There are a lot of same examples that you may have worked before, where the mass on a spring uses a classics when it comes to mechanics. So in this system, always put in your mind that there is an enormous quantum standard that one can use in the equation. It should be 2.10x10 raise to a negative sixth. J.
Answer:
a) T = (2,375 ± 0.008) s
, b) When comparing this interval with the experimental value we see that it is within the possible theoretical values.
Explanation:
a) The period of a simple pendulum is
T = 2π √ L / g
Let's calculate
T = 2π √1.40 / 9.8
T = 2.3748 s
The uncertainty of the period is
ΔT = dT / dL ΔL
ΔT = 2π ½ √g/L 1/g ΔL
ΔT = π/g √g/L ΔL
ΔT = π/9.8 √9.8/1.4 0.01
ΔT = 0.008 s
The result for the period is
T = (2,375 ± 0.008) s
b) the experimental measure was T = 2.39 s ± 0.01 s
The theoretical value is comprised in a range of [2,367, 2,387] when we approximate this measure according to the significant figures the interval remains [2,37, 2,39].
When comparing this interval with the experimental value we see that it is within the possible theoretical values.
You would use h=1/2*g*t^2
Therefore,
g= 9.81
h= 1/2*9.81*6^2
h= 176.58m
Answer:
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