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CH-04 physics notes

# CH-04 physics notes - Chapter 4 1(a The magnitude of r is...

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Chapter 4 1. (a) The magnitude of r is 5.0 + ( + 2.0 = 6.2 m. 2 2 30 2 . ) (b) A sketch is shown. The coordinate values are in meters. 2. Wherever the length unit is not specified (in this solution), the unit meter should be understood. (a) The position vector, according to Eq. 4-1, is ˆ ˆ = ( 5.0 m) i + (8.0 m)j r r . (b) The magnitude is 2 2 2 2 2 2 | | = + + = ( 5.0) (8.0) 0 = 9.4 m. r x y z r (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, we are using Eq. 3-6: 1 8.0 = tan = 58 or 122 5.0 where we choose the latter possibility (122° measured counterclockwise from the + x direction) since the signs of the components imply the vector is in the second quadrant. (d) In the interest of saving space, we omit the sketch. The vector is 32° counterclockwise from the + y direction, where the + y direction is assumed to be (as is standard) +90° counterclockwise from + x , and the + z direction would therefore be “out of the paper.” (e) The displacement is r r r r ' r where r is given in part (a) and r ' ˆ = 3.0i. Therefore, ˆ ˆ = 8.0i 8.0j r r (in meters). 115

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CHAPTER 4 (f) The magnitude of the displacement is 2 2 | | = (8.0) + ( 8.0) = 11 m. r r (g) The angle for the displacement, using Eq. 3-6, is found from 1 8.0 tan = 45 or 135 8.0 where we choose the former possibility (-45°, which means 45° measured clockwise from + x , or 315° counterclockwise from + x ) since the signs of the components imply the vector is in the fourth quadrant. 3. The initial position vector r o satisfies r r r o , which results in o ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ (3.0j 4.0k) (2.0i 3.0j 6.0k) 2.0i 6.0 j 10k r r r     r r r where the understood unit is meters. 4. We choose a coordinate system with origin at the clock center and + x rightward (towards the “3:00” position) and + y upward (towards “12:00”). (a) In unit-vector notation, we have (in centimeters) r r 1 2 10 10  . i and j Thus, Eq. 4-2 gives 2 1 ˆ ˆ 10i 10j . r r r   r r r Thus, the magnitude is given by 2 2 | | ( 10) ( 10) 14 cm. r   r (b) The angle is 1 10 tan 45 or 135 . 10 We choose 135 since the desired angle is in the third quadrant. In terms of the magnitude-angle notation, one may write 2 1 ˆ ˆ 10i 10j (14 135 ). r r r     r r r (c) In this case, r r r 1 2 10 10 20  j and j, and j cm. Thus, | | 20 cm. r r (d) The angle is given by 1 20 tan 90 . 0 116
(e) In a full-hour sweep, the hand returns to its starting position, and the displacement is zero.

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CH-04 physics notes - Chapter 4 1(a The magnitude of r is...

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