Physics 140 HOMEWORK Chapter 1B 16. Time Standards Are Now Based

Physics 140 HOMEWORK Chapter 1B 16. Time standards are now based on atomic clocks. A promising second standard is based on pulsars, which are rotating neutron stars (highly compact stars consisting only of neutrons). Some rotate at a rate that is highly stable, sending out a radio beacon that sweeps briefly across Earth once with each rotation, like a lighthouse beacon. Pulsar PSR 1937+21 is an example; it rotates once every 1.55780644887275 ± 3 ms, where the trailing ±3 indicates the uncertainty in the last decimal place (it does not mean ±3 ms). (a) How many rotations does PSR 1937+21 make in 7.00 days? (b) How much time does the pulsar take to rotate exactly one million times and (c) what is the associated uncertainty? (a) The number of rotations in 7 days N is given by: N =(7 days)/(time per rotation). I will convert to seconds: 7 days = (7 days) ×86400 s/da = 604800 s N = (6.048 × 105 s)/(1.5578 × 10−3 s/rev) = 3.88 × 108 rev. (3 sigdig because “7.00 days”) (b) Time for 106 rotations = (106)(1.55780644887275 × 10−3 s) = 1557.806 448 872 75 s. Note the conversion from milliseconds to seconds. (c) The uncertainty is 3 in the last digit, or ±3 × 10−11 s. 18. Because Earth’s rotation is gradually slowing, the length of each day increases: The day at the end of 1.0 century is 1.0 ms longer than the day at the start of the century. In 20 centuries, what is the total of the daily increases in time? The length of the day is uniformly increasing, and one day is 20 ms longer after 2000 years. That means the average of the last 2000 years is that each day was 10 ms longer than the reference day at the beginning. Therefore, the total discrepancy would be (10 ms/day) × (365.25days/yr) × (2000 yr) = 7300 s. An application would be determining the region on Earth from which a historical solar eclipse was a total eclipse. 19. Suppose that, while lying on a beach near the equator watching the Sun set over a calm ocean, you start a stopwatch just as the top of the Sun disappears. You then stand, elevating your eyes by a height H =1.70 m, and stop the watch when the top of the Sun again disappears. If the elapsed time is t = 11.1 s, what is the radius r of Earth? Sketch this triangle: Vertex A at the center of the Earth; vertex B at the point where you were lying down; at this point, draw a dashed horizontal line (perpendicular to segment AB) toward the Sun (put the Sun on the right); extend this dashed line a bit to the left; the third vertex C is at an angle θ to the left of line segment AB and on the left extension of the dashed line, because the Sun is staying fixed while the Earth rotates. The left extension of the dashed line represents the locus of points from which the Sun is just disappearing below the horizon. Now, triangle ABC is a right triangle with the right angle at B. Side AB has a length of r, while side AC has a length of r + H. Hence cos θ = r/(r + H). Solve this for r to get r = H cos θ/(1 − cos θ) Eq 1 We have H =1.70 m but need θ. But θ can be easily calculated from Earth’s rotation and the elapsed time, noting that one day is 86 400 s: θ = (360◦) × (11.1 s)/(86400 s) = 0.04625◦. Plug numbers to get r = 5.22 × 106 m. In plugging numbers, note that cos θ = 0.9999996742 is very close to 1. You can approximate it by 1 in the numerator of Eq 1, but definitely not in the denominator. Finally, note that you are about 20% off, but at least in the ballpark. 22. Gold, which has a density of 19.32 g/cm3, is the most ductile metal and can be pressed into a thin leaf or drawn out into a long fiber. (a) If a sample of gold, with a mass of 27.63 g, is pressed into a leaf of 1.000 µm thickness, what is the area of the leaf? (b) If, instead, the gold is drawn out into a cylindrical fiber of radius 2.500 µm, what is the length of the fiber? (a) We will work in grams and centimeters. 1 µm = 10−4 cm. We need the volume of gold. Since ρ = m/V , V = m/ρ = (27.63 g)/(19.32 g/cm3)=1.4301 cm3. Since Volume = (Area)(thickness), A = V/t = (1.4301 cm3)/(10−4 cm) = 14 300 cm2 (b) Now, V = πr2L, so L = V/(πr2)=(1.4301 cm3)/(3.14 · (2.5 × 10−4 cm)2)=7.28 × 106 cm = 72.8 km. 29. On a spending spree in Malaysia, you buy an ox with a weight of 28.9 piculs in the local unit of weights: 1 picul = 100 gins, 1 gin = 16 tahils, 1 tahil = 10 chees, and 1 chee = 10 hoons. The weight of 1 hoon corresponds to a mass of 0.3779 g. When you arrange to ship the ox home to your astonished family, how much mass in kilograms must you declare on the shipping manifest? (Hint: Set up multiple chain-link conversions.) Note that 1 hoon = 3.779 × 10−4 kg. 100 gin 16 tahil 10 chee 10 hoon 3.779 × 10−4 kg Mass = 28.9 picul × × × × × = 1747 kg. picul gin tahil chee hoon 47. An astronomical unit (AU) is the average distance between Earth and the Sun, approximately 1.50 × 108 km. The speed of light is about 3.0 × 108 m/s. Express the speed of light in astronomical units per minute Convert the AU from km to meters, then: c = (3 × 108 m/s) × (1 AU)/(1.5 × 1011 m) × (60 s)/(1 min) = 0.12 AU/min..

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