physics - Chapter 1 Page 1 CHAPTER 1 - Introduction,...

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Unformatted text preview: 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 = 1 . 2 × 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. 10. We assume an uncertainty of 0.1 × 10 4 cm. We compare the area for the specified radius to Chapter 1 Page 2 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 ....
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physics - Chapter 1 Page 1 CHAPTER 1 - Introduction,...

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