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Unformatted text preview: wolz (cmw2833) – HW #3 – Antoniewicz – (57420) 1 This printout should have 19 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points Vector vector A has x and y components of − 20 cm and 7 . 8 cm , respectively; vector vector B has x and y components of 5 . 92 cm and − 20 cm , respec tively. If vector A − vector B +3 vector C = 0, what is the x component of vector C ? Correct answer: 8 . 64 cm. Explanation: Let : ( A x , A y ) = ( − 20 cm , 7 . 8 cm) and ( B x , B y ) = (5 . 92 cm , − 20 cm) . The sum of the vectors is the sum of indi vidual components of each vector: 3 vector C = vector B − vector A vector C = 1 3 parenleftBig vector B − vector A parenrightBig C x = 1 3 ( B x − A x ) = 1 3 [5 . 92 cm − ( − 20 cm)] = 8 . 64 cm . 002 (part 2 of 2) 10.0 points What is the y component of vector C ? Correct answer: − 9 . 26667 cm. Explanation: C y = 1 3 ( B y − A y ) = 1 3 [( − 20 cm − (7 . 8 cm)] = − 9 . 26667 cm . 003 10.0 points Initially (at time t = 0) a particle is mov ing vertically at 6 . 3 m / s and horizontally at 0 m / s. Its horizontal acceleration is 1 . 4 m / s 2 . At what time will the particle be traveling at 57 ◦ with respect to the horizontal? The acceleration due to gravity is 9 . 8 m / s 2 . Correct answer: 0 . 52694 s. Explanation: Let : v y = 6 . 3 m / s , g = 9 . 8 m / s 2 , v x = 0 , and a = 1 . 4 m / s 2 , and θ = 57 ◦ . v x t v y t v t 57 ◦ The vertical velocity is v y t = v y − g t and the horizontal velocity is v x t = v y + a t = a t . The vertical component is the opposite side and the horizontal component the adjacent side to the angle, so tan θ = v y t v x t = v y − g t a t a t tan θ = v y − g t a t tan θ + g t = v y t = v y a tan θ + g = 6 . 3 m / s (1 . 4 m / s 2 ) tan(57 ◦ ) + 9 . 8 m / s 2 = . 52694 s . wolz (cmw2833) – HW #3 – Antoniewicz – (57420) 2 004 (part 1 of 2) 10.0 points A particle undergoes three displacements. The first has a magnitude of 15 m and makes an angle of 31 ◦ with the positive x axis. The second has a magnitude of 6 . 2 m and makes an angle of 161 ◦ with the positive x axis. (see the figure below). After the third displacement the particle returns to its initial position. 161 ◦ 31 ◦ 1 5 m 6 . 2 m Find the magnitude of the third displace ment. Correct answer: 11 . 9951 m. Explanation: Let : bardbl vector A bardbl = 15 m , θ a = 31 ◦ , bardbl vector B bardbl = 6 . 2 m , and θ B = 161 ◦ . θ C θ A θ B A B C − C vector A + vector B + vector C = 0 , so vector C = − vector A − vector B C x = − A x − B x = − A cos θ A − B cos θ b = − (15 m) cos31 ◦ − (6 . 2 m) cos161 ◦ = − 6 . 99529 m and C y = − A y − B x = − A sin θ A − B sin θ b = − (15 m) sin31 ◦ − (6 . 2 m) sin161 ◦ = − 9 . 74409 m , so the magnitude of...
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This note was uploaded on 05/02/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics

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