Homework 36 - portillo (gdp347) homework 36 Turner (58185)...

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Unformatted text preview: portillo (gdp347) homework 36 Turner (58185) 1 This print-out should have 9 questions. Multiple-choice 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 9.97 s. The acceleration of gravity is 9 . 81 m / s 2 . What is the height of the tower? Correct answer: 24 . 7001 m. Explanation: Basic Concept: T = 2 radicalBigg L g Given: T = 9 . 97 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 . 97 s 2 parenrightbigg 2 = 24 . 7001 m 002 (part 1 of 2) 10.0 points A simple pendulum has a period of 3 . 03 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: 2 . 27904 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 = (3 . 03 s) 2 (9 . 8 m / s 2 ) 4 2 = 2 . 27904 m ....
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Homework 36 - portillo (gdp347) homework 36 Turner (58185)...

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