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01_InstSolManual_PC

01_InstSolManual_PC - DOING PHYSICS 1 EXERCISES Section 1.2...

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DOING PHYSICS E XERCISES Section 1.2 Measurements and Units 10. SI prefixes are explained in Table 1.1. (a) 3 3 6 9 10 MW 10 (10 W) 10 W. = = (b) and (c) 9 6 10 W 10 kW 1 GW. = = 11. I NTERPRET We interpret this as a problem involving the comparison of the diameter of two objects (a hydrogen and a proton) expressed in different units. D EVELOP Before any comparison can be made, the quantity of interest must first be expressed in the same units. From Table 1.1, we see that a nanometer is 9 10 m, - and a femtometer (fm) is 15 10 m. - E VALUATE Using the conversion factors, the diameter of a hydrogen atom is 10 H 0.1 nm 10 m, d - = = and the diameter of a proton is 15 p 1 fm 10 m. d - = = Therefore, the ratio of the diameters of a hydrogen atom and a proton (its nucleus) is 10 5 H 15 p 10 m 10 10 m d d - - = = A SSESS The hydrogen atom is about 100,000 times larger than its nucleus. 12. The current definition of the meter is such that the speed of light in vacuum is exactly 299,792,458 m/s. The distance traveled is the speed multiplied by the travel time, 9 1 ns 10 s - = in this case. Thus 9 (299,792,458 m/s) 10 s 0.299792458 m, d - = × = or approximately 30 cm. 13. I NTERPRET We interpret this as a problem involving expressing the period of cesium radiation in different units. D EVELOP By definition, 1 s 9,192,631,770 = periods of a cesium atomic clock. In addition, we know that 9 1 ns 10 s. - = E VALUATE One period of cesium radiation is 10 1 s 1.087827757 10 s 0.1087827757 ns 9,192,631,770 - = × = A SSESS Since one nanosecond corresponds to about 9 periods of the cesium radiation, each period is about 1 9 of a nanosecond. Note that there exists an alternative definition based on the frequency of the cesium-133 hyperfine transition, which is the reciprocal of the period. 14. The prefix eka equals 18 10 , so 18 3 16 14 Eg 14 10 (10 kg) 1.4 10 kg. - = × = × 15. I NTERPRET We interpret this as a problem involving expressing 1-cm line as multiples of the diameter of hydrogen atoms. D EVELOP We first express the quantities of interest (diameter of a hydrogen atom and1-cm line) in the same units. Since a nanometer is 9 10 m - (Table 1.1), we see that 10 H 0.1 nm 10 m. d - = = In addition, 2 1cm 10 m. - = E VALUATE The desired number of atoms is the length of the line divided by the diameter of one atom: 2 8 10 10 m 10 10 m N - - = = A SSESS If 1 cm corresponds to 8 10 hydrogen atoms, then each atom would correspond to 8 10 10 cm 10 - - = m = 0.1 nm. 1.1 1

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1.2 Chapter 1 16. s (8.1 cm)(1.4) 11.3 cm. r θ = = = ( Note : Radians are pure numbers with no units; for this reason, they are the simplest angular measure mathematically.) 17. I NTERPRET This is a problem that involves the definition of an angle subtended by a circular arc. D EVELOP The angle in radians is the circular arc length s divided by the radius R , or / . sR θ = E VALUATE Using the equation above, we find the angle to be 2.1 km 0.62 rad 3.4 km s R θ = = = Using the conversion factor 1 rad 180 , π = ° the result can be expressed as 180 0.62 rad (0.62 rad) 35 rad θ π ° = = ° A SSESS Since a complete turn corresponds to 360 , 35 ° ° would be roughly 1/10 of a circle. The circumference of a circle of radius R = 3.4 km is 2 (3.4 km) 21.4 km. C π = = Therefore, we expect the jetliner to fly approximately 1/10 of C , or 2.1 km, in complete agreement with the problem statement.
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