HW #3-solutions

# HW #3-solutions - wolz(cmw2833 HW#3 Antoniewicz(57420 This...

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wolz (cmw2833) – HW #3 – Antoniewicz – (57420) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001(part1of2)10.0points 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(part2of2)10.0points 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 10.0points 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 0 = 6 . 3 m / s , g = 9 . 8 m / s 2 , v x 0 = 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 0 g t and the horizontal velocity is v x t = v y 0 + 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 0 g t a t a t tan θ = v y 0 g t a t tan θ + g t = v y 0 t = v y 0 a tan θ + g = 6 . 3 m / s (1 . 4 m / s 2 ) tan(57 ) + 9 . 8 m / s 2 = 0 . 52694 s .

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wolz (cmw2833) – HW #3 – Antoniewicz – (57420) 2 004(part1of2)10.0points 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 15 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) cos 31 (6 . 2 m) cos 161 = 6 . 99529 m and C y = A y B x = A sin θ A B sin θ b = (15 m) sin 31 (6 . 2 m) sin 161 = 9 . 74409 m , so the magnitude of vector C is bardbl vector C bardbl = radicalBig C 2 x + C 2 y = radicalBig ( 6 . 99529 m) 2 + ( 9 . 74409 m) 2 = 11 . 9951 m .
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