Lesson_1.5_Homework_and_Solutions

Lesson_1.5_Homework_and_Solutions - Lesson 1.5 Homework...

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Lesson 1.5 Homework Solutions 1. l = α l 1 ∆θ = 11 × 10 -6 × 100 (40 – (-10)) = 0.011 Percent error = 0.011 100 10 1 0 0.01 % × = 2. α copper = 17 × 10 -6 K -1 , l 1 = 500 m, ∆θ = 50 o C. l = α l 1 ∆θ = 17 × 10 -6 × 500 × 50 = 0.425 m 3. (i) ....( 4 ) 1 261.6 2 1 261.4 ...( ) 2( ) ( ) ( ) 261.6 1 261.4 261.6 1 261.4 261.6 261.4 261.4 0.2 261.4 1 0.2 261 7.65 10 .4 i T l T ii l l Dividing i by ii l l l l l l l l l l l t m Bu l l μ - = = + ∆ + ∆ = = + - = - = = = ∆ = × = (ii) 1 4 6 1 7.65 10 69.5 11 10 1 l l l l K α θ - - ∆ = × ∆ = = = × ×
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4. Period T = 1 2 2 6 5 2 1 2 9.8 4 1 9.8 9.8 0.25 4 11 10 0.25 30 8.25 10 ' 0.250083 ' 0.250083 ' 2 2 1.0037 9.8 l T g l l l m l l m l l l m l T s g π α θ - - = = = = = ∆ = ∆ = × × × = × = + ∆ = = = = Since T’ is greater than T, the clock is loosing time. It looses 0.0037 s every second. In a day it loses 0.0037
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This note was uploaded on 05/01/2011 for the course PHY 2049 taught by Professor George during the Spring '11 term at Edison State College.

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Lesson_1.5_Homework_and_Solutions - Lesson 1.5 Homework...

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