... The experiments are conducted in a closed barn to avoid wind disturbances. The barn has windows on the roof and on the sides so that daylight illuminates the scene. The set up is sketched in Fig.\ref{Set_Up}, with the reference frame used hereinafter. A $15m\times3m$ white tarp sets the background. On the ground, a mesh of ($1\times1m^2$) panels is installed, covering an area of $3\times16m^2$. The blower is a Kuhn Primotor 3570 , positioned $3$ meters from the first row of panels. The camera used is a Canon $EOS1100D$, with $EFS \, 18-55mm$ objective, and it is installed on a podium at $3m$ from the ground, and $30m$ from the image plane. The field of view is about $12m$ large. \begin{figure}[h] \centering \includegraphics[width=0.5\textwidth]{Set_UP.png} \caption{Sketch of the experimental set up.}\label{Set_Up} \end{figure} Fig.\ref{CALI} shows a typical calibration image. The test section is empty and a calibrator is aligned with the axis of the nozzle at the blower outlet. The support of the calibrator indicates the location of the ground line in the local $XZ$ plane, so the image is cropped to have $Z=0$ in the last pixel row. The calibrator has a pattern of circles of $22cm$ diameter, positioned in the $Z=-0.5m$ and $Z=0.5m$ plane. This has been used to compute the magnification factor and its uncertainty, estimated as twice the standard deviation of the circles diameter measurements in pixels. The result is $M=2.78\pm0.08 mm/pixel$. The mesh of panels has been...
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...Factor x line 3) Weight Measurements 5 Weight of tare + wet soil from hole 6 Weight of tare 7 Weight of wet soil, W Moisture content 8 Weight of moisture tin 9 Weight of tin + wet soil 10 Weight of tin + dry soil 11 Weight of water (line 9 – line 10) , Ww 12 Weight of dry soil (line 10 – line 8), Ws 13 Moisture content (line 11 /line 12) OR 14 Moisture content determination by Calcium Carbide Method Density Calculations 15 Wet density, γ = (line 7/line 4) 16 Dry density, γd = (line 14/(1+line...
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...Student Name: Andreas Sippl Student ID: 4445491 Date: 04 January 4, 2015 Course and Section Number: SCIN131 Lesson 6 Lab: Titrations and Natural Acid/Base Indicators Begin by viewing the following Thinkwell videos 15.1.7 CIA Demonstration: Titrations 11.2.2 CIA Demonstration: Natural Acid-Base Indicators (NOTE: The second video is sort of a supplement to the first, and shows how this applies to your daily life. Feel free to try some of the things listed in the video, but be sure to report back to the class regarding the results of your experiments!) After you watch the above video, answer the questions below in sufficient detail: (a) (1 pts.) In YOUR OWN WORDS, what is a titration? What type of reaction is it? Be detailed and specific...
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...1. Which of the variables in the second flux equation, J = 푃퐴(퐶퐼 − 퐶2), will be manipulated during the lab? In this lab, the concentration gradient across the membrane, (퐶퐼 − 퐶2), is the variable. Since we are using the same solute and membrane manipulated, P, the permeability coefficient of the solute across the membrane is constant. A, the membrane area, is also constant since we are using the same cell for all different concentrations. In this case, the concentration gradient across the membrane, (퐶퐼 − 퐶2), is the only variable in the experiment. 2. What do you expect to observe during the lab? How might your data be used to determine the solute concentration of the unknown solution in the dialysis tubes? I expect to see a positive...
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...School of Science and Technology Title: Lab 1-6 Student Name: Manuel MikewaOduor Student No: M00519008 Supervisor: Mr. Kiama Date: November 26th 2014 AIMS • Explain how data is represented and basic operations on data; • Answer all the questions in this Lab exercise OBJECTIVES • Understand binary numbers and their various calculation methods • Understand the basic concepts of processors, memory, and I/O devices, and how they are interconnected; • Understanding the functions of operating systems and BIOS Lab Unit 1 1. What is 62 in binary? 2 into 62= 31 remainder 0 2 into 31= 15 remainder 1 2 into 15= 7 remainder 1 2 into 7= 3 remainder 1 2 into 3= 1 remainder 1 Answer= Binary 11110 2. What is 63 in Hexadecimal? 63/16=...
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...For the first round of the stimulation was random mating. After the 20 students drew out two new cards from starting genotype, the new genetic pool had five pairs of dominant homozygous, ten pairs of heterozygous, and five pairs of recessive homozygous. The new generation had the same number of the starting gene pool of the known allele frequency. Random selection frequency of alleles does not change over time, due to which all individuals have an equal chance of being selected. The Hardy-Weinberg equilibrium equation, p2 + 2pq + q2 = 1, is used to define a population in which to understand both allele and genotype frequencies. The equation only occurs when mating is at random on a large population, and the relative genotype and allele frequencies persist in being constant. In all possible combination defined by the Punnett square, 60% or 0.6 of the gene pool is the A allele, and 40% or 0.4 of the a allele, totaling in for 100%. When adding the probability, the results are 0.36 for AA, 0.24 for aA, 0.24 for Aa, and 0.16 aa. In total, will equal 1. The AA homozygotes will have the frequency of p2, and similarly, aa homozygotes will have the frequency of q2 when the population is in equilibrium. Then lastly, the Aa heterozygotes will have the relative genotype frequency of 2pq. The calculation would remain the same throughout generations, but only with random mating. The Hardy-Weinberg equilibrium will not equal 1 when it is disturbed by nonrandom mating, mutations, migration, and...
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...only a small amount of power. Step 3. Traffic density metric • Select the path that is lightly loaded from the three selected paths Algorithm 1.1 Algorithm for broadcasting RReq 1.2. Algorithm for node failures reduction Step1: Source broadcasts Route Request packets which are heard by neighbor nodes within the coverage area. Step 2: all node those get the RReq will calculate their remaining battery power by using: Rp = Einitial - Eci Step 3: Also all nodes will calculate their transition power needed to transmit the RReq packet to next-hop by using the following formula: Step 4: Check that they have enough power for transmitting the packet to next-hop. Step 5: The neighboring nodes re-broadcast the route request. Step 6: The destination will select all available path that has enough power 1.3. Algorithm for reducing traffic density Step 1: once all the path those have enough power have been selected the destination nodes can count the number of hop for all the available paths and order them based on the shortest path first. Step 2: The destination node will select the path that is lightly loaded from the ordered list. Step 3: The destination will send the RRep to the source node through the selected path. Step 4: The destination sends Route Reply only to the first received Route Request. ...
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...Adedamola Iyiola M.E. LAB: Lab View Section 06 Lab View Adedamola Iyiola Department of Mechanical and Aerospace Engineering Rutgers University, Piscataway, New Jersey 08854 A Data acquisition board, thermocouple and a BNC terminal block were used simultaneously with Lab view to obtain measurements of different waveforms with varying frequencies of 500, 100, and 3000 Hz. Additionally, we varied sampling rates, at an input of 1000 Hz, at 500 Hz and 2500 Hz (0.5x and 2.5x the input wave frequency). Furthermore for the second set of measurements, we obtained data from running a thermocouple data acquisition program. We acquired temperature measurements of a room for a period of 60 seconds and an individual’s finger for 50 seconds....
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...Probability of getting no more than 0 heads: 0.0625 Probability of getting no more than 1 head: 0.3125 Probability of getting no more than 2 heads: 0.6875 Probability of getting no more than 3 heads: 0.9375 Probability of getting no more than 4 heads: 1.000 4. True 5. > pbinom(1, size=3, prob=.5) [1] 0.5 6. > round(pbinom(3, size=5, prob=.5), 2) [1] 0.81 7. > lmb expectation expectation [1] 0.25 or 15 seconds 8. The answer is .06 for minutes or 225 for seconds. 9. > round(pexp(15.5/60, rate=4), 2) # using minutes [1] 0.64 10. > pexp(40.2/60, rate=4) - pexp(10.7/60, rate=4) [1] 0.421445 11. > round(qexp(.95, rate=4), 2) # minutes [1] 0.75 > round(qexp(.95, rate=4/60), 2) # seconds [1] 44.94 12. The R command pexp(1.2, rate=3) shows the estimated probability of randomly selecting a value less than or equal to 1.2 from an exponential distribution that has a rate of 3. 13. > round(1 - pnorm(9, mean=7, sd=3),...
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...Schematic drawing of the particle passed through SEC. The colored boxes represent the SEC modules with energy deposit and white boxes are the modules without energy deposit. The group of green and orange boxes are the reconstructed clusters created due to the particle The energy of a neutral track (y-axis) is plotted against the angle between two photon candidates (x-axis) in the laboratory frame. (e) represents the data at 1.50~GeV beam energy. (c), (d), (e) and (f) represent the Monte Carlo simulation for the reactions $pd$$\to$$^3$He$~\eta$($\eta$$\to$$\gamma\gamma$), $pd\to\text{$^{3}$He}~\pi^0\pi^0\pi^0$, $pd\to\text{$^{3}$He}~\pi^0\pi^0\pi^0$ and $pd\to\text{$^{3}$He}\omega(\omega\to\pi^0\gamma)$, respectively, at 1.50~GeV beam energy. The events coming from the split-offs can be seen as a small tail near 20$^{\circ}$. % (b) is the statistical significance estimated for 1.45~GeV and 1.50~GeV beam energy, respectively. The optimal cut chosen for two energies are shown as dashed lines. The solid black hyperbola ($E > \frac{1.52}{\angle(\gamma1,\gamma2)}$) in all plots illustrates the final optimal cut used to reject the split-off tail. %cut used to reject the events from the neutral split-offs. } \label{Nsplitoff} \end{figure} % \subsubsection{Neutral Split-offs}\label{pi0gdeteffect} %\subsubsection{Neutral Split-offs {\textcolor{red}{This section need to be re-written more scientifically :-(}}}\label{pi0gdeteffect} % Fig.~\ref{Nsplitoffclus} Apart from the kinematics...
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...CHAPTER-6 EXPERIMENTAL SETUP AND RESULTS This chapter provides an overview of network simulation and different VANET simulators that can be used to simulate different VANET algorithms to analyze the performance of the network without the need of real systems. This not only saves cost but also provides opportunity to test new protocols and algorithms in a controlled environment which otherwise would have not been possible. 6.I INTRODUCTION A network simulator is a software program that models the working of a computer network and its communications. It can be a software or hardware that helps to predict the behavior of a network, without need of any actual network. It imitates the working of a network such that the performance of the network can be analyzed without...
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...files and their individual roles. 4. Opening a file and different modes using which we can open a file. 5. Programs to implement various operations on files. 15.1 Introduction Most programs written in C++ handle large volumes of data that needs to be stored permanently on secondary storage devices such as hard disk. The data resides in these devices in the structure of files. A file is basically a collection of bytes stored on a secondary storage device, such as disk. Programs are written to carry out the read/write operations on these files. Following are the types of data communication a program is typically involved in. These two data communication types may be used in individual or together. 1. Transfer of data among console unit and program. Figure 15.1 Program-file-console interaction 2. Transfer of data among program and a disk file. In this lesson, we are basically concerned with various methods provided by C++ for storing and retrieving the data from files. The I/O system of C++ is capable of managing the file operations just as managed by the console input and output. The file stream acts as an interface between the program and the file. Input stream provides data to the program and output stream gets data from the program. The concept is demonstrated in the figure given below: Figure 15.2 Program-file interaction The read operation is used to create input stream between the program and the input file. And, the write operation is used to create output...
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...CHAPTER-6 EXPERIMENT RESULT 6.1 System Model For experimental purposes we assume that total 5000 jobs will be scheduled on the grid consists of 7 clusters. In general, each resource/cluster contains 80 computing nodes (Machines), and each computing node contains 1 Processing Elements (PE). The processors of computing nodes in different resources have same processing power (i.e. 1GHz). Every computing node consist RAM of 53GB approx. Network speeds among the computing node are also assumed same for different resources. The characteristics of 7resources used in our experiment, shown in Table 6.1. Parameter Name Value Total Number of Jobs 5000 Number of Cluster 07 Number of CPU per Cluster 80 RAM 53 GB Baud Rate 10000 CPU Speed 1GHz...
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...Katie Carroll Mr. Miller APP1 November 15, 2015 1. Background a. Equations: i. Σ F = T-mhg = mha ii. fk=ukN 2. Materials a. Before assembling the materials, prepare a clean, dry workspace (away from food) with all necessary materials. i. Wooden board that can be adjusted to various angles ii. Spring scale iii. Wooden block (with tape on one side) iv. Pulley v. Paperclips (weights) vi. Safety goggles 3. Procedure a. (Part 1)-determine kinetic friction between the wooden block and the wooden board b. Read through the entire procedure and prepare to carefully collect your data before you begin. c. Before assembling the materials, prepare a clean, dry workspace (away from food) with all necessary materials. d. Weigh and record the mass of the wooden block. e. Make sure the surfaced of the wooden plank and the block are clean of dirt so that nothing will disrupt the experimental friction coefficient. f. Put the wooden block (with tape side down) on the end of the wooden board and attach the string to the block. Then attach the spring scale to the other end of the pulley so that it hangs over the far end of the wooden board. g. Put mass on the spring scale and start the block with a small push. i. If the wooden block speeds up, take some of the mass off. ii. If the wooden block slows down, add more mass. iii. Once the block moves across the whole wooden board with a constant speed, record the total hanging mass and its weight. h. (Part 2)- Determine the difference surface area...
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...The attractive force between the positive charge in the nucleus and the valence electrons decreases because these electrons are farther from the nucleus. 4. a) The elements that are found at the main peaks of my graph are Lithium, Sodium, Potassium, and Rubidium. These elements are all found in the same group. They are all Alkali Metals. b) The elements that are found at the main valleys of my graph are Neon, Argon, Krypton, and Helium. These elements are all found in the same group. They are all part of the Noble Gases. 5. I predict Rubidium would be the largest atom in the atomic table because it has an atomic radius of 248pm. I predict the smallest atom in the atomic table would be Hydrogen because it has an atomic radius of 32pm. 6. There is a significant jump in the size of the nucleus (protons + neutrons) each time you move from period to period down a group. Additionally, new energy levels of elections are added to the atom as you move from period to period down a group, making the each atom significantly more massive, both is mass and volume. This makes the atomic radii bigger. Part 2 – Ionization Energy 7. The ionization energy is the exact quantity of energy that it takes to remove the outermost electron from the atom. 9. a) The elements that are found at the main peaks of my graph are Helium, Neon, Argon, and Krypton. These elements are all found in the same group. They are all part of the Noble Gases. b) The elements that are found at the main valleys of my...
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