\caption{Separations r for quadruple star systems} \label{tab:threecols} \end{table} \subsection{Flux ratios} Finally, I determined the flux ratios, the ratios of the peaks in flux, for the components of the sources. To calculate these flux ratios, I have to subtract the background from the peak values before dividing both. I calculated the flux ratios of $2$ sources A and B fr via \begin{equation} fr = \frac{\text{peak in flux of source A} - \text{background}}{\text{peak in flux of source B} - \text{background}} \end{equation} and calculated the errors of fr via the Gaussian error propagation.\\ I measured the following flux ratios fr for the UY Aur A system: \begin{table}[h] \centering \begin{tabular}{l|ll} UY Aur A & fr$_{A/B}$ & $\Delta$fr$_{A/B}$ \\ …show more content…
\hline LI 30$\%$& 12.8663 & 5.2105 \\ HI 30$\%$& 75.8071 & 7.0055 \\ \end{tabular} \caption{Flux ratios fr for UY Aur A system} \label{tab:threecols} \end{table} \\ I have done the same calculations for my other observed sources and I got the following flux ratios: \paragraph{Binary sources} $\;$ \begin{table}[H] \centering \begin{tabular}{l|ll||l|ll} GG Tau A & fr$_{Aa/Ab}$ & $\Delta$fr$_{Aa/Ab}$ & ISTau & fr$_{A/B}$ & $\Delta$fr$_{A/B}$ \\ \hline
The absorbance of these mixtures is measured at a suitable wavelength. If 'x' mole/litre are added to (1-x) mole/litre of M and if C1, C2
I have to pull two alleles (two straws) from the bag to represent one fish because fishes like humans get two alleles one from their father and one from their mother.
Because it is a way of knowing the pressure that the blood is putting on the walls of arteries and veins.
Okay, if our lithium weight is going to be 6.941 g/moL Then that means we have to take 24.6g of Lithium and multiply it by 1 mol of Lithium over 6.941 g of Lithium. This would equal to be 3.544 mol of Lithium. Then we have to take that 3.544 and multiply it by 1 mol of hydrogen gas over 2 mol of lithium. Which would then equal into 1.772 mol of hydrogen gas. We can then figure out that 1.772 is our “n”. The “T” is our 301 Kelvin, the “P” is our 1.01 atm and the “R” is our 0.0820 which would be the L atm over mol k. And we can’t forget about our “V” which would be V equals nRT over P which equals 1.772 mol divided by 0.0820 L atm over mol kelvin multiplied by 301 kelvin over 1.01 atm which equals to our final answer of: 43.33 of H2
2.1 What are the coordinates for the White House in Degrees, Minutes and Seconds? 38°53'51.47"N 77° 2'11.64"W
In the process of this experiment, there were a total of two bilingual aphasics and eight monolingual aphasics who were tested through nine different EF test batteries to measure their level of EF which includes behavioural inhibition (response inhibition & interference control), working memory, planning/problem solving and reconstitution. The nine EF test batteries consists of the Stroop Color Word Test, Trail Making Test, Self-Ordered Pointing Test, Complex Figures, Wisconsin Card Sorting Test, Tower of London, Raven’s Progressive Matrices, Five Point Test and Design Fluency. The main focus of the experimentation was to test these 10 different individuals through conversation to investigate their EF profiles. To attain these results, each
As the Fig 1 showing, the node which generates the packet is the source node. There are many sensor nodes in a 3-D interest area. The packet is delivered among these sensor nodes and finally try to reach one sink node. Sink nodes are deployed on the water surface. In the figure, it is a multiple-sink topology. Multiple sinks equipped both radio-frequency and acoustic modems are fixed on the water surface.The packet is assumed delivered successfully if it reaches any one of the sink nodes because sink nodes use radio-frequency channel to communicate with each other which is several orders of magnitudes faster than acoustic channel.
It is an attack by our best friends, …… and these attacks on mostly in randomly generated user name sites it was easy to short.
The first thing to do is to find the initial concentration (C2) of cobalt isopropanol:
For the most part, the probability matrix for $P^2$ is the same as the probability matrix for $B^2$; however, there is one important distinction to be made. Which is that while $B^2[5,0] = \frac{1}{3}$ in the quantum simulation $P^2[5,0] = 0$. On a mathematical basis, this is trivially written as
The goal of this lab is to configure AD DC and PSO on the Windows 2012 previously installed. The main tasks in the lab are to create a group policy object, join the domain, and create a user and apply policy object to that user. In order to do so, I had to add a Windows 7 client to test the functionality of the GPO. The last task in the lab is to create a Password Setting Object (PSO) where we can define the policy of the passwords for all users in a certain group or the whole domain. PSOs are used to define the password requirements such as complexity, age, and repetition. By the end of the lab, we should have Active Directory installed and configured in the infrastructure with GPOs and PSOs defined and tested using a client with a domain
Obesity a risk factor in which excess body fat accumulates and can have negative effects on your health. Here we identify how the hormone insulin reacts in 3T3-L1 fibroblasts and its role on adipogenesis. Adipogenesis is the development of fat cells from pre adipocytes. Insulin is an important factor in the differentiation of 3T3-L1 pre adipocytes to mature adipocytes. Oil Red O (ORO) is used to demonstrate the presence of lipids in each different treatment. A spectrophotometer is used to get the optical density of liquid at the different insulin concentrations. One factor CREB is revealed from preadipocytes to mature adipocytes. By demonstrating how insulin triggers transcription factors. When cells are insulin induced CREB is activated in differentiation. Insulin increased the rate of differentiation and the amassing of triglycerides in 3T3-L1 cells . Insulin was able to induce adipogenesis by observing cell morphology and optical density of liquid from ORO stain. Insulin at 1 µg/ml had the optimal rate of differentiation compared to the other insulin concentrations. Morphology of cells changed significantly from Day 0 to Day 7 at 1 µg/ml and appeared larger and
The purpose of the experiment is to determine the ID of an unknown diprotic acid by establishing its pKa values. The first phase is to determine the unknown diprotic acid by titration, which is a technique where a solution of known concentration is used to determine the molecular weight. While the second phase involved seeing how much NaOH needed to standardize diprotic acid.
The amplitude, or A in the equation is 1/2 |max-min| of the height of the graph.
Statistic images and landscapes, or know as fractal landscapes, and the way that this component works is that these statistic images...