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CHAPTER 1 - 5

# CHAPTER 1 - 5 - 38 (a x=vt 2at...

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38. We test to see if each term has the same dimensions. ( a ) x  =  vt 2  + 2 at ; [ L ] =? [ L / T ][ T ] 2  + [ L / T 2 ][ T ]; [ L ] ? [ LT ] + [ L / T ];  therefore, this is        not correct . ( b ) x  =  v 0 t  +  ! at 2 ; [ L ] =? [ L / T ][ T ] + [ L / T 2 ][ T ] 2 ; [ L ] = [ L ] + [ L ];  therefore, this        may be correct . ( c ) x  =  v 0 t  + 2 at 2 ; [ L ] =? [ L / T ][ T ] + [ L / T 2 ][ T ] 2 ; [ L ] = [ L ] + [ L ];  therefore, this        may be correct . 39. ( a ) 1.0  Å  = (1.0 × 10 –10  m)/(10 –9  m/nm) =          0.10 nm . ( b ) 1.0  Å  = (1.0 × 10 –10  m)/(10 –15  m/fm) =          1.0 × 10 5  fm . ( c ) 1.0 m = (1.0 m)/(10 –10  m/ Å ) =          1.0 × 10 10   Å . ( d ) From the result for Problem 23, we have 1.0 ly = (9.5 × 10 15  m)/(10 –10  m/ Å ) =          9.5 × 10 25   Å . 40. We use the values for the masses from Table 1–3. ( a ) N bacterium  = (10 –15  kg)/(10 –27  kg/proton) =          10 12  protons (or neutrons).

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CHAPTER 1 - 5 - 38 (a x=vt 2at...

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