T1a10s - Assignment 10 solutions(Fall term Physics 102...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
10.12 (a) As the temperature drops by 20°C, the length of the pendulum changes by () i LLT α ∆= Thus, the final length of the rod is ()( ) ( ) 1 64 19 10 C 1.3 m 20 C 4.9 10 m 0.49 mm −− ⎡⎤ ° °= −× = ⎣⎦ 1.3 m 0.49 mm f L = Assignment 10 solutions (Fall term Physics 102) (b) From the expression for the period, 2 TL π = g , we see that as the length decreases the period decreases. Thus, the pendulum will swing too rapidly and the clock will run fast . 10.15 From ( fi i LL L L T ) , the final value of the linear dimension is − = ( ) i LL LT =+ . To remove the ring from the rod, the diameter of the ring must be at least as large as the diameter of the rod. Thus, we require that ( ) ( ) ff brass al LL = , or ( ) ( ) brass brass brass T L L T + ∆= + This gives T αα . (a) If 10 L = , .01 cm ()( ) ()( ) 11 -6 -6 10.01 10.00 199 C 19 10 C 10.00 24 10 C 10.01 T = − ° ×° , so 20.0 C 199 C= TT T °− ° 179 C which is attainable ° . (b) If 10 L = , .02 cm -6 -6 10.02 10.00 396 C 19 10 C 10.00 24 10 C 10.02 T = , and = 376 C which is below absolute zero and unattainable −° 347
Background image of page 1

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

View Full DocumentRight Arrow Icon
CHAPTER 11 10.55
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 3

T1a10s - Assignment 10 solutions(Fall term Physics 102...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online