SOLUTION MANUAL FOR c2011 VOLUME 1. 1 Measurement. 2 Motion Along a Straight Line.
4 Motion in Two and Three Dimensions. 5 Force and Motion — I. 6 Force and Motion — II. 7 Kinetic Energy and Work. 8 Potential Energy and Conservation of Energy. 9 Center of Mass and Linear Momentum. 11 Rolling, Torque, and Angular Momentum.
12 Equilibrium and Elasticity. 13 Gravitation. 15 Oscillations. 16 Waves — I.
17 Waves — II. 18 Temperature, Heat, and the First Law of Thermodynamics. 19 The Kinetic Theory of Gases. 20 Entropy and the Second Law of Thermodynamics.
21 Electric Charge. 22 Electric Fields. 23 Gauss’ Law. 24 Electric Potential. 25 Capacitance. 26 Current and Resistance.
28 Magnetic Fields. 29 Magnetic Fields Due to Currents. 30 Induction and Inductance.
31 Electromagnetic Oscillations and Alternating Current. 32 Maxwell’s Equations; Magnetism of Matter. 33 Electromagnetic Waves. 35 Interference. 36 Diffraction.
37 Relativity. 38 Photons and Matter Waves. 39 More About Matter Waves. 40 All About Atoms.
41 Conduction of Electricity in Solids. 42 Nuclear Physics. 43 Energy from the Nucleus. 44 Quarks, Leptons, and the Big Bang. Various geometric formulas are given in Appendix E. (a) Expressing the radius of the Earth as R = ( 6.37 × 106 m )(10−3 km m ) = 6.37 × 103 km, its circumference is s = 2π R = 2π (6.37 × 103 km) = 4.00 × 104 km.
(b) The surface area of Earth is A = 4π R 2 = 4π ( 6.37 × 103 km ) = 5.10 × 108 km 2. 2 (c) The volume of Earth is V = 4 π 3 4π R = 6.37 × 103 km 3 3 ( ) 3 = 1.08 × 1012 km3. The conversion factors are: 1 gry = 1/10 line, 1 line = 1/12 inch and 1 point = 1/72 inch. The factors imply that 1 gry = (1/10)(1/12)(72 points) = 0.60 point.
Thus, 1 gry2 = (0.60 point)2 = 0.36 point2, which means that 0.50 gry 2 = 0.18 point 2. The metric prefixes (micro, pico, nano, ) are given for ready reference on the inside front cover of the textbook (see also Table 1–2). (a) Since 1 km = 1 × 103 m and 1 m = 1 × 106 μm, ( )( ) 1km = 103 m = 103 m 106 μ m m = 109 μ m. The given measurement is 1.0 km (two significant figures), which implies our result should be written as 1.0 × 109 μm. (b) We calculate the number of microns in 1 centimeter. Since 1 cm = 10−2 m, ( )( ) 1cm = 10−2 m = 10−2 m 106 μ m m = 104 μ m.
We conclude that the fraction of one centimeter equal to 1.0 μm is 1.0 × 10−4. (c) Since 1 yd = (3 ft)(0.3048 m/ft) = 0.9144 m, 1 2 CHAPTER 1 ( ) 1.0 yd = ( 0.91m ) 106 μ m m = 9.1 × 105 μ m. (a) Using the conversion factors 1 inch = 2.54 cm exactly and 6 picas = 1 inch, we obtain ⎛ 1 inch ⎞ ⎛ 6 picas ⎞ 0.80 cm = ( 0.80 cm ) ⎜ ⎟⎜ ⎟ ≈ 1.9 picas. ⎝ 2.54 cm ⎠ ⎝ 1 inch ⎠ (b) With 12 points = 1 pica, we have ⎛ 1 inch ⎞ ⎛ 6 picas ⎞ ⎛ 12 points ⎞ 0.80 cm = ( 0.80 cm ) ⎜ ⎟⎜ ⎟⎜ ⎟ ≈ 23 points.
⎝ 2.54 cm ⎠ ⎝ 1 inch ⎠ ⎝ 1 pica ⎠ 5. Given that 1 furlong = 201.168 m, 1 rod = 5.0292 m and 1 chain = 20.117 m, we find the relevant conversion factors to be 1 rod 1.0 furlong = 201.168 m = (201.168 m ) = 40 rods, 5.0292 m and 1 chain 1.0 furlong = 201.168 m = (201.168 m ) = 10 chains. 20.117 m Note the cancellation of m (meters), the unwanted unit. Using the given conversion factors, we find (a) the distance d in rods to be d = 4.0 furlongs = ( 4.0 furlongs ) 40 rods = 160 rods, 1 furlong (b) and that distance in chains to be d = 4.0 furlongs = ( 4.0 furlongs ) 10 chains = 40 chains. We make use of Table 1-6. (a) We look at the first (“cahiz”) column: 1 fanega is equivalent to what amount of cahiz? We note from the already completed part of the table that 1 cahiz equals a dozen fanega.
1 Thus, 1 fanega = 12 cahiz, or 8.33 × 10−2 cahiz. Similarly, “1 cahiz = 48 cuartilla” (in the already completed part) implies that 1 cuartilla = 1 48 cahiz, or 2.08 × 10−2 cahiz. Continuing in this way, the remaining entries in the first column are 6.94 × 10−3 and 3.47 ×10−3. 3 (b) In the second (“fanega”) column, we find 0.250, 8.33 × 10−2, and 4.17 × 10−2 for the last three entries. (c) In the third (“cuartilla”) column, we obtain 0.333 and 0.167 for the last two entries. (d) Finally, in the fourth (“almude”) column, we get 1 2 = 0.500 for the last entry.
(e) Since the conversion table indicates that 1 almude is equivalent to 2 medios, our amount of 7.00 almudes must be equal to 14.0 medios. (f) Using the value (1 almude = 6.94 × 10−3 cahiz) found in part (a), we conclude that 7.00 almudes is equivalent to 4.86 × 10−2 cahiz. (g) Since each decimeter is 0.1 meter, then 55.501 cubic decimeters is equal to 0.055501 7.00 7.00 m3 or 55501 cm3.
Thus, 7.00 almudes = 12 fanega = 12 (55501 cm3) = 3.24 × 104 cm3. We use the conversion factors found in Appendix D. 1 acre ⋅ ft = (43,560 ft 2 ) ⋅ ft = 43,560 ft 3 Since 2 in. = (1/6) ft, the volume of water that fell during the storm is V = (26 km 2 )(1/6 ft) = (26 km 2 )(3281ft/km) 2 (1/6 ft) = 4.66 ×107 ft 3. Thus, 4.66 × 107 ft 3. × 103 acre ⋅ ft.
V = = 11 4 3 4.3560 × 10 ft acre ⋅ ft 8. 1-4, we see that 212 S is equivalent to 258 W and 212 – 32 = 180 S is equivalent to 216 – 60 = 156 Z. The information allows us to convert S to W or Z. (a) In units of W, we have ⎛ 258 W ⎞ 50.0 S = ( 50.0 S) ⎜ ⎟ = 60.8 W ⎝ 212 S ⎠ (b) In units of Z, we have ⎛ 156 Z ⎞ 50.0 S = ( 50.0 S) ⎜ ⎟ = 43.3 Z ⎝ 180 S ⎠ 9. The volume of ice is given by the product of the semicircular surface area and the thickness. The area of the semicircle is A = πr2/2, where r is the radius. Therefore, the volume is 4 CHAPTER 1 V = π 2 r z 2 where z is the ice thickness.
Fundamentals Of Physics Solutions Pdf
Since there are 103 m in 1 km and 102 cm in 1 m, we have ⎛ 103 m ⎞ r = ( 2000 km ) ⎜ ⎟ ⎝ 1km ⎠ ⎛ 102 cm ⎞ 5 ⎜ ⎟ = 2000 × 10 cm. ⎝ 1m ⎠ In these units, the thickness becomes ⎛ 102 cm ⎞ 2 z = 3000 m = ( 3000 m ) ⎜ ⎟ = 3000 × 10 cm 1m ⎝ ⎠ which yields V = π 2000 × 105 cm 2 ( ) ( 3000 × 10 2 2 ) cm = 1.9 × 1022 cm3. Since a change of longitude equal to 360° corresponds to a 24 hour change, then one expects to change longitude by 360° / 24 = 15° before resetting one's watch by 1.0 h. (a) Presuming that a French decimal day is equivalent to a regular day, then the ratio of weeks is simply 10/7 or (to 3 significant figures) 1.43.
(b) In a regular day, there are 86400 seconds, but in the French system described in the problem, there would be 105 seconds. The ratio is therefore 0.864. A day is equivalent to 86400 seconds and a meter is equivalent to a million micrometers, so 3.7 m 106 μ m m = 31. 14 day 86400 s day b b gc gb h g 13. The time on any of these clocks is a straight-line function of that on another, with slopes ≠ 1 and y-intercepts ≠ 0. From the data in the figure we deduce tC = 2 594 tB +, 7 7 tB = 33 662 tA −.
40 5 These are used in obtaining the following results. (a) We find t B′ − t B = when t'A − tA = 600 s. 33 ( t ′A − t A ) = 495 s 40 5 (b) We obtain t C′ − t C = b g b g 2 2 495 = 141 s. T B′ − t B = 7 7 (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.
The metric prefixes (micro (μ), pico, nano, ) are given for ready reference on the inside front cover of the textbook (also Table 1–2). ⎛ 100 y ⎞ ⎛ 365 day ⎞ ⎛ 24 h ⎞ ⎛ 60 min ⎞ (a) 1 μ century = (10−6 century ) ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ = 52.6 min.
1 century 1 y 1 day 1 h ⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠ (b) The percent difference is therefore 52.6 min − 50 min = 4.9%. A week is 7 days, each of which has 24 hours, and an hour is equivalent to 3600 seconds. Thus, two weeks (a fortnight) is 1209600 s.
Solutions Manual Halliday Fundamentals Of Physics
By definition of the micro prefix, this is roughly 1.21 × 1012 μs. We denote the pulsar rotation rate f (for frequency). F = 1 rotation 1.7275 × 10−3 s (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: ⎛ ⎞ 1 rotation N =⎜ ⎟ ( 604800 s ) = 388238218.4 −3 × 1.7275 10 s ⎝ ⎠ which should now be rounded to 3.88 × 108 rotations since the time-interval was specified in the problem to three significant figures. (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 ⎛ 1 rotation 1 × 106 = ⎜ −3 ⎝ 1.7275 × 10 ⎞ ⎟t s⎠ 6 CHAPTER 1 which yields the result t = 157275 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 ± 3 ×10− 17 s.
We therefore expect that as a result of one million revolutions, the uncertainty should be ( ± 3 × 10−17 )(1× 106 )= ± 3 ×10− 11 s. None of the clocks advance by exactly 24 h in a 24-h period but this is not the most important criterion for judging their quality for measuring time intervals.
What is important is that the clock advance by the same amount in each 24-h period. The clock reading can then easily be adju.
ISBN: 813. Publisher: Wiley. Authors: Halliday, Resnick, Walker Find your textbook below for step-by-step solutions to every problem. Fundamentals Of Physics Extended, 9th. Fundamentals of Physics, 9th Edition Fundamentals Of Physics, Volume 1 Fundamentals of Physics, 8th Edition Fundamentals of Physics, Volume 1, 8th. Fundamentals of Physics Volume 2, 9th. Fundamentals of Physics, Chapters 1-11.
Fundamentals of Physics (Part 3), 9th. Principles of Physics (International. Fundamentals of Physics, Volume 1. Fundamentals of Physics Extended, 9th. Fundamentals of Physics, 7th Edition Fundamentals of Physics: Extended. Physics Volume 1, 5th Edition Fundamentals of Physics (Volume 1), 6th.
Fundamentals of Physics Extended, 6th. Physics, Volume 2, 5th Edition Fundamentals of Physics, 8th Edition Modern Physics, 3rd Edition Fundamentals of Physics, 9th Edition Fundamentals of Physics Extended, 10th.
Fundamentals Of Physics, 10th Edition Fundamentals of Physics Extended, 10th. Fundamentals of Physics Volume 2, 10th. Principles of Physics (International. Fundamentals of Physics Volume 1, 10th.
Fundamentals of Physics Extended, 10th. Fundamentals of Physics, 11th Edition Fundamentos de Fisica. Mecanica-Volume 1. Fundamentos de Fisica. Gravitacao, Ondas.
Fundamentos de Fisica. Fundamentos de Fisica. Optica e Fisica. Don't see your textbook?
Step-by-step instructions and exploded views are included to make your repairs simple and quick. General Information 2. 2016 honda xr 100 maintenance manual. EACH SECTION INCLUDED: 1. This manual has everything you need to do repairs, service, and maintenance. For all 2012-2015 Honda NC700X / XA / XD / SA-C / XD-E and NC 750XA / SA-E models.