S08P2AChap1Sol

S08P2AChap1Sol - Chapter 1 Problem Solutions (Easy) 1.7 The...

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Chapter 1 Problem Solutions (Easy) 1.7 The desired number of atoms is the length of the line divided by the diameter of one atom, or (1 cm) = (0.1 nm) = (10 ! 2 m) = (10 ! 10 m) = 10 8 . (See Table 1-1 for SI prefixes.) 1.10 (a) (35.0 mi/h)(1609 m/mi)(1 h = 3600 s) = 15.6 m/s. (b) (35.0 mi/h)(5280 ft/mi)(1 h = 3600 s) = 51.3 ft/s. 1.12 The number of seconds in a year is approximately (365.24 d)(86,400 s/d) = 3.1557 ! 10 7 s (see Appendix C), which differs from the mnemonic ! " 10 7 s in the third decimal place. The percent difference, to two significant figures, is 100(3.1557 ! 3.1416) = 3.1416 = 0.45%. 1.13 1 m 3 = (10 2 cm) 3 = 10 6 cm 3 . 1.20 With reference to Appendix C, 550 ! 10 9 bbl y " # $ % & 42 gal bbl " # $ % & 1 y 3.156 ! 10 7 s " # $ % & = 7.32 ! 10 5 gal s " # $ % & 3.786 L gal " # $ % & = 2.77 ! 10 6 L s 1.25 We can first solve for a = 2 x = t 2 to find its dimensions (which are usually denoted by square brackets). Thus [ a ] = [ x ] = [ t 2 ] = LT ! 2 . 1.35 Substitute the given values into the equation for the temperature T = S 4 " # $ % & 1 = 4 = 1.4 ( 10 3 kg ) s * 3 4 ( 5.7 ( 10 * 8 kg ) s * 3 ) K * 4 " # $ % & 1 = 4 = 2.8 ( 10 2 K The result is expressed in scientific notation with two significant figures.
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(Medium) 1.27
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This note was uploaded on 06/12/2008 for the course PHYS 2A taught by Professor Hicks during the Spring '07 term at UCSD.

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S08P2AChap1Sol - Chapter 1 Problem Solutions (Easy) 1.7 The...

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