# hw36 - This print-out should have 9 questions...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 9.75 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 π radicalBigg L g Given: T = 9 . 75 s g = 9 . 81 m / s 2 Solution: parenleftbigg T 2 π parenrightbigg 2 = L g L = g parenleftbigg T 2 π parenrightbigg 2 = ( 9 . 81 m / s 2 ) parenleftbigg 9 . 75 s 2 π 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 φ θ What is its length? Correct answer: 0 . 273681 m. Explanation: Basic Concepts: Angular frequency of simple pendulum, is ω = radicalbigg g ℓ , and its period is T = 2 π ω . Solution: The period is T = 2 π ω = 2 π radicalBigg ℓ g , so we can solve for ℓ to find ℓ = T 2 g 4 π 2 = (1 . 05 s) 2 (9 . 8 m / s 2 ) 4 π 2 = 0 . 273681 m ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 4

hw36 - This print-out should have 9 questions...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online