Problem_Set_2_Solutions

Problem_Set_2_Solutions - 22. The desired result is the...

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22. The desired result is the displacement vector, in units of km, A = (5.6 km), 90º (measured counterclockwise from the + x axis), or ˆ (5.6 km)j A = G , where ˆ j is the unit vector along the positive y axis (north). This consists of the sum of two displacements: during the whiteout, (7.8 km), 50 B G , or ˆˆ ˆ ˆ (7.8 km)(cos50 i sin50 j) (5.01 km)i (5.98 km)j B + ° = + G and the unknown C G . Thus, ABC =+ GG G . (a) The desired displacement is given by ( 5.01 km) i (0.38 km) j CAB =−=− G . The magnitude is 22 ( 5.01 km) ( 0.38 km) 5.0 km. −+ = (b) The angle is 1 tan [( 0.38 km)/( 5.01 km)] 4.3 , −− = ° south of due west.
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44. The two vectors are written as, in unit of meters, 11 1 2 2 2 ˆˆˆˆ 4.0i+5.0j i j, 3.0i+4.0j i j x yx y dd d d d d == + = −= + GG (a) The vector (cross) product gives 12 1 2 1 2 ˆˆ ˆ ( )k [(4.0)(4.0) (5.0)( 3.0)]k=31 k xy dd d d dd ×= = (b) The scalar (dot) product gives 12 12 (4.0)( 3.0) (5.0)(4.0) 8.0. xx yy dd dd dd ⋅= + = + = (c) 22 2 1221 ( ) 8.0 ( 3.0) (4.0) 33. ddd ddd +⋅ = += + + = GGG G G (d) Note that the magnitude of the d 1 vector is 16+25 = 6.4. Now, the dot product is (6.4)(5.0)cos θ = 8. Dividing both sides by 32 and taking the inverse cosine yields = 75.5 ° . Therefore the component of the d 1 vector along the direction of the d 2 vector is 6.4cos 1.6.
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(h) The y component of the net displacement net d G is ,123 ( 0.28 m) (0 m) (0.52 m) 0.24 m. net y y y y dd d d =++= + + = (i) The magnitude of the net displacement is 22 2 2 ,, (0.52 m) (0.24 m) 0.57 m. net net x net y d =+ = + = (j) The direction of the net displacement is , 11 , 0.24 m tan tan 25 (north of east) 0.52 m net y net x d d θ −− §· == = ° ¨¸ ©¹ If the ant has to return directly to the starting point, the displacement would be net d G . (k) The distance the ant has to travel is | | 0.57 m. net d −= G (l) The direction the ant has to travel is 25 (south of west) ° . 72. The ant’s trip consists of three displacements: 1 2 3 ˆˆ ˆ ˆ (0.40 m)(cos225 i sin 225 j) ( 0.28 m)i ( 0.28 m) j ˆ (0.50 m)i ˆ ˆ (0.60 m)(cos60 i sin 60 j) (0.30 m)i (0.52 m)j, d d d + ° = + = + ° = + G G G
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This note was uploaded on 11/16/2009 for the course PHYSICS 11a taught by Professor Silvera during the Fall '07 term at Harvard.

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Problem_Set_2_Solutions - 22. The desired result is the...

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