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Unformatted text preview: This printout should have 9 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points A visitor to a lighthouse wishes to determine the height of the tower. The visitor ties a spool of thread to a small rock to make a simple pendulum, then hangs the pendulum down a spiral staircase in the center of the tower. The period of oscillation is 9.75 s. The acceleration of gravity is 9 . 81 m / s 2 . What is the height of the tower? Correct answer: 23 . 6221 m. Explanation: Basic Concept: T = 2 radicalBigg L g Given: T = 9 . 75 s g = 9 . 81 m / s 2 Solution: parenleftbigg T 2 parenrightbigg 2 = L g L = g parenleftbigg T 2 parenrightbigg 2 = ( 9 . 81 m / s 2 ) parenleftbigg 9 . 75 s 2 parenrightbigg 2 = 23 . 6221 m 002 (part 1 of 2) 10.0 points A simple pendulum has a period of 1 . 05 s. The acceleration of gravity is 9 . 8 m / s 2 . g m v negationslash = 0 r a r a t a What is its length? Correct answer: 0 . 273681 m. Explanation: Basic Concepts: Angular frequency of simple pendulum, is = radicalbigg g , and its period is T = 2 . Solution: The period is T = 2 = 2 radicalBigg g , so we can solve for to find = T 2 g 4 2 = (1 . 05 s) 2 (9 . 8 m / s 2 ) 4 2 = 0 . 273681 m ....
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 Spring '08
 Turner
 Work, Light

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