Posted by **hania** on Thursday, April 5, 2012 at 3:54pm.

A clock has an aluminum pendulum with a period of 1.000 s at 20.3 °C. Suppose the clock is moved to a location where the average temperature is 28.1 °C. (The linear expansion coefficient for aluminum is 2.20 10-5 °C−1.)

(a) Determine the new period of the clock's pendulum. (Enter your answer to six siginficant figures.)

(b)Determine how much time the clock will lose in 2 weeks.

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**bobpursley**, Thursday, April 5, 2012 at 4:03pm
There are two standard equations here: what is it you have a question about?

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**hania**, Thursday, April 5, 2012 at 4:12pm
what equations do i use?

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**Elena**, Thursday, April 5, 2012 at 4:19pm
a.

Moment of inertia respectively the axis

that passes through the end of the pendulum is

I = Io+mx^2 = mL^2/12 + mL^2/4 = mL^2/3.

T1 = sqrt(I/m•g•x) = sqrt( mL^2/3m•g•x)

=sqrt( L^2/3•g•x) = L/sqrt(3•g•x) =1 s.

L = sqrt(3•g•x),

α =ΔL/(L•ΔT),

ΔL = α •sqrt(3•g•x) • ΔT,

T2 = sqrt ((L+ ΔL)^2/3•g•x) =

=(L+ΔL)/ sqrt(3•g•x)=

= (L + α •sqrt(3•g•x) • ΔT)/ sqrt(3•g•x) =

= ( sqrt(3•g•x) + α •sqrt(3•g•x) • ΔT)/ sqrt(3•g•x)=

= 1+7.8•2.2•10^-5 = 1.000172 s.

b.

Δto= 0.000172 s.

Δt =0.000172•3600•24•14 = 208.0512 s.

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**hania**, Thursday, April 5, 2012 at 4:35pm
can you help more, i didnt get the right answer..i didnt get where you got 7.8 from

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**Elena**, Thursday, April 5, 2012 at 4:51pm
ΔT = 28.1 -20.3 = 7.8 oC

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**hania**, Thursday, April 5, 2012 at 4:56pm
i didnt get the right answer

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**Elena**, Thursday, April 5, 2012 at 5:12pm
What is the right answer?

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**hania**, Thursday, April 5, 2012 at 5:24pm
i dont know

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**Elena**, Thursday, April 5, 2012 at 5:27pm
Why do you believe that this answer is incorrect?

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**hania**, Thursday, April 5, 2012 at 5:46pm
i get it wring when i enter it

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