Lab Report M0

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School

University at Buffalo *

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151

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Physics

Date

Dec 6, 2023

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docx

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4

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Lab Report M0 Diameter of bob with ruler: d= 2.2cm, 2.4cm, 2.7cm The range of the values is maximum – minimum= 2.7cm – 2.2cm= 0.5cm Diameter of bob with caliper: d = 2.47cm, 2.45cm, 2.48cm The range of the values is max – min = 2.48cm – 2.45 cm = 0.03 cm Using a caliper instead of a ruler decreases the range of measurements immensely. Measuring an object without straight edges (like a ball) is very difficult to do with a standard ruler but a caliper can hug each end to get the perfect measurement. The range of measurement with the ruler was 16x more than the range using a caliper. Thickness of 100g weight using ruler: t = 1.0cm, 0.9cm, 0.9cm The range of the values max – min= 1.0cm – 0.9cm = 0.1 cm Thickness of 100g weight using caliper: t = 0.89cm, 0.89cm, 0.9cm The range of the values max – min= 0.9cm – 0.89cm = 0.01 cm Even though a different object is being measured, a caliper is still more of a useful, pinpoint measuring tool because the range is so much smaller on the caliper compared to the ruler. The range of measurement was 10x more using a ruler compared to using a caliper. The difference of ranges of measurements with a caliper instead of a ruler is very noticeable. Measuring different objects can also skew up results. Measuring the thickness of the weight with a ruler isn’t too different than with a caliper because the weight has sharp edges that can be measured easily, however measuring a ball, which is completely round, can have skewed answers depending on where you measure it. VI-2 Measured values of T in seconds 2.35 2.45 2.0 4 1.97 1.61 1.8 5 2.05 1.81 1.8 9 2.25 1.68 1.8 9 2.09 1.83 1.8 5
Excel returns the average value of T = 1.974 seconds Excel returns the standard deviation of T as σT= 0.2359 seconds The standard deviation of the mean is given as σ(T with line on top)= σT/(square root of)n = 0.2359/(square root of)15 = 0.061 seconds Therefore: T (+/-) σT(average) = 2.00(+/-)0.06 seconds VI-3 X(small 0)= 64.4 cm -- xm(cm) x(cm) m (g) -- 61.0 4.4 20 -- 61.2 4.2 20 -- 61.3 4.1 20 -- 57.4 7.0 40 -- 57.3 7.1 40 -- 57.5 6.9 40 -- 54.0 10.4 60 -- 53.9 10.5 60 -- 54.0 10.4 60 -- 49.0 15.4 80 -- 49.3 15.1 80 -- 49.0 15.4 80 -- 46.2 18.2 100 -- 46.1 18.3 100 -- 46.0 18.4 100 -- 42.0 22.4 120 -- 42.5 21.9 120 -- 42.0 22.4 120 -- Slope 5.407146 0.157703 Intercept Slope unc 0.099072 1.423782 Unc int X= X(small 0) – Xm For example, when m = 20g, 64.4cm – 61.0cm = 4.4 cm
0 5 10 15 20 25 0 20 40 60 80 100 120 140 f(x) = 5.41 x + 0.16 Mass v Elongation of Spring Elongation, cm Mass, grams S (+/-) σs = 5.41 (+/-) 0.09 g/cm b (+/-) σb = 0.16 (+/-) 1.4 g Since m = k/g(x) The slop S = k/g Therefore, k = gs k = (980 cm/s(squared))(5.4071 g/cm) = 5298.9 g/s(squared) From eq 10 of the prologue, with k = C, g = A and s = B, we see that (CHECK SCREENSHOT AND COPY FROM THERE INCLUDING DIGITS WHICH ARE K = (5298.9 g/s(squared) σs = 0.0991 g/cm s = 5.4071 g/cm VI-4 X(small 0)= 64.4 cm -- xm(cm) x(cm) m (g) -- 61.0 4.4 20 -- 61.2 4.2 20 -- 61.3 4.1 20 -- 57.4 7.0 40 --
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57.3 7.1 40 -- 57.5 6.9 40 -- 54.0 10.4 60 -- 53.9 10.5 60 -- 54.0 10.4 60 -- 49.0 15.4 80 -- 49.3 15.1 80 -- 49.0 15.4 80 -- 46.2 18.2 100 -- 46.1 18.3 100 -- 46.0 18.4 100 -- 42.0 22.4 120 -- 42.5 21.9 120 -- 42.0 22.4 120 -- Slope 5.407146 0.157703 Intercept Slope unc 0.099072 1.423782 Unc int Values of x are determined the same as in VI-3 The value of the slope calculated with the parallax error is the same as was determined in VI-3 The calculation of k is carried out exactly as in VI-3, therefore, k = 5298.9 g/cm(squared) The error analysis is also the same σk = (answer to sigma k above) (CHECK OTHER SCREENSHOT) The value of k is the same as determined in VI-3