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HW #3-solutions

# HW #3-solutions - rathbun(rar2954 – HW#3 – Antoniewicz...

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Unformatted text preview: rathbun (rar2954) – HW #3 – Antoniewicz – (57380) 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 (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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HW #3-solutions - rathbun(rar2954 – HW#3 – Antoniewicz...

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