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Unformatted text preview: 77 Chapter 3 1. A vector a r can be represented in the magnitudeangle notation ( a , θ ), where 2 2 x y a a a = + is the magnitude and 1 tan y x a a θ = is the angle a r makes with the positive x axis. (a) Given A x =  25.0 m and A y = 40.0 m, 2 2 ( 25.0 m) (40.0 m) 47.2 m A = + = (b) Recalling that tan θ = tan ( θ + 180°), tan –1 [(40.0 m)/ (– 25.0 m)] = – 58° or 122°. Noting that the vector is in the third quadrant (by the signs of its x and y components) we see that 122° is the correct answer. The graphical calculator “shortcuts” mentioned above are designed to correctly choose the right possibility. 2. The angle described by a full circle is 360° = 2 π rad, which is the basis of our conversion factor. (a) ( 29 2 rad 20.0 20.0 0.349 rad 360 π ° = ° = ° . (b) ( 29 2 rad 50.0 50.0 0.873 rad 360 π ° = ° = ° . (c) ( 29 2 rad 100 100 1.75 rad 360 π ° = ° = ° . (d) ( 29 360 0.330 rad = 0.330 rad 18.9 2 rad π ° = ° . (e) ( 29 360 2.10 rad = 2.10 rad 120 2 rad π ° = ° . (f) ( 29 360 7.70 rad = 7.70 rad 441 2 rad π ° = ° . CHAPTER 3 78 3. The x and the y components of a vector r a lying on the xy plane are given by cos , sin x y a a a a θ θ = = where   a a = r is the magnitude and θ is the angle between r a and the positive x axis. (a) The x component of r a is given by a x = 7.3 cos 250° = – 2.5 m. (b) and the y component is given by a y = 7.3 sin 250° = – 6.9 m. In considering the variety of ways to compute these, we note that the vector is 70° below the – x axis, so the components could also have been found from a x = – 7.3 cos 70° and a y = – 7.3 sin 70°. In a similar vein, we note that the vector is 20° to the left from the – y axis, so one could use a x = – 7.3 sin 20° and a y = – 7.3 cos 20° to achieve the same results. 4. (a) The height is h = d sin θ , where d = 12.5 m and θ = 20.0°. Therefore, h = 4.28 m. (b) The horizontal distance is d cos θ = 11.7 m. 5. The vector sum of the displacements r d storm and v d new must give the same result as its originally intended displacement o ˆ (120 km)j d = r where east is $ i , north is $ j . Thus, we write storm new ? ? (100 km) i , i j. d d A B = = + r r (a) The equation storm new o d d d + = r r r readily yields A = –100 km and B = 120 km. The magnitude of r d new is therefore equal to 2 2 new   156 km d A B = + = r . (b) The direction is tan –1 ( B / A ) = –50.2° or 180° + ( –50.2°) = 129.8°. We choose the latter value since it indicates a vector pointing in the second quadrant, which is what we expect here. The answer can be phrased several equivalent ways: 129.8° counterclockwise from east, or 39.8° west from north, or 50.2° north from west....
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This note was uploaded on 08/09/2010 for the course PHY 202 101A2 taught by Professor Prof.yang during the Summer '10 term at National Taiwan University.
 Summer '10
 Prof.Yang
 Physics

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