Physics Solution Manual for 1100 and 2101

# These two equations will allow us to find the

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Unformatted text preview: hat the x-component of this equation must be zero (Ax + Bx + Cx = 0) and the y-component must be zero (Ay + By + Cy = 0). These two equations will allow us to find the magnitudes of B and C. +y (north) A −x 145 units 35.0° B 65.0° +x (east) 15.0° C −y SOLUTION The x- and y-components of A, B, and C are given in the table below. The plus and minus signs indicate whether the components point along the positive or negative axes. Vector x Component y Component A –(145 units) cos 35.0° = –119 units +(145 units) sin 35.0° = +83.2 units B C +B sin 65.0° = +B (0.906) –C sin 15.0° = –C (0.259) +B cos 65.0° = +B (0.423) –C cos 15.0° = –C (0.966) A+B+C –119 units + B (0.906) – C (0.259) +83.2 units + B (0.423) – C (0.966) Setting the separate x- and y- components of A + B + C equal to zero gives x-component (–119 units) + B (0.906) – C (0.259) = 0 y-component (+83.2 units) + B (0.423) – C (0.966) = 0 Solving these two equations simultaneously, we find that a. B = 178 units b. C = 164 units Chapter 1 Problems 35 56. REASONING The following table shows the components of the individual displacements and the components of the resultant. The directions due east and due north are taken as the positive directions. East/West Component Displacement (1) (2) (3) (4) North/South Component –27.0 cm 0 –(23.0 cm) cos 35.0° = –18.84 cm –(23.0 cm) sin 35.0° = –13.19 cm (28.0 cm) cos 55.0° = 16.06 cm –(28.0 cm) sin 55.0° = –22.94 cm (35.0 cm) cos 63.0° = 15.89 cm (35.0 cm) sin 63.0° = 31.19 cm –13.89 cm Resultant –4.94 cm SOLUTION a. From the Pythagorean theorem, we find that the magnitude of the resultant displacement vector is 13.89 cm R = (13.89 cm) 2 + (4.94 cm) 2 = 14.7 cm R b. The angle θ is given by θ = tan −1 F4.94 cm I = G cm J HK 13.89 θ 4.94 cm 19 .6° , south of west 57. REASONING AND SOLUTION The following figure (not drawn to scale) shows the geometry of the situation, when the observer is a distance r from the base of the arch. The angle θ is related to r and h by tan θ = h / r . Solving for r, we find h = 192 m θ h 192 m r= = = 5.5 × 10 3 m = 5.5 km r tanθ tan 2.0° ______________________________________________________________________________ 58. REASONING AND SOLUTION In the diagram below, θ = 14.6° and h = 2830 m. We know that sin θ = H/h and, therefore, h H = h sin θ = (2830 m) sin 14.6° = 713 m H θ 36 INTRODUCTION AND MATHEMATICAL CONCEPTS 59. SSM REASONING AND SOLUTION In order to determine which vector has the largest x and y components, we calculate the magnitude of the x and y components explicitly and compare them. In the calculations, the symbol u denotes the units of the vectors. Ax = (100.0 u) cos 90.0° = 0.00 u 2 Bx = (200.0 u) cos 60.0° = 1.00 × 10 u Cx = (150.0 u) cos 0.00° = 150.0 u 2 Ay = (100.0 u) sin 90.0° = 1.00 × 10 u By = (200.0 u) sin 60.0° = 173 u Cy = (150.0 u) sin 0.00° = 0.00 u a. C has the largest x component. b. B has the largest y component. ______________________________________________________________________________ 60. REASONING Multiplying an equation by a factor of 1 does not alter the equation; this is the basis of our solution. We will use factors of 1 in the following forms: 1 gal = 1 , since 1 gal = 128 oz 128 oz 3.785 ×10−3 m3 = 1 , since 3.785 × 10−3 m3 = 1 gal 1 gal 1 mL = 1 , since 1 mL = 10−6 m3 −6 3 10 m SOLUTION The starting point for our solution is the fact that Volume = 1 oz Multiplying this equation on the right by factors of 1 does not alter the equation, so it follows that 1 gal 3.785 ×10−3 m3 1 mL Volume = (1 oz )(1)(1)(1) = 1 oz = 29.6 mL 128 oz 10−6 m3 1 gal ( ) Note that all the units on the right, except one, are eliminated algebraically, leaving only the desired units of milliliters (mL). Chapter 1 Problems 37 61. REASONING AND SOLUTION The east and north components are, respectively a. Ae = A cos θ = (155 km) cos 18.0° = b. An = A sin θ = (155 km) sin 18.0° = 147 km 47.9 km _____________________________________________________________________________________________ 62. REASONING We will use the scalar x and y components of the resultant vector to obtain its magnitude and direction. To obtain the x component of the resultant we will add together the x components of each of the vectors. To obtain the y component of the resultant we will add together the y components of each of the vectors. +y SOLUTION The x and y components of the resultant vector R are Rx and Ry, respectively. In terms of these components, the magnitude R and the B directional angle θ (with respect to the x axis) A for the resultant are R= 2 Rx 2 + Ry and Ry θ = tan −1 R x 20.0° 35.0° (1) C The following table summarizes the components of the individual vectors shown in the drawing: Vector +x 50.0° x component D y component A Ax = − (16.0 m ) cos 20.0° = −15.0 m Ay = (16.0 m ) sin 20.0° = 5.47 m B Bx = 0 m B y = 11.0 m C C x = − (12.0 m ) cos 35.0° = −9.83 m C y = − (12.0 m ) sin 35.0° = −6.88 m D Dx = ( 26.0 m ) cos 50.0° = 16.7 m Dy = − ( 26.0 m ) sin 50.0...
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## This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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