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· Ax= Part II: Vectors À (Ax,y) -(b₁₁by) = A + B = ((x,y) Cx = Ax+Bx = Ay+By 1. A vector A has a magnitude of 10.0 m and points in a direction that makes an angle of 30.0° above the positive x-axis. A second vector B, has a magnitude of 15.0 m and points in a direction 60.0" below the negative x-axis. a) Draw the two vectors and their components. Al cos 30°: 8.66m Ay = 141 sing05m Bx=-181 (05600-7.5m By=-161i0-12m Ax if @ & IAI given A=(Ax, Ag) if given Ax A IATA A Ay tong = 10m A 230° 60° x Ax = Al-case A: AI-sine 15m B b) Find the magnitude of the vector's components. COS300 A 10m (10m) cos 30° 8.66 - A A sin 500= 0 10 m (10m) Sin (30) = 0 S 0 Sin 60° 15m (15m) sin (60°) O -12.99m-0 C0160°= 4. (15m) Cos(60) A -7.5=A Ax = 8.66m Ay = 5.0m By = -7.5m By= -133 c) Using the component method to find the magnitude (II) and direction (Dir()) of the resultant vector C = A + B. C = ((x, y) Cy: Ay+By S+ (-13) -8 Cx = Ax+Bx (x 8.66m+(-7.5m) = 1.16 (1.16)²+(-8)2 65.3456 0 -8 8.08 - 8.08 1.16 tan (11) = = 0 |c|= 8.08m Dir()=87 below tx

· Ax= Part II: Vectors À (Ax,y) -(b₁₁by) = A + B = ((x,y) Cx = Ax+Bx = Ay+By 1. A vector A has a magnitude of 10.0 m and points in a direction that makes an angle of 30.0° above the positive x-axis. A second vector B, has a magnitude of 15.0 m and points in a direction 60.0" below the negative x-axis. a) Draw the two vectors and their components. Al cos 30°: 8.66m Ay = 141 sing05m Bx=-181 (05600-7.5m By=-161i0-12m Ax if @ & IAI given A=(Ax, Ag) if given Ax A IATA A Ay tong = 10m A 230° 60° x Ax = Al-case A: AI-sine 15m B b) Find the magnitude of the vector's components. COS300 A 10m (10m) cos 30° 8.66 - A A sin 500= 0 10 m (10m) Sin (30) = 0 S 0 Sin 60° 15m (15m) sin (60°) O -12.99m-0 C0160°= 4. (15m) Cos(60) A -7.5=A Ax = 8.66m Ay = 5.0m By = -7.5m By= -133 c) Using the component method to find the magnitude (II) and direction (Dir()) of the resultant vector C = A + B. C = ((x, y) Cy: Ay+By S+ (-13) -8 Cx = Ax+Bx (x 8.66m+(-7.5m) = 1.16 (1.16)²+(-8)2 65.3456 0 -8 8.08 - 8.08 1.16 tan (11) = = 0 |c|= 8.08m Dir()=87 below tx

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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· Ax= Part II: Vectors À (Ax,y) -(b₁₁by) = A + B = ((x,y) Cx = Ax+Bx = Ay+By 1. A vector A has a magnitude of 10.0 m and points in a direction that makes an angle of 30.0° above the positive x-axis. A second vector B, has a magnitude of 15.0 m and points in a direction 60.0" below the negative x-axis. a) Draw the two vectors and their components. Al cos 30°: 8.66m Ay = 141 sing05m Bx=-181 (05600-7.5m By=-161i0-12m Ax if @ & IAI given A=(Ax, Ag) if given Ax A IATA A Ay tong = 10m A 230° 60° x Ax = Al-case A: AI-sine 15m B b) Find the magnitude of the vector's components. COS300 A 10m (10m) cos 30° 8.66 - A A sin 500= 0 10 m (10m) Sin (30) = 0 S 0 Sin 60° 15m (15m) sin (60°) O -12.99m-0 C0160°= 4. (15m) Cos(60) A -7.5=A Ax = 8.66m Ay = 5.0m By = -7.5m By= -133
Transcribed Image Text:· Ax= Part II: Vectors À (Ax,y) -(b₁₁by) = A + B = ((x,y) Cx = Ax+Bx = Ay+By 1. A vector A has a magnitude of 10.0 m and points in a direction that makes an angle of 30.0° above the positive x-axis. A second vector B, has a magnitude of 15.0 m and points in a direction 60.0" below the negative x-axis. a) Draw the two vectors and their components. Al cos 30°: 8.66m Ay = 141 sing05m Bx=-181 (05600-7.5m By=-161i0-12m Ax if @ & IAI given A=(Ax, Ag) if given Ax A IATA A Ay tong = 10m A 230° 60° x Ax = Al-case A: AI-sine 15m B b) Find the magnitude of the vector's components. COS300 A 10m (10m) cos 30° 8.66 - A A sin 500= 0 10 m (10m) Sin (30) = 0 S 0 Sin 60° 15m (15m) sin (60°) O -12.99m-0 C0160°= 4. (15m) Cos(60) A -7.5=A Ax = 8.66m Ay = 5.0m By = -7.5m By= -133
c) Using the component method to find the magnitude (II) and direction (Dir()) of the resultant vector C = A + B. C = ((x, y) Cy: Ay+By S+ (-13) -8 Cx = Ax+Bx (x 8.66m+(-7.5m) = 1.16 (1.16)²+(-8)2 65.3456 0 -8 8.08 - 8.08 1.16 tan (11) = = 0 |c|= 8.08m Dir()=87 below tx
Transcribed Image Text:c) Using the component method to find the magnitude (II) and direction (Dir()) of the resultant vector C = A + B. C = ((x, y) Cy: Ay+By S+ (-13) -8 Cx = Ax+Bx (x 8.66m+(-7.5m) = 1.16 (1.16)²+(-8)2 65.3456 0 -8 8.08 - 8.08 1.16 tan (11) = = 0 |c|= 8.08m Dir()=87 below tx
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