Introduction
The cross product of two vectors can be defined as the binary operation that is done on the two vectors in the three dimensional space. This operation is defined by the symbol ×. If the two vectors, suppose a and b are perpendicular to each other than their product will be also a vector quantity. Also this true for the normal plane containing these vectors. Cross product has many applications in engineering, physics and mathematics.
If two vectors are in a same direction or one of these vectors has zero magnitude then their result of cross product will be zero. Generally the magnitude of the product of these two vectors is the area of parallelogram with the sides of vectors. In a simple way we can state that the magnitude of product two vectors is a product of their lengths.
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The actual experimental errors were close to the theoretical errors. The difference between the errors was not so high. Rather it was in acceptable region.
The value for percent difference was calculated using the following formula.
Percent Difference =Absolute Difference Average ×100%=|E2-E1|(E2+E1)/2 ×100%
Discussion
Following results were obtained from the experiment.
Mass kg Force N Radius mm Torque
1 1.8 .024 .432
0.135 1.3203 0.022 .029026
0.115 1.1247 0.012 .01349
Residual Average torque Percent results
0.000657 0.028581 2.2987
The physics behind this experiment is very simple. The purpose of this experiment was to observe the results from the experiment of vector cross product and torque. In the experiment, the experiment of cross product and torque was performed with the different masses and length of arms. Radius was changed according to the need of experiment and masses
Marwah Alabbad Post lab 10/21/15 Question 1: 1. Experiment 1: Number of trails NaOH concentration (M) Volume of HCl solution (mL) Initial volume of NaOH(mL) final volume of NaOH(mL) The volume of NaOH to titrate HCl (mL) Concentration of HCl (M) 1st 0.1023 25.0 10.05 36.12 26.07 0.085 2nd 0.1023 25.0 5.74 31.40 25.66 0.105 3rd 0.1023 25.0 9.84 35.52 25.68 0.105 First trail calculation: 0.02607L× (0.1023mole NaOH/1L)×(1 mol of HCL/1 mol of NaOH)×(1/0.025)= 0.085M of HCl
Calculation: Initial Mass(g)-Final Mass (g)=Change in Mass (g) Trial 1 74.5-62.0=12.5(g) Trial 2 272.7-271.5=1.2(g) Percent Error: 272.7-271.5 x 100 272.7 =0.440% Percent Change: 74.5-62.0 x 100 74.5 (Trial 1) =16.778% 272.7-271.5 x 100 272.7 (Trial 2) =0.440%
The predicted and experimental responses are compared in order to validate the model and to calculate the prediction error. The prediction error was found to be below 7% indicating that the observed responses were very close to the predicted values. Percentage prediction error is useful in constituting the validity of generated equations and describes how close the predicted responses to that of actual values. The values of <15 are desirable to have closeness of the predicted values with the actual values
In this lab there were five different stations. For the first station we had to determine an unknown mass and the percent difference. To find the unknown mass we set up the equation Fleft*dleft = Fright*dright. We then substituted in the values (26.05 N * 41cm = 34cm * x N) and solved for Fright to get (320.5g). To determine the percent difference we used the formula Abs[((Value 1 - Value 2) / average of 1 & 2) * 100], substituted the values (Abs[((320.5 - 315.8) /
The measured tensions were normalized relative to the maximum tension and the lengths were normalized relative to the length at which maximum force was generated (Ln = 29 mm). The measured data and expected data were plotted together (Figure 3). Correlation between measured and expected tensions at the same normalized lengths was determined for the three expected segments of the length-tension relationship.2,3,4 This was done using piecewise linear regression and yielded r2isometric = 0.956033. The fitted curve for the isotonic experiment resulted in r2isotonic = 0.960557. The F0 was 19.5 N for the fit and 6.35 N for the guess.
Purpose: The purpose of the experiment was to understand how strong a bessbug by using weights and observing the time the best bug takes to travel to a certain distance. Background Information: The horned Passalus; also known as the Bess Beetle, is widely known beetle that is easily recognized. The Bessbug is a shiny black insect with a hard shell.
Porter´s Five Forces is the analytical framework chosen to analyse GE´s Playbook. GE is one of the world´s most diverse companies spanning a wide range of businesses (Grant, 2005), including appliances and lighting, aviation, capital (commercial lending and leasing, consumer, real estate, energy financial services, aviation financial services), energy management, healthcare, oil & gas, power & water, and transportation (General Electric, 2015). Some of their customers are: - Aviation, Commercial Engines: Boeing - Capital Inventory Financing: P.C. Richard and Son - Distributed Energy: Songas - Healthcare: Wheaton Franciscan
3. In this experiment, the percent yield was 90%. This number implies that there was little error in this experiment. However, this result could have been caused by certain external factors.
However, any doubts regarding the results may be traced to a few elements of the experiment that lend themselves to possible error. The following factors may have contributed to potential errors in the experiment; the need to zero the machine between each of the readings in obtaining the absorption spectrum and the resulting peak wavelength, the precision with which a person can accurately adjust the needle on the spectrophotometer to zero is limited, not putting in the inaccurate amount of cobalt chloride or water into the substance, and getting oil from our fingers onto the
The final product weight for percent yield was only the solid E product, which missed one half of the final product produce. If both products were weight, the percent yield would have been larger that it was. Instead of 22.33%, it could have been 44.66%. To prove that both products were obtained, but only one of the two products was analyze, a TLC plate of the DCM layer, that contains both products, and of the final product, was obtain.
The actual data is the result on our experiment vs theoretical, which is based on the calculations above. I have also learned to pay more attention to draining out all of the product completely before continuing to test the experiment, as any small drop of contaminant can veer our results into a different
But the difference was no bigger than 0.08, and after the values were rounded the same empirical formula was deduced. So the experiment can be concluded as successful. Evaluation: The method used was simple and easy to follow; however, it did not include how much oxygen was needed to react completely. Also it didn 't mention what magnesium oxide looked like after it finished reacting, so it was a guesswork of determining whether the reaction was finished or not.
Porter’s five forces is a framework that provides analysts with knowledge of the external factors regarding their company and the development of business strategy. These shows people how attractive a company is in a certain industry. I have chosen to develop the porter’s five forces strategy regarding Cisco and the information received. I will evaluate the competiveness, threat of substation, buyer power, supplier power and the threat of new entry.
6.1 Marketing Mix Marketing mix is a set of controllable marketing tactics used by business to promote their product and achieve its marketing objectives. (L. Lake, 15 June 2017) Marketing mix is also called the 4Ps which consist of Promotion, Place, Product and Price. (M. J. Baker, 2001, p.54) 6.1.1 Product
3.2 Industry conditions (Porter 's Five Forces Analysis) Five forces which would impact an organization 's behavior in the market. Understanding the nature of these forces provides organizations the required insights to enable them to formulate the appropriate strategies to be successful in their market (Thurlby, 1998). 3.2.1 Threat of new entrants (high entry barriers) High capital investment for competitor entry into telecommunication industry. Companies in this industry maintain development, spend fairly large amount of capital on network equipment and incurred high fixed costs. Besides, technologies are also considered as barriers for new companies to enter the market.