Chapter 1 - Solutions Chapter 1 1.1 Substituting dimensions...

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2 T g = l π [ ] [ ] [ ] T g = l 2 2 L T T T L T = = = 1 1 2 2 0 2 2 m m mgh = + v v 2 2 2 2 0 2 L ML [ ] [ ] M T T m m = = = v v 1 2 2 L M L M L T T mgh = = 2 0 at = + v v 0 L [ ] [ ] T = = v v ( 29 2 2 2 2 L [ ] [ ][ ] T L T at a t = = = 2 ma = v 2 2 L ML [ ] [ ][ ] M T T ma m a = = = 2 2 2 2 L L [ ] T T = = v ( 29 ( 29 21.3 0.2 cm 9.8 0.1 cm A w   = = ± ±   l Solutions Chapter 1 1.1 Substituting dimensions into the given equation , recognizing that 2 π is a dimensionless constant, we have or Thus, the dimensions are consistent . 1.4 In the equation , while . Thus, the equation is dimensionally incorrect. In , but Hence, this equation is dimensionally incorrect. In the equation , we see that while 1.8 Multiplying out this product of binominals gives
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( 29 ( 29 ( 29 ( 29 2 21.3 9.8 21.3 0.1 0.2 9.8 0.2 0.1 cm A = ± ± + 2
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This note was uploaded on 09/11/2011 for the course PHYSICS 2A taught by Professor Vu during the Spring '11 term at Evergreen Valley.

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Chapter 1 - Solutions Chapter 1 1.1 Substituting dimensions...

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