hw2b_part2 - HW#2b part 2 8. Page 1 of 3 9. 10. Practice...

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HW#2b part 2 Page 1 of 3 8. 9. 10. Practice with vectors: (important!) For each of the vectors below, I.) do a rough sketch of it II.) convert it to the form it is not already in (size&direction, or components) (Note: cw=clockwise, ccw=counterclockwise) a.) = 5 m, 30 o ccw from +x axis D x = m D y = m y x A r A y A x θ
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HW#2b part 2 Page 2 of 3 Answers. Here and later we will use the relations between the (magnitude, angle) and component representations of a vector: 22 cos sin tan xy x y y x AA A A A θ =+ = ⎬⎨ = = So for part (a) , () cos 5 m cos30 4.33 m D sin 5 m sin30 2.5 m x y DD D == ° = ° = b.) = 15 m/s, 10 o cw from –y axis v av x = -2.604 m/s v av y = -14.77 m/s Answer. Look at the diagram - the angle, measured in the usual direction from the positive x axis, is equal to 260 ° . And both x and y components of the vector are negative. Let's see if the trig functions get this right automatically. cos 15 m/s cos260 2.604 m/s sin 15 m/s sin 260 14.77 m/s x y ° = ° = c.) Q x = -6 m, Q y = +2 m = 6.325m, 18.43 o cw from –x axis Answer. Here we make the transformation back from components to magnitude and angle. ( ) 6 m 2 m 6.325 m QQ Q =− + = ( ) tan .33333
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This note was uploaded on 08/25/2010 for the course PHYS 111 taught by Professor Jamesm.lockhart during the Spring '09 term at S.F. State.

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hw2b_part2 - HW#2b part 2 8. Page 1 of 3 9. 10. Practice...

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