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Unformatted text preview: rathbun (rar2954) – HW #3 – Antoniewicz – (57380) 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 A has x and y components of 20 cm and 1 . 5 cm , respectively; vector B has x and y components of 18 . 8 cm and 20 cm , respec tively. If A B +3 C = 0, what is the x component of C ? Correct answer: 12 . 9333 cm. Explanation: Let : ( A x , A y ) = ( 20 cm , 1 . 5 cm) and ( B x , B y ) = (18 . 8 cm , 20 cm) . The sum of the vectors is the sum of indi vidual components of each vector: 3 C = B A C = 1 3 B A C x = 1 3 ( B x A x ) = 1 3 [18 . 8 cm ( 20 cm)] = 12 . 9333 cm . 002 (part 2 of 2) 10.0 points What is the y component of C ? Correct answer: 7 . 16667 cm. Explanation: C y = 1 3 ( B y A y ) = 1 3 [( 20 cm (1 . 5 cm)] = 7 . 16667 cm . 003 10.0 points Initially (at time t = 0) a particle is mov ing vertically at 7 . 5 m / s and horizontally at 0 m / s. Its horizontal acceleration is 1 . 9 m / s 2 . At what time will the particle be traveling at 42 ◦ with respect to the horizontal? The acceleration due to gravity is 9 . 8 m / s 2 . Correct answer: 0 . 651564 s. Explanation: Let : v y = 7 . 5 m / s , g = 9 . 8 m / s 2 , v x = 0 , and a = 1 . 9 m / s 2 , and θ = 42 ◦ . v x t v y t v t 42 ◦ 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 = 7 . 5 m / s (1 . 9 m / s 2 ) tan(42 ◦ ) + 9 . 8 m / s 2 = . 651564 s . rathbun (rar2954) – HW #3 – Antoniewicz – (57380) 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 45 ◦ with the positive x axis. The second has a magnitude of 8 . 8 m and makes an angle of 149 ◦ with the positive x axis. (see the figure below). After the third displacement the particle returns to its initial position. 149 ◦ 45 ◦ 1 5 m 8 . 8 m Find the magnitude of the third displace ment. Correct answer: 15 . 4458 m. Explanation: Let : A = 15 m , θ a = 45 ◦ , B = 8 . 8 m , and θ B = 149 ◦ . θ C θ A θ B A B C C A + B + C = 0 , so C = A B C x = A x B x = A cos θ A B cos θ b = (15 m) cos45 ◦ (8 . 8 m) cos149 ◦ = 3 . 06353 m and C y = A y B x = A sin θ A B sin θ b = (15 m) sin45 ◦ (8 . 8 m) sin149 ◦ = 15 . 1389 m , so the magnitude of C is C = C...
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 Spring '08
 Turner
 Physics, Acceleration, Correct Answer, rathbun

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