ch04-p002

# ch04-p002 - 2. (a) The position vector, according to Eq....

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2. (a) The position vector, according to Eq. 4-1, is ˆˆ = ( 5.0 m) i + (8.0 m)j r G . (b) The magnitude is 222 2 2 2 | | + + ( 5.0 m) (8.0 m) (0 m) 9.4 m. rx y z == + + = G (c) Many calculators have polar rectangular conversion capabilities which make this computation more efficient than what is shown below. Noting that the vector lies in the xy plane and using Eq. 3-6, we obtain: 1 8.0 m tan 58 or 122 5.0 m θ §· ° ° ¨¸ ©¹ where the latter possibility (122° measured counterclockwise from the + x direction) is chosen since the signs of the components imply the vector is in the second quadrant. (d) The sketch is shown on the right. The vector is 122° counterclockwise
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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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