Ch 1Work - Chapter 1 CHAPTER 1 - Introduction, Measurement,...

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Chapter  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 
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Ch 1Work - Chapter 1 CHAPTER 1 - Introduction, Measurement,...

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