# Chap 1 - DOING PHYSICS 1 EXERCISES Section 1.2 Measurements...

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1.1 DOING PHYSICS EXERCISES Section 1.2 Measurements and Units 10. SI prefixes are explained in Table 1.1. (a) 10 10 10 10 33 6 9 MW W W. == () (b) and (c) 10 10 1 96 Wk W G W . 11. INTERPRET We interpret this as a problem involving the comparison of the diameter of two objects (a hydrogen and a proton) expressed in different units. DEVELOP 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 is10 9 m, and a femtometer (fm) is10 15 m. EVALUATE Using the conversion factors, the diameter of a hydrogen atom is d H nm m, 01 10 10 . and the diameter of a proton is d p fm 11 0 15 Therefore, the ratio of the diameters of a hydrogen atom and a proton (its nucleus) is d d H p m m 10 10 10 10 15 5 ASSESS 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, 0 9 ns s = in this case. Thus d = (,, / ) 299 792 458 10 9 m s s 0.299792458 m, or approximately 30 cm. 13. INTERPRET We interpret this as a problem involving expressing the period of cesium radiation in different units. DEVELOP By definition, 1 9 192 631 770 s = ,,, periods of a cesium atomic clock. In addition, we know that 0 9 ns s. = EVALUATE One period of cesium radiation is 1 9 192 631 770 1 087827757 10 10 s s 0.1087 . = 827757 ns ASSESS 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 10 18 , so 14 Eg 14 10 (10 kg) 1.4 10 kg. 18 3 16 = × 15. INTERPRET We interpret this as a problem involving expressing 1-cm line as multiples of the diameter of hydrogen atoms. DEVELOP We first express the quantities of interest (diameter of a hydrogen atom and1-cm line) in the same units. Since a nanometer is 10 9 m (Table 1.1), we see that d H nm m. 10 . In addition, 0 2 cm = EVALUATE The desired number of atoms is the length of the line divided by the diameter of one atom: N 10 10 10 2 10 8 m m ASSESS If 1 cm corresponds to10 8 hydrogen atoms, then each atom would correspond to 10 cm 8 −− = 10 10 m = 0.1 nm. 1

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1.2 Chapter 1 16. sc m c m . == = r θ (. ) (.) . 81 14 113 ( Note : Radians are pure numbers with no units; for this reason, they are the simplest angular measure mathematically.) 17. INTERPRET This is a problem that involves the definition of an angle subtended by a circular arc. DEVELOP The angle in radians is the circular arc length s divided by the radius R , or = sR /. EVALUATE Using the equation above, we find the angle to be = s R 2.1 km 3.4 km 0.62 rad Using the conversion factor 1 180 π rad , the result can be expressed as ° ≈° 0.62 rad 0.62 rad) 180 (3 5 ASSESS 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 C 23 4 2 1 4 ) . km km. 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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## This note was uploaded on 10/21/2010 for the course PHYSICS 2131441 taught by Professor Pheong during the Fall '10 term at Berkeley.

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Chap 1 - DOING PHYSICS 1 EXERCISES Section 1.2 Measurements...

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