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Unformatted text preview: bello (rtb473) – hw1 – Demkov – (59910) 1 This printout should have 51 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A human hair is approximately 51 μ m in di ameter. Express this diameter in meters. Correct answer: 5 . 1 × 10 − 5 m. Explanation: Basic Concept: 1 μ m = 1 × 10 − 6 m Let: diameter = 51 μ m Solution: diameter = 51 μ m = ( 5 . 1 × 10 1 μ m ) · 1 × 10 − 6 m 1 μ m = 5 . 1 × 10 − 5 m keywords: /* If you use any of these, fix the comment symbols. 002 (part 1 of 3) 10.0 points A hydrogen atom has a diameter of about 10 nm. a) Express this diameter in meters. Correct answer: 1 × 10 − 8 m. Explanation: Basic Concept: 1 nm = 1 × 10 − 9 m 1 μ m = 1 × 10 − 6 m 1 mm = 1 × 10 − 3 m Let: diameter = 10 nm Solution: diameter = 10 nm = ( 1 × 10 1 nm ) · 1 × 10 − 9 m 1 nm = 1 × 10 − 8 m 003 (part 2 of 3) 10.0 points b) Express this diameter in millimeters. Correct answer: 1 × 10 − 5 mm. Explanation: Solution: diameter = ( 1 × 10 − 8 m ) · 1 mm 1 × 10 − 3 m = 1 × 10 − 5 mm 004 (part 3 of 3) 10.0 points c) Express this diameter in micrometers. Correct answer: 0 . 01 μ m. Explanation: Solution: diameter = ( 1 × 10 − 8 m ) · 1 μ m 1 × 10 − 6 m = 1 × 10 − 2 μ m keywords: 005 10.0 points A typical radio wave has a period of 1 . 4 μ s. Express this period in seconds. Correct answer: 1 . 4 × 10 − 6 s. Explanation: Basic Concept: 1 μ s = 1 × 10 − 6 s Let: period = 1 . 4 μ s Solution: period = (1 . 4 μ s) · 1 × 10 − 6 s 1 μ s = 1 . 4 × 10 − 6 s keywords: /* If you use any of these, fix the comment symbols. bello (rtb473) – hw1 – Demkov – (59910) 2 006 10.0 points The standard of time is based on 1. a precision pendulum clock. 2. the yearly revolution of the earth about the sun. 3. None of these correct 4. the frequency of light emitted by 86 Kr. 5. the daily rotation of the earth. Explanation: The second is defined from oscillations of cesium133 atoms. 007 10.0 points You can obtain a rough estimate of the size of a molecule by the following simple exper iment. Let a droplet of oil spread out on a smooth water surface. The resulting oil slick will be approximately one molecule thick. Given an oil droplet of mass of 9 . 03 × 10 − 7 kg and density of 935 kg / m 3 that spreads out into a circle of radius 41 . 2 cm on the water surface, estimate of magnitude of the diameter of an oil molecule. Correct answer: 1 . 81018 × 10 − 9 m. Explanation: Given : m = 9 . 03 × 10 − 7 kg , ρ = 935 kg / m 3 , and r = 41 . 21 cm · 1 m 100 cm = 0 . 4121 m . The volume of the oil slick is V = A base h = ( π r 2 ) d and its density is ρ = m V = m ( π r 2 ) d d = m ρ π r 2 = 9 . 03 × 10 − 7 kg (935 kg / m 3 ) π (0 . 4121 m) 2 = 1 . 81018 × 10 − 9 m ....
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This note was uploaded on 11/10/2008 for the course PHY 301 taught by Professor Swinney during the Spring '07 term at University of Texas.
 Spring '07
 Swinney
 mechanics

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