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Physics and Measurement

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Q1.1 Atomic clocks are based on electromagnetic waves which atoms emit. Also, pulsars are highly regular astronomical clocks. Q1.2 Density varies with temperature and pressure. It would be necessar... y to measure both mass and volume very accurately in order to use the density of water as a standard. Q1.3 People have different size hands. Defining the unit precisely would be cumbersome. Q1.4 (a) 0.3 millimeters (b) 50 microseconds (c) 7.2 kilograms Q1.5 (b) and (d). You cannot add or subtract quantities of different dimension. Q1.6 A dimensionally correct equation need not be true. Example: 1 chimpanzee = 2 chimpanzee is dimensionally correct. If an equation is not dimensionally correct, it cannot be correct. Q1.7 If I were a runner, I might walk or run 101 miles per day. Since I am a college professor, I walk about 100 miles per day. I drive about 40 miles per day on workdays and up to 200 miles per day on vacation. Q1.8 On February 7, 2001, I am 55 years and 39 days old. 55 365 25 1 39 20 128 86 400 1 yr 1 74 109 109 d yr d d s d s s . . ~ FH G IK J + = FH G IK J = × . Many college students are just approaching 1 Gs. Q1.9 Zero digits. An order-of-magnitude calculation is accurate only within a factor of 10. Q1.10 The mass of the forty-six chapter textbook is on the order of 100 kg . Q1.11 With one datum known to one significant digit, we have 80 million yr + 24 yr = 80 million yr. 1 2 Physics and Measurement SOLUTIONS TO PROBLEMS Section 1.1 Standards of Length, Mass, and Time No problems in this section Section 1.2 Matter and Model-Building P1.1 From the figure, we may see that the spacing between diagonal planes is half the distance between diagonally adjacent atoms on a flat plane. This diagonal distance may be obtained from the Pythagorean theorem, Ldiag = L2 + L2 . Thus, since the atoms are separated by a distance L = 0.200 nm, the diagonal planes are separated by 1 2 L2 + L2 = 0.141 nm . Section 1.3 Density and Atomic Mass *P1.2 Modeling the Earth as a sphere, we find its volume as 4 3 4 3 3 6 37 106 1 08 10 3 21 3 π r = π e . × mj = . × m . Its density is then ρ = = × × m = × V 5 98 10 1 08 10 5 52 10 24 21 3 . 3 3 . . kg m kg m . This value is intermediate between the tabulated densities of aluminum and iron. Typical rocks have densities around 2 000 to 3 000 kg m3 . The average density of the Earth is significantly higher, so higher-density material must be down below the surface. P1.3 With V = abase areafbheightg V = eπ r 2 jh and ρ = m V , we have ρ π π ρ = = FH G IK J = × m r 2h 2 9 4 3 1 19 5 39 0 10 1 2 15 10 kg mm mm mm m kg m 3 3 . . . . a fa f *P1.4 Let V represent the volume of the model, the same in ρ = m V for both. Then ρ iron = 9.35 kg V and ρ gold gold = m V . Next, ρ ρ gold iron gold kg = m 9.35 and mgold 3 3 3 kg 19.3 10 kg / m kg / m = kg × × FH G IK J 9 35 = 7 86 10 23 0 3 . . . . P1.5 V = Vo −Vi = r − r 4 3 2 3 1 π e 3 [Show More]

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