The confidence interval for the first group mean is thus (4.1, 13.9). Similarly for the second group, the confidence interval for the mean is (12.1, 21.9). Notice that the two intervals overlap. However, the t-statistic for comparing two means is: t= 17 − 9 2.5 2 + 2.5 2 = 2.26 which reflects that the null hypothesis, that the means of the two groups are the same, should be rejected at the α = 0.05 level. To verify the above conclusion, consider the 95 percent confidence interval for the difference between the two group means: (17 − 9 ) ± 1.96 × 2.5 2 + 2.5 2 which yields (1.09, 14.91).
In a gas chromatogram, the peak areas are considered to be directly proportional to the amound of sample passing the detector in the gas chromatogram. The table below shows the areas for each peak in each of our printed Chromatograms and the temperature at which we collected each fraction. Fraction # Peak # Area of Peak Temperature at which Fraction was collected 1 1 2353548.74 29.9-37.7 degrees C 2 1 No Peak Recorded 38.6-39.9 degrees C 3 1 153621.26 39.9-49.2 degrees C 2 1431547.82 39.9-49.2 degrees C Left Over 1 2896057.41 Room Temperature These peaks were taken with a Hexane-Heptane Mixture We can read several things from these results. First is that the largest area of a peak that we recorded was the material that was left over from our distillation. This tells us that there was still a lot of mixture that was not distilled.
We calculated the average time and standard deviation using Excel. We used Newton’s 2nd law to calculate the period, the stopper mass, the radius, and the acceleration due to gravity. We drew a confidence interval to show that our measurements were inconsistent. | Time (T) [s] | Texp (T/# of revolutions(10)) [s] | 1 | 5.42 s | 0.547 s | 2 | 5.27 s | 0.527 s | 3 | 4.86 s | 0.486 s | 4 | 5.28 s | 0.528 s | 5 | 5.05 s | 0.505 s | 6 | 5.33 s | 0.533 s | 7 | 4.92 s | 0.492 s | 8 | 5.29 s | 0.529 s | 9 | 5.16 s | 0.516 s | 10 | 5.11 s | 0.511 s | 11 | 5.16 s | 0.516 s | 12 | 5.20 s | 0.520 s | 13 | 5.21 s | 0.521 s | 14 | 5.13 s | 0.513 s | 15 | 5.31 s | 0.531 s | | | | Rubber stopper mass [mr] = 8.2 gRadius of circle [r] = 0.4633 mWeight mass = [m] = 50.1 g | | Tavg = 0.518 sσT = .0152 sTexpt = 0.518 s ± .0152 s | | | Tthy = .5541 s V = 5.255 m/s2 | Our average time was approximately 2.374 standard deviations from the theoretical time. In analysis, we consider our results not consistent with the theoretical time.
(15 marks) Answers (a) C1 and C2 in series = 2uF Cupper = 2uF + 2uF + 2uF = 6uF Clower = 3uF + 3uF = 6uF Cequ = 3uF (b) Qupper = Qequ = Cequ V = 3uF100V = 300uC Vupper = Qupper/Cupper = 300uC/6uF = 50V Charge stored in C3 = 2uF50V = 100uC (15 marks) (15 marks) Question 2 (27.53) A charge Q is placed on a capacitor of capacitance C. The capacitor is connected into the circuit as shown in the figure below, with an open switch, a resistor, and an initially uncharged capacitor of capacitance 4C. The switch is then closed and the circuit comes to an equilibrium. In terms of Q and C, (a) (b) (c) (d) Find the final potential difference between the plates of each capacitor (10 marks) Find the charge on each capacitor. (7 marks) Find the final energy stored in each capacitor. (8 marks) Find the energy dissipated in the resistor.
– final | 36 | 33 | 46 | Delta T (oC change | 12C | 12C | 1 C | Questions: A. Which of the foods tested contains the most energy per gram? Peanut B. How do your experimental results compare to published? They were not the same, they actually were completely different.
In order to have valid results, your percent error must be equal to or below 3% and ours was below. Once we found our results to be valid, we compared our average density to the density of the given metals. The density of zinc, 7.00 g/cm3, was the closest density in range of our calculated density. Our procedure had to be adjusted three times in order to find accurate data. Our original procedure didn’t measure mass or volume correctly.
What is the energy of a photon with a frequency of 6·1014 Hz? Answer | | 4·10-20 J | | | 4·10-19 J | | | 4 J | | | 40 J | 1 points Question 6 1. Which of the following statements (related to the photoelectric effect) is FALSE? Answer | | There is a minimum amount of energy which is necessary to remove an electron from the surface. | | | The energy of a single photoelectron depends only on the frequency of the incident light.
KSCN ionizes into K+ and SCN-, and in the presence of the H+ ion supplied by nitric acid, the H+ and SCN- will combine to form the weak acid HSCN. Since there is a large excess of nitric acid compared to KSCN, we can assume that all of the SCN- will be in the form of HSCN. Procedure part 2: Six test tubes were filled with various amounts of 0.003M sodium thiocynate. A constant amount of 10 ml of 0.003 M Fe(NO3)3. The first tube had no sodium thiocynate.
On the 0 mark on the scale. 13. With the dosimeter installed on the charger, the indicating fiber can be seen moving upscale away from 0. What does this indicate? D. Either the dosimeter or the charger is dirty or faulty.
Given: F - 36 V R = 12 f l V 36 V , , Unknown: I = ? = Original equation: V ^ IR This 3 A is the current through the entire circuit. Use this current to f i n d the potential difference across the parallel combination. Remember, the potential difference across resistors wired i n parallel is the same regardless of w h i c h path is taken. Because the resistors i n parallel have a combined resistance of 6 O, y o u f i n d the potential difference across the parallel branch as follows.