Asked by Justin
A 0.810 m long brass pendulum experiences a 18.0°C temperature increase.
(a) Calculate the original period(s).
(b) What is the change in the period(s)?
(c) Find the ratio of the change in period to the original period, and calculate how long (h) a clock run by this pendulum would take to fall 1.00 s behind the correct time.
(a) Calculate the original period(s).
(b) What is the change in the period(s)?
(c) Find the ratio of the change in period to the original period, and calculate how long (h) a clock run by this pendulum would take to fall 1.00 s behind the correct time.
Answers
Answered by
Elena
T₀=2πsqrt(L/g) =
=2πsqrt(0.81/9.8)=
=1.8064 s.
ΔL=α•L•ΔT=19•10⁻⁶•0.81•18
=2.77•10⁻⁴m
L₁=L+ ΔL =0.81+2.77•10⁻⁴=
=0.810277
T₁=2πsqrt(L₁/g)=
=2πsqrt(0.810277/9.8)=
=1.8067 s
T₁-T=1.8067-1.8064=0.0003 s.
(T₁-T)/T=0.0003/1.8064=1.66•10⁻⁴
(t/T₁)0.0003= 1 s
t= T₁/0.0003 = 1.8067/0.0003 =6022 s=1.67 h
=2πsqrt(0.81/9.8)=
=1.8064 s.
ΔL=α•L•ΔT=19•10⁻⁶•0.81•18
=2.77•10⁻⁴m
L₁=L+ ΔL =0.81+2.77•10⁻⁴=
=0.810277
T₁=2πsqrt(L₁/g)=
=2πsqrt(0.810277/9.8)=
=1.8067 s
T₁-T=1.8067-1.8064=0.0003 s.
(T₁-T)/T=0.0003/1.8064=1.66•10⁻⁴
(t/T₁)0.0003= 1 s
t= T₁/0.0003 = 1.8067/0.0003 =6022 s=1.67 h
Answered by
Henry
L1 = 0.810 m.
Temp. Change = +18o C.
T.C. = 1.9*10^-5/C
a. T^2 = 4pi^2(L/g)
T^2 = 39.4(0.81/9.8) = 3.25653
T = 1.8046 s.
b. L2 = L1 + L1*(T.C.)*18
L2=0.81 + 0.81(1.9*10^-5)*18=0.81027702
m.
T^2 = 39.4(0.81027702/9.8)=3.25764
T = 1.8049 s.
Change in period = 1.8049-1.8046 = 0.00003 s.
c. 3*10^-5/1.8046 = 1.6624*10^-5
Temp. Change = +18o C.
T.C. = 1.9*10^-5/C
a. T^2 = 4pi^2(L/g)
T^2 = 39.4(0.81/9.8) = 3.25653
T = 1.8046 s.
b. L2 = L1 + L1*(T.C.)*18
L2=0.81 + 0.81(1.9*10^-5)*18=0.81027702
m.
T^2 = 39.4(0.81027702/9.8)=3.25764
T = 1.8049 s.
Change in period = 1.8049-1.8046 = 0.00003 s.
c. 3*10^-5/1.8046 = 1.6624*10^-5
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