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Experiment 1 - Determination of Local Free-Fall Acceleration Name Lab Day Lab Time wiiO Partners Du Omoled Lab Instructor piup e Procedure A Initial distance of fall rn ± 0.002 m Proportional error in distance 부' Times of Fall (in seconds): Drop l Drop 5 Average time of fall Sample standard deviation in time-of-fall data Standard error in the mean for time of fall Table 1-1 value for η Uncertainty in the time of fall Proportional error in time Calculated value of g Proportional error in g Uncertainty in g for a level of confidence of_ - - m/s m/s 1-5

Answer 1

(i) The average time of fall which will be given by -

$\bar{t}$_avg = (t₁ + t₂ + t₃ + t₄ + t₅) / (N)

$\bar{t}$_avg = [(0.6047 s) + (0.6045 s) + (0.6043 s) + (0.6046 s) + (0.5989 s)] / (5)

$\bar{t}$_avg = [(3.017 s) / (5)]

$\bar{t}$_avg = 0.6034 s

(ii) The sample standard deviation in time-of-fall data which will be given by -

$\sigma$_t = _$\sqrt{}$(1/N) _{$\sum_{i=1}^{N}$} (t_i - $\bar{t}$_avg)²

where, (t₁ - $\bar{t}$_avg)² = [(0.6047 s) - (0.6034 s)]² = 1.69 x 10^-6 s²

(t₂ - $\bar{t}$_avg)² = [(0.6045 s) - (0.6034 s)]² = 1.21 x 10^-6 s²

(t₃ - $\bar{t}$_avg)² = [(0.6043 s) - (0.6034 s)]² = 8.1 x 10^-7 s²

(t₄ - $\bar{t}$_avg)² = [(0.6046 s) - (0.6034 s)]² = 1.44 x 10^-6 s²

(t₅ - $\bar{t}$_avg)² = [(0.5989 s) - (0.6034 s)]² = 2.025 x 10^-5 s²

then, we get

$\sigma$_t = _$\sqrt{}$(1/5) [(1.69 x 10^-6 s²) + (1.21 x 10^-6 s²) + (8.1 x 10^-7 s²) + (1.44 x 10^-6 s²) + (2.025 x 10^-5 s²)]

$\sigma$_t = _$\sqrt{}$(1/5) (2.54 x 10^-5 s²)

$\sigma$_t = _$\sqrt{}$5.08 x 10^-6 s²

$\sigma$_t = 0.00225 s

(iii) The standard error in mean for time-of-fall which will be given by -

$\sigma$_avg = $\sigma$_t / _$\sqrt{}$N

$\sigma$_avg = (0.00225 s) / _$\sqrt{}$5

$\sigma$_avg = [(0.00225 s) / (2.236)]

$\sigma$_avg = 0.001 s

(v) The uncertainty in time-of-fall which will be given as -

Uncertainty = $\bar{t}$_avg$\pm$$\sigma$_t

Uncertainty = (0.6034 s) _$\pm$(0.00225 s)

(vi) The proportional error in time which will be given by -

Error = $\sigma$_t / $\bar{t}$_avg

Error = [(0.00225 s) / (0.6034 s)]

Error = 0.00372