Physics and Measurement TESTBANK 2021 GRADED A

Physics and Measurement 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 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 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 j ρ = m V , so m V r r r r = = FH G IK J − = − ρ ρ π 4 π ρ 3 4 2 3 3 1 3 2 3 1 3 e j e j . Chapter 1 3 P1.6 For either sphere the volume is V = r 4 3 π 3 and the mass is m = ρV = ρ 4π r 3 3. We divide this equation for the larger sphere by the same equation for the smaller: m m r r r s s rs