Physics hw 3 solutions - mata (jpm2873) HW #3 Antoniewicz...

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Unformatted text preview: mata (jpm2873) 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 (part 1 of 2) 10.0 points Vector vector A has x and y components of 20 cm and 8 . 1 cm , respectively; vector vector B has x and y components of 18 . 9 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: 12 . 9667 cm. Explanation: Let : ( A x , A y ) = ( 20 cm , 8 . 1 cm) and ( B x , B y ) = (18 . 9 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 [18 . 9 cm ( 20 cm)] = 12 . 9667 cm . 002 (part 2 of 2) 10.0 points What is the y component of vector C ? Correct answer: 9 . 36667 cm. Explanation: C y = 1 3 ( B y A y ) = 1 3 [( 20 cm (8 . 1 cm)] = 9 . 36667 cm . 003 10.0 points Initially (at time t = 0) a particle is mov- ing vertically at 4 . 8 m / s and horizontally at 0 m / s. Its horizontal acceleration is 1 . 8 m / s 2 . At what time will the particle be traveling at 32 with respect to the horizontal? The acceleration due to gravity is 9 . 8 m / s 2 . Correct answer: 0 . 439369 s. Explanation: Let : v y = 4 . 8 m / s , g = 9 . 8 m / s 2 , v x = 0 , and a = 1 . 8 m / s 2 , and = 32 . v x t v y t v t 32 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 = v y t v x t = v y g t a t a t tan = v y g t a t tan + g t = v y t = v y a tan + g = 4 . 8 m / s (1 . 8 m / s 2 ) tan(32 ) + 9 . 8 m / s 2 = . 439369 s . 004 (part 1 of 2) 10.0 points mata (jpm2873) HW #3 Antoniewicz (57420) 2 A particle undergoes three displacements. The first has a magnitude of 15 m and makes an angle of 24 with the positive x axis. The second has a magnitude of 8 . 4 m and makes an angle of 143 with the positive x axis. (see the figure below). After the third displacement the particle returns to its initial position. 143 24 1 5 m 8 . 4 m Find the magnitude of the third displace- ment. Correct answer: 13 . 1677 m. Explanation: Let : bardbl vector A bardbl = 15 m , a = 24 , bardbl vector B bardbl = 8 . 4 m , and B = 143 . C A 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 A B cos b = (15 m) cos 24 (8 . 4 m) cos 143 = 6 . 99464 m and C y = A y B x = A sin A B sin b = (15 m) sin24 (8 . 4 m) sin143 = 11 . 1563 m , so the magnitude of vector...
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Physics hw 3 solutions - mata (jpm2873) HW #3 Antoniewicz...

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