = 72.84 km. nr up Free Foron 30this Days Sign to vote title & The New York Times Useful Not useful 12. THINK This problem consists of two parts: in the first part, we are asked to find the mass of 4
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