Solutions - Chapter 1 CHAPTER 1 - Introduction,...

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Chapter 1 Page 1 CHAPTER 1 - Introduction, Measurement, Estimating 1. ( a ) Assuming one significant figure, we have 10 billion yr = 10 × 10 9 yr = 1 × 10 10 yr . ( b ) (1 × 10 10 yr)(3 × 10 7 s/yr) = 3 × 10 17 s . 2. ( a ) 4 significant figures . ( b ) Because the zero is not needed for placement, we have 4 significant figures . ( c ) 3 significant figures . ( d ) Because the zeros are for placement only, we have 1 significant figure . ( e ) Because the zeros are for placement only, we have 2 significant figures . ( f ) 4 significant figures . ( g ) 2, 3, or 4 significant figures , depending on the significance of the zeros. 3. ( a ) 1,156 = 1.156 × 10 3 . ( b ) 21.8 = 2.18 × 10 1 . ( c ) 0.0068 = 6.8 × 10 –3 . ( d ) 27.635 = 2.7635 × 10 1 . ( e ) 0.219 = 2.19 × 10 –1 . ( f ) 22 = 2.2 × 10 1 . 4. ( a ) 8.69 × 10 4 = 86,900 . ( b ) 7.1 × 10 3 = 7,100 . ( c ) 6.6 × 10 –1 = 0.66 . ( d ) 8.76 × 10 2 = 876 . ( e ) 8.62 × 10 –5 = 0.000 086 2 . 5. % uncertainty = [(0.25 m)/(3.26 m)] 100 = 7.7% . Because the uncertainty has 2 significant figures, the % uncertainty has 2 significant figures. 6. We assume an uncertainty of 1 in the last place, i. e., 0.01, so we have % uncertainty = [(0.01 m)/(1.28 m)] 100 = 0.8% . Because the uncertainty has 1 significant figure, the % uncertainty has 1 significant figure. 7. We assume an uncertainty of 0.2 s. ( a ) % uncertainty = [(0.2 s)/(5 s)] 100 = 4% . Because the uncertainty has 1 significant figure, the % uncertainty has 1 significant figure. ( b ) % uncertainty = [(0.2 s)/(50 s)] 100 = 0.4% . Because the uncertainty has 1 significant figure, the % uncertainty has 1 significant figure. ( c ) % uncertainty = [(0.2 s)/(5 min)(60 s/min)] 100 = 0.07% . Because the uncertainty has 1 significant figure, the % uncertainty has 1 significant figure. 8. For multiplication, the number of significant figures in the result is the least number from the multipliers; in this case 2 from the second value. (2.079 × 10 2 m)(0.072 × 10 –1 ) = 0.15 × 10 1 m = 1.5 m . 9. To add, we make all of the exponents the same: 9 . 2 × 10 3 s + 8.3 × 10 4 s + 0.008 × 10 6 s = 0.92 × 10 4 s + 8.3 × 10 4 s + 0.8 × 10 4 s = 10.02 × 10 4 s = 1.0 × 10 5 s . Because we are adding, the location of the uncertain figure for the result is the one furthest to the left. In this case, it is fixed by the third value.
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Chapter 1 Page 2 10. We assume an uncertainty of 0.1 × 10 4 cm. We compare the area for the specified radius to the area for the extreme radius. A 1 = p R 1 2 = p(3.8 × 10 4 cm) 2 = 4.54 × 10 9 cm 2 ; A 2 = p R 2 2 = p[(3.8 + 0.1) × 10 4 cm] 2 = 4.78 × 10 9 cm 2 , so the uncertainty in the area is ? A = A 2 A 1 = 0.24 × 10 9 cm 2 = 0.2 × 10 9 cm 2 . We write the area as
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This note was uploaded on 02/05/2010 for the course PHYSICS 37,38,39 taught by Professor Fielding during the Spring '09 term at Los Angeles Valley College.

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Solutions - Chapter 1 CHAPTER 1 - Introduction,...

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