ch03-p028 - 28. Many of the operations are done efficiently...

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(j) The two possibilities presented by a simple calculation for the angle between the vector described in part (i) and the + x direction are tan –1 [(–11 m)/(–2.0 m)] = 80°, and 180° + 80° = 260°. The latter possibility is the correct answer (see part (k) for a further observation related to this result). (k) Since G G G G ab ba −=− () ( ) 1 , they point in opposite (anti-parallel) directions; the angle between them is 180°. 28. Many of the operations are done efficiently on most modern graphical calculators using their built-in vector manipulation and rectangular polar “shortcuts.” In this solution, we employ the “traditional” methods (such as Eq. 3-6). (a) The magnitude of G a is 22 (4.0 m) ( 3.0 m) 5.0 m. a =+ = (b) The angle between G a and the + x axis is tan –1 [(–3.0 m)/(4.0 m)] = –37°. The vector is 37° clockwise from the axis defined by # i . (c) The magnitude of G b is (6.0 m) (8.0 m) 10 m. b = (d) The angle between G b and the + x axis is tan –1 [(8.0 m)/(6.0 m)] = 53°. (e) ˆˆ ˆ ˆ (4.0 m 6.0 m) i [( 3.0 m) 8.0 m]j
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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