ch01 - CHAPTER 1 INTRODUCTION AND MATHEMATICAL CONCEPTS...

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CHAPTER 1 INTRODUCTION AND MATHEMATICAL CONCEPTS ANSWERS TO FOCUS ON CONCEPTS QUESTIONS 1. (d) The resultant vector R is drawn from the tail of the first vector to the head of the last vector. 2. (c) Note from the drawing that the magnitude R of the resultant vector R is equal to the shortest distance between the tail of A and the head of B . Thus, R is less than the magnitude (length) of A plus the magnitude of B . 3. (a) The triangle in the drawing is a right triangle. The lengths A and B of the two sides are known, so the Pythagorean theorem can be used to determine the length R of the hypotenuse. 4. (b) The angle is found by using the inverse tangent function, 1 4.0 km tan 53 3.0 km θ - = = ° ÷ . 5. (b) In this drawing the vector – C is reversed relative to C , while vectors A and B are not reversed. 6. (c) In this drawing the vectors – B and – C are reversed relative to B and C , while vector A is not reversed. 7. (e) These vectors form a closed four-sided polygon, with the head of the fourth vector exactly meeting the tail of the first vector. Thus, the resultant vector is zero. 8. (c) When the two vector components A x and A y are added by the tail-to-head method, the sum equals the vector A . Therefore, these vector components are the correct ones. 9. (b) The three vectors form a right triangle, so the magnitude of A is given by the Pythagorean theorem as 2 2 x y A A A = + . If A x and A y double in size, then the magnitude of A doubles: ( 29 ( 29 2 2 2 2 2 2 2 2 4 4 2 2 . x y x y x y A A A A A A A + = + = + = 10. (a) The angle is determined by the inverse tangent function, 1 tan y x A A - = ÷ . If A x and A y both become twice as large, the ratio does not change, and remains the same.
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2 INTRODUCTION AND MATHEMATICAL CONCEPTS 11. (b) The displacement vector A points in the – y direction. Therefore, it has no scalar component along the x axis ( A x = 0 m) and its scalar component along the y axis is negative. 12. (e) The scalar components are given by A x = - (450 m) sin 35.0 ° = - 258 m and A y = - (450 m) cos 35.0 ° = - 369 m. 13. (d) The distance (magnitude) traveled by each runner is the same, but the directions are different. Therefore, the two displacement vectors are not equal. 14. (c) A x and B x point in opposite directions, and A y and B y point in the same direction. 15. (d) 16. A y = 3.4 m, B y = 3.4 m 17. R x = 0 m, R y = 6.8 m 18. R = 7.9 m, θ = 2 1 degrees
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Chapter 1 Problems 3 CHAPTER 1 INTRODUCTION AND MATHEMATICAL CONCEPTS PROBLEMS ____________________________________________________________________________________________ 1. REASONING We use the fact that 1 m = 3.28 ft to form the following conversion factor: (1 m)/(3.28 ft) = 1. SOLUTION To convert ft 2 into m 2 , we apply the conversion factor twice: 2 Area = 1330 ft ( 29 1 m 3.28 ft 1 m 3.28 ft ÷ 2 124 m = ÷ _____________________________________________________________________________ 2. REASONING The word “per” indicates a ratio, so “0.35 mm per day” means 0.35 mm/d, which is to be expressed as a rate in ft/century. These units differ from the given units in both length and time dimensions, so both must be converted. For length, 1 m = 10 3 mm, and 1 ft = 0.3048 m. For time, 1 year = 365.24 days, and 1 century = 100 years. Multiplying the
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This note was uploaded on 03/30/2012 for the course PHYSICS 201 taught by Professor Rollino during the Fall '11 term at Rutgers.

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ch01 - CHAPTER 1 INTRODUCTION AND MATHEMATICAL CONCEPTS...

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