physics HW3

# physics HW3 - rose(kmr2427 – HW#3 – Antoniewicz...

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Unformatted text preview: rose (kmr2427) – 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 vector A has x and y components of − 20 cm and 7 . 9 cm , respectively; vector vector B has x and y components of 19 . 3 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: 13 . 1 cm. Explanation: Let : ( A x , A y ) = ( − 20 cm , 7 . 9 cm) and ( B x , B y ) = (19 . 3 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 [19 . 3 cm − ( − 20 cm)] = 13 . 1 cm . 002 (part 2 of 2) 10.0 points What is the y component of vector C ? Correct answer: − 9 . 3 cm. Explanation: C y = 1 3 ( B y − A y ) = 1 3 [( − 20 cm − (7 . 9 cm)] = − 9 . 3 cm . 003 10.0 points Initially (at time t = 0) a particle is moving vertically at 7 m / s and horizontally at 0 m / s. Its horizontal acceleration is 2 m / s 2 . At what time will the particle be traveling at 40 ◦ with respect to the horizontal? The acceleration due to gravity is 9 . 8 m / s 2 . Correct answer: 0 . 609852 s. Explanation: Let : v y = 7 m / s , g = 9 . 8 m / s 2 , v x = 0 , and a = 2 m / s 2 , and θ = 40 ◦ . v x t v y t v t 40 ◦ 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 m / s (2 m / s 2 ) tan(40 ◦ ) + 9 . 8 m / s 2 = . 609852 s . rose (kmr2427) – 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 46 ◦ with the positive x axis. The second has a magnitude of 8 . 1 m and makes an angle of 157 ◦ with the positive x axis. (see the figure below). After the third displacement the particle returns to its initial position. 157 ◦ 46 ◦ 1 5 m 8 . 1 m Find the magnitude of the third displace- ment. Correct answer: 14 . 2663 m. Explanation: Let : bardbl vector A bardbl = 15 m , θ a = 46 ◦ , bardbl vector B bardbl = 8 . 1 m , and θ B = 157 ◦ . 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) cos46 ◦ − (8 . 1 m) cos157 ◦ = − 2 . 96379 m and C y = − A y − B x = − A sin θ A − B sin θ b = − (15 m) sin46 ◦ − (8 . 1 m) sin157 ◦ = − 13 . 955 m , so the magnitude of vector C is bardbl vector C...
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physics HW3 - rose(kmr2427 – HW#3 – Antoniewicz...

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