ism_ch03 - Chapter 3 Vectors in Physics Solutions to...

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Chapter 3 Vectors in Physics Solutions to Even-numbered Conceptual Questions 2. Vectors , , and are all equal to one another. Also, vector G A G G G J G I is the same as vector G . L 4. No. The component and the magnitude can be equal if the vector has only a single component. If the vector has more than one nonzero component, however, its magnitude will be greater than either of its components. 6. No. If a vector has a nonzero component, the smallest magnitude it can have is the magnitude of the component. 8. (a) The magnitude of 1.4 G A is equal to the magnitude of 2.2 G B . These vectors point in differen directions, however. (b) The x component of 1. t 4 G A is less than the y component of G because the x component is negative. The two components have the same magnitude, however. 2.2 B 10. The vector can point in the following directions: 45 ˚ , 135 ˚ , 225 ˚ , and 315 ˚ . In each of these directions G A A x = A y . 12. The vectors and must point in the same direction. G A G B 14. The direction angle for this vector must be greater 180 ˚ and less than 270 ˚ . 16. There are two possible direction angles for this vector, 135 ˚ and 315 ˚ . 18. When sailing upwind, your speed relative to the wind is greater than the speed of the wind itself. If you sail downwind, however, you move with the wind, and its speed relative to you is decreased. 20. (a) Relative to the ground, the aircraft being refueled has a velocity of 125 m/s due east, the same as the KC-10A. (b) Relative to the KC-10A, the aircraft being refueled has zero velocity. Solutions to Problems 1. y = height of the press box x = distance to second base tan y x θ = 38.0 ft 142 ft tan tan15.0 y x == = ° 36
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Physics: An Introduction Chapter 3: Vectors in Physics 2. y = elevation x = distance driven (a) () 11 5280 ft mi 550 ft sin sin (1.7 mi) 3.5 y x θ −− == (b) 150 ft 1 mi sin sin3.51 5280 ft 0.46 mi y x   °  = 3. 6 ft tan tan 3.43 100 ft y x = ° 4. (a) cos (75 m)cos35.0 61 m x rr = ° sin (75 m)sin35.0 43 m y = ° (b) (75 m)cos85.0 6.5 m x r = (75 m)sin85.0 75 m y r = 5. (a) ˆˆ (90 ft) +(90 ft) xy (b) ˆ (90 ft) y (c) (0 ft) +(0 ft) 6. h = height of the lighthouse r = distance from the base to the edge of the rocky cliff s = height of the sailor d = distance of the ship from the rocks y = h s x = d + r tan tan yh xd r s + = tan 49 ft –14 ft 19 ft tan30.0 42 ft hs dr =− ° = 37
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Chapter 3: Vectors in Physics Physics: An Introduction 38 ° 7. The lengths between the atoms form an isosceles triangle. By drawing a line segment between the center of the oxygen atom and the midpoint of the hydrogen atoms, two right triangles are formed. The opposite angle to the segment between a hydrogen atom and the midpoint is 104.5°/2 = 52.25°. Half the distance between hydrogen atoms is given by and the total distance is 2 y = 2(0.759 Å) = 1.5 Å. sin (0.96 Å)sin52.25 0.759 Å yr θ == = 8. (a) 11 9.5 m tan tan 14 m 34 y x r r −−    =− ° (b) 22 2 (14 m) ( 9.5 m) 17 m xy rrr =+ = + = 2 (c) The direction does not change. The magnitude is doubled.
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This note was uploaded on 04/07/2008 for the course PHYS 105 taught by Professor Klie during the Spring '08 term at Ill. Chicago.

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ism_ch03 - Chapter 3 Vectors in Physics Solutions to...

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