CHAPTER 3 7 - z V D H y North x 16 D =H =4580sin32.4=2454m x y z x y East D D =H = 4580cos32.4= 3867m =V= 2450m D =(D D D x y z 2 2 2 1/2

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y z φ V H D North East x y A 1 A 2 A y 16. For the components we have D x H x  = – 4580 sin 32.4° =         – 2454 m ;   D y H y  = + 4580 cos 32.4° =         + 3867 m D z =                                                  + 2450 m . By extending the Pythagorean theorem,  we find the magnitude from D = ( D x 2  +  D y 2  +  D z 2 ) 1/2   = [(– 2454 m) 2  + (3867 m) 2   + (2450 m) 2 ] 1/2    =          5194 m . 17. ( a ) We find the  x -component from A 2  =  A x 2  +  A y 2 ; (90.0) 2  =  A x 2  + (– 35.0) 2 ; which gives         A x  = ± 82.9 . ( b ) If we call the new vector  B , we have R x A x  +  B x  ; – 80.0 = + 82.9 +  B x  , which gives  B x  = – 162.9; R y A y  +  B y  ; 0 = – 35.0 +  B y  , which gives   B y  = + 35.0. We find the resultant from
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This note was uploaded on 03/08/2010 for the course PHYSICS 7A/7B taught by Professor All during the Fall '08 term at University of California, Berkeley.

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CHAPTER 3 7 - z V D H y North x 16 D =H =4580sin32.4=2454m x y z x y East D D =H = 4580cos32.4= 3867m =V= 2450m D =(D D D x y z 2 2 2 1/2

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