Instructor Manual Fundamentals of Physics 10th Edition Halliday, Resnick, Walker Instant download and all chapters Instructor Manual Fundamentals of Physics 10th Edition Halliday, Resnick, Walker https://testbankdata.com/download/instructor-manual-fundamentals-physics-10thedition-halliday-resnick-walker/
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(b) and in chains to be d = 4.0 furlongs =(4.0 furlongs)
10
chains
1
furlong
40 chains.
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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. Sun. -Mon.
-16
-15
-17
-3
+5
-10
+5
+6
-58
-58
+67
+67
+67
+67
+2
+20
+10
+10
+67 +70
Wed. -Thurs.
+67 +55
Fri. -Sat. -15 -7 -58
8
8
in -
D E
Tues. -Wed.
8
-16
B C
Mon. -Tues.
in -
A
Thurs . -Fri. -15 in -
CLOCK
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)/2 = 0.01 s longer than the first day. Since the increase occurs uniformly, the cumulative effect T is T = ( average increase in length of a day)( number of days) ='
V 265^]( 2000 y )
= 7305 s
or roughly two hours.
8. When the Sun first disappears while lying down, your line of sight to the top of the Sun is tangent to the Earth’s surface at point A shown in the figure. As you stand, elevating your eyes by a height h, the line of sight to the Sun is tangent to the Earth’s surface at point B.
Let dbe the distance from point B to your eyes. From the Pythagorean theorem, we have d2 + r2 = (r + h ) 2 = r2 + 2 rh + h2
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or d2 = 2rh + h2, where r is the radius of the Earth. Since r » h, the second term can be dropped, leading to d2 « 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 t = 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) Using d = rtan0, we have d2 = r2 tan2 0 = 2rh, or 2h r = ------ ;— tan 0
Using the above value for 0and h = 1.7 m, we have r = 5.2 x 106 m. 9.
(a) We find the volume in cubic centimeters 3
193 gal = (193 gal)
231 in 1
2.54 cm v 1in J
= 7.31 x 105 cm3
gal and subtract this from 1 x 10 cm to obtain 2.69 x 10 5 cm3. The conversion gal ^ in 3 is given in Appendix D (immediately below the table of Volume conversions). 6
3
(b) The volume found in part (a) is converted (by dividing by (100 cm/m) 3) to 0.731 m3, which corresponds to a mass of (1000 kg/m3) (0.731 m2) = 731 kg
using the density given in the problem statement. At a rate of 0.0018 kg/min, this can be filled in -------------------- = 4.06 x 10 5 min = 0.77 y 0.0018 kg/min after dividing by the number of minutes in a year (365 days)(24 h/day) (60 min/h). 10. If ME is the mass of Earth, m is the average mass of an atom in Earth, and N is the number of atoms, then ME = Nm or N = ME/m. We convert mass m to kilograms using Appendix D (1 u = 1.661 x 10-27 kg). Thus,
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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 leaf 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 -> V = — = 1.430 cm3. P We convert the volume to SI units: V v
1m
= (1.430 cm ) | 3
100
/
X
3
cm J = 1.430 x 10 6 m3.
Since V = A z with z = 1 x 10 -6 m (metric prefixes can be found in Table 1-2), we obtain 1.430 x 106 m3 A = --- ----------- ------1 x 10 6 m
= 1.430 m2.
(b) The volume of a cylinder of length t is V = At where the cross-section area is that of a circle: A = wr2. Therefore, with r = 2.500 x 10 _6 m and V = 1.430 x 10 _6 m3, we obtain V „ £ = —- =7.284 x 104 m = 72.84 km. nr 12. THINK This problem consists of two parts: in the first part, we are asked to find the mass of water, given its volume and density; the second part deals with the mass flow rate of water, which is expressed as kg/s in SI units. EXPRESS From the definition of density: p = m / V, we see that mass can be calculated as m = pV, the product of the volume of water and its density. With 1 g = 1 x 10 _3 kg and 1 cm3 = (1 x 10_2 m) 3 = 1 x 10_6 m3, the density of water in SI units (kg/m 3) is 1 / 3 | 1 g V10 kg V cm 1 x 10 k ^m p = 1 g/cm = I —%■ --------- s—_ = g/ . 1 cm J^ g J^ 10 m J 3
3
3
/3
To obtain the flow rate, we simply divide the total mass of the water by the time taken to drain it.
ANALYZE (a) Using m = pV, the mass of a cubic meter of water is
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