On Sunday, January 19, 2014 12:20:28 PM UTC-5, wrote: I have the instructor solution manuals to accompany mathematical, engineering, physical, chemical, financial textbooks, and others. These solution manuals contain a clear and concise step-by-step solution to every problem or exercise in these scientific textbooks.
They are all in PDF format. If you are interested in any one, simply send me an email to cartermathatgmaildotcom. Please this service is NOT free.
Here are the titles. Solution Manual Fundamentals of Signa.
[email protected] 20/5/2014, 8:03 น. axed: Can you please send me Solution Manual Fundamentals of Physics 9th Ed by Resnick, Walker, Halliday. thanks Sure can, it's why were all here. Please check your mailbox everyday. Also, check your neighbors mailbox everyday, we may not have your correct address.
If you see something in a plain brown wrapper, that's it! Whatever you do, don't look at the pictures.
Oh, and wear gloves. Oh, and don't let anyone see you. Other than that, have fun and drink Dos Equis my friend. ASSHOLE#104 Len '73XLH '16FLTRU 'Barack' BS#90 Wolf#21 PH#3 SENS MANS [email protected] 14/2/2018, 19:40 น.
![Physics resnick halliday pdf 5th edition Physics resnick halliday pdf 5th edition](/uploads/1/2/3/7/123731573/909984677.jpg)
Set the following eddies spiralling through the space-time continuum: I have the instructor solution manuals to accompany mathematicalengineering, physical, chemical, financial textbooks, and others. These solution manuals contain a clear and concise step-by-step solution to every problem or exercise in these scientific textbooks.
They are all in PDF format. If you are interested in any one, simply send me an email to cartermathatgmaildotcom. Please this service is NOT free. Here are the titles. Solution Manual A Course in Modern Mathematical Physics by Peter Szekeres Solution Manual 2500 Solved Problems in Fluid Mechanics & Hydraulics & mucel mo vpon þis wyse And the Tolkien connection? - ξ: ) Proud to be curly Interchange the alphabetic letter groups to reply Stan Brown 19/1/2014, 17:07 น.
On Sun, 19 Jan 2014 22:03:48 +0000, Curlytop wrote: set the following eddies spiralling through the space-time continuum: spam, snipped And the Tolkien connection? Spammers don't read the newsgroups that they spam. When you reply to them, you don't change their behavior one iota; all you do is add another off-topic article to the newsgroup. Stan Brown, Oak Road Systems, Tompkins County, New York, USA Tolkien FAQs: (Steuard Jensen's site) Tolkien letters FAQ: FAQ of the Rings: Encyclopedia of Arda: more FAQs: Paul S. Person 20/1/2014, 9:48 น. On Sun, 19 Jan 2014 20:07:43 -0500, Stan Brown wrote: On Sun, 19 Jan 2014 22:03:48 +0000, Curlytop wrote: set the following eddies spiralling through the space-time continuum: spam, snipped And the Tolkien connection? Spammers don't read the newsgroups that they spam.
When you reply to them, you don't change their behavior one iota; all you do is add another off-topic article to the newsgroup. Perhaps if there were some (not 'more', since there aren't any, so 'some') on-topic articles in the newsgroup, the occasional irrelevant spam wouldn't be such an attractive target. Are we /really/ going to get nothing but the autoposts and Troels monthly summaries (which latter, to be sure, are well worth the posting) until the final Hobbit movie appears? - 'Nature must be explained in her own terms through the experience of our senses.' [email protected] 20/4/2014, 14:05 น.
On Sunday, January 19, 2014 11:22:10 AM UTC-6, wrote: I have the instructor solution manuals to accompany mathematical, engineering, physical, chemical, financial textbooks, and others. These solution manuals contain a clear and concise step-by-step solution to every problem or exercise in these scientific textbooks. They are all in PDF format. If you are interested in any one, simply send me an email to cartermathatgmaildotcom. Please this service is NOT free. Here are the titles.
ok. [email protected] 29/4/2014, 22:37 น.
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. 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. 4 CHAPTER 1 V = π 2 r z 2 where z is the ice thickness.
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. 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⎠ 7 T = ( average increase in length of a day )( number of days ) ⎛ 0.01 s ⎞ ⎛ 365.25 day ⎞ =⎜ ⎟⎜ ⎟ ( 2000 y ) y ⎝ day ⎠ ⎝ ⎠ = 7305 s or roughly two hours. 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 d be the distance from point B to your eyes. From the Pythagorean theorem, we have d 2 + r 2 = (r + h) 2 = r 2 + 2rh + h 2 or d 2 = 2rh + h 2, 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 θ, which is also the angle through which the Sun moves about Earth during the time interval t = 11.1 s. The value of θ can be obtained by using θ 360 ° This yields θ= = t. 24 h (360°)(11.1 s) = 0.04625°.
(24 h)(60 min/h)(60 s/min) Using d = r tan θ, we have d 2 = r 2 tan 2 θ = 2rh, or r= 2h tan 2 θ Using the above value for θ and h = 1.7 m, we have r = 5.2 ×106 m. CHAPTER 1 8 20.
(a) We find the volume in cubic centimeters 3 ⎛ 231 in 3 ⎞ ⎛ 2.54 cm ⎞ 5 3 193 gal = (193 gal ) ⎜ ⎟⎜ ⎟ = 7.31 × 10 cm 1 gal 1in ⎠ ⎝ ⎠⎝ and subtract this from 1 × 106 cm3 to obtain 2.69 × 105 cm3. The conversion gal → in3 is given in Appendix D (immediately below the table of Volume conversions). (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 c1000 kg m h c0.731 m h = 731 kg 3 2 using the density given in the problem statement.
At a rate of 0.0018 kg/min, this can be filled in 731kg = 4.06 × 105 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). 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 × 10−27 kg). Thus, 5.98 × 1024 kg ME N = = = 9.0 × 1049. −27 m × 10 kg u 40 u 1661. The density of gold is ρ= m 19.32 g = = 19.32 g/cm3. 3 V 1 cm (a) We ta.
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.
Halliday Resnick Solutions Manual Pdf
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. 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.
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(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.
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 adjusted to give the correct interval. If the clock reading jumps around from one 24-h period to another, it cannot be corrected since it would impossible to tell what the correction should be. The following gives the corrections (in seconds) that must be applied to the reading on each clock for each 24-h period. The entries were determined by subtracting the clock reading at the end of the interval from the clock reading at the beginning. CLOCK A B C D E Sun.
−16 −3 −58 +67 +70 Mon. −16 +5 −58 +67 +55 Tues. −15 −10 −58 +67 +2 Wed. −17 +5 −58 +67 +20 Thurs. −15 +6 −58 +67 +10 Fri.
−15 −7 −58 +67 +10 Clocks C and D are both good timekeepers in the sense that each is consistent in its daily drift (relative to WWF time); thus, C and D are easily made “perfect” with simple and predictable corrections. The correction for clock C is less than the correction for clock D, so we judge clock C to be the best and clock D to be the next best. The correction that must be applied to clock A is in the range from 15 s to 17s. For clock B it is the range from -5 s to +10 s, for clock E it is in the range from -70 s to -2 s. After C and D, A has 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. 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 7 T = ( average increase in length of a day )( number of days ) ⎛ 0.01 s ⎞ ⎛ 365.25 day ⎞ =⎜ ⎟⎜ ⎟ ( 2000 y ) y ⎝ day ⎠ ⎝ ⎠ = 7305 s or roughly two hours. 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 d be the distance from point B to your eyes.
From the Pythagorean theorem, we have d 2 + r 2 = (r + h) 2 = r 2 + 2rh + h 2 or d 2 = 2rh + h 2, 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 θ, which is also the angle through which the Sun moves about Earth during the time interval t = 11.1 s. The value of θ can be obtained by using θ 360 ° This yields θ= = t.
24 h (360°)(11.1 s) = 0.04625°. (24 h)(60 min/h)(60 s/min) Using d = r tan θ, we have d 2 = r 2 tan 2 θ = 2rh, or r= 2h tan 2 θ Using the above value for θ and h = 1.7 m, we have r = 5.2 ×106 m. CHAPTER 1 8 20. (a) We find the volume in cubic centimeters 3 ⎛ 231 in 3 ⎞ ⎛ 2.54 cm ⎞ 5 3 193 gal = (193 gal ) ⎜ ⎟⎜ ⎟ = 7.31 × 10 cm 1 gal 1in ⎠ ⎝ ⎠⎝ and subtract this from 1 × 106 cm3 to obtain 2.69 × 105 cm3. The conversion gal → in3 is given in Appendix D (immediately below the table of Volume conversions).
(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 c1000 kg m h c0.731 m h = 731 kg 3 2 using the density given in the problem statement. At a rate of 0.0018 kg/min, this can be filled in 731kg = 4.06 × 105 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). 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 × 10−27 kg). Thus, 5.98 × 1024 kg ME = 9.0 × 1049. = N = −27 m × 10 kg u. 40 u 1661 b gc h 22.
The density of gold is ρ= m 19.32 g = = 19.32 g/cm3. 3 V 1 cm (a) We take the volume of the leaf to be its area A multiplied by its thickness z. With density ρ = 19.32 g/cm3 and mass m = 27.63 g, the volume of the leaf is found to be V = We convert the volume to SI units: m ρ = 1430. 9 V = (1.430 cm 3 ) 3 ⎛ 1m ⎞ 3 −6 ⎜ ⎟ = 1.430 × 10 m.
⎝ 100 cm ⎠ Since V = Az with z = 1 × 10-6 m (metric prefixes can be found in Table 1–2), we obtain A=. 1430 × 10−6 m3.
= 1430 −6 1 × 10 m (b) The volume of a cylinder of length is V = A where the cross-section area is that of a circle: A = πr2. Therefore, with r = 2.500 × 10−6 m and V = 1.430 × 10−6 m3, we obtain = V = 7.284 × 104 m = 72.84 km. We introduce the notion of density: ρ= m V and convert to SI units: 1 g = 1 × 10−3 kg. (a) For volume conversion, we find 1 cm3 = (1 × 10−2m)3 = 1 × 10−6m3. Thus, the density in kg/m3 is −3 3 ⎛ 1 g ⎞ ⎛ 10 kg ⎞ ⎛ cm ⎞ 3 3 1 g cm3 = ⎜ 3 ⎟ ⎜ ⎟ ⎜ −6 3 ⎟ = 1 × 10 kg m. ⎝ cm ⎠ ⎝ g ⎠ ⎝ 10 m ⎠ Thus, the mass of a cubic meter of water is 1000 kg.
(b) We divide the mass of the water by the time taken to drain it. The mass is found from M = ρV (the product of the volume of water and its density): ( M = 5700 m3 ) (1 × 10 3 ).
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