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.
9 answers
There are two standard equations here: what is it you have a question about?
what equations do i use?
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.
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.
can you help more, i didn't get the right answer..i didn't get where you got 7.8 from
ΔT = 28.1 -20.3 = 7.8 oC
i didn't get the right answer
What is the right answer?
Why do you believe that this answer is incorrect?
i get it wring when i enter it
1 answer