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March 31, 2015

March 31, 2015

Posted by **Anonymous** on Monday, September 3, 2012 at 4:43am.

- physics -
**Elena**, Monday, September 3, 2012 at 3:26pmCooling from 23ºC to -32ºC

ΔL1= α•L•ΔT1 =12•12•10^-3•[23-(-32)]=

=12•12•10^-3•55= 7.92•10^-3 m

Heating from 23ºC to +55ºC

ΔL2= α•L•ΔT2 =12•12•10^-3•[55-23]=

=12•12•10^-3•32= 4.61•10^-3 m

Chamge in the beam’s length is

ΔL=ΔL1+ ΔL2=7.92•10^-3+ 4.61•10^-3=

=1.253•10^-2 m.

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