...Kristin’s Cookie Company Solution This is what the process looks like. [pic] 1. The capacity of the system is based on the output of the bottleneck. The bottleneck is the oven, which can process 60min/10min = 6 dozen orders per hour. This assumes one-dozen order sizes. 2. The throughput time is the sum of the processing times for each of the steps. This equals 8+10+5+2+1=26 minutes. 3. In a four-hour shift you can produce the hourly capacity for each hour. This means that you can produce 4*6=24 dozen in the four hours. However, you have a startup time of 8 minutes and a cleanup time of 8 minutes that must be taken away from the four hours. Thus, you have 60min*4hours= 240 minutes – 16 minutes = 224 minutes (3.73 hours) of available processing time. This means that you can produce 3.73*6 = 22.4 dozen per hour. Rounding down is fine here. 4. You are working 6+2 minutes for each dozen. Over the hour, you do this 6 times so you are working 48 minutes. Your utilization would be 48/60=80%. Your roommate is working 1+2+1 = 4 minutes per dozen. So, 4*6=24 minutes per hour. Your roommate’s utilization would be 24/60=40%. 5. Assuming 6 dozen per hour you can gross 6*5=$30. The material expense is 6*(.1+.6)=$4.20. This leaves $25.80 for you and your roommate. 6. You can do this alone. This would mean that you would have to take over your roommates responsibilities of 4 minutes per dozen. Your new total time would be 12 minutes...
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...Jhalak Dikhhla Jaa Show Type : Dance Reality Show, Show Timing : Sat - Sun 9:00 pm Created by : Synergy Communications Judges : Madhuri Dixit, Karan Johar, Remo D'Souza , Director: Priya Wagal, Producer(s) : BBC India, Production company(s) : Big Synergy. Dance Ke Superkids – Battle of the Baaps Show Type : Reality, Dance, Show Timing : Sat - Sun 9:00 pm Created by : TRP (The Right Picture) ,Judges : Geeta, Marzi & Farah Khan. Anchor : Jay Bhanushali & Shreya Acharya Producer(s) : TRP (The Right Picture), Production company(s) : TRP (The Right Picture). | Z Zee_TV'S Zingers | Serial No. | Rank | Date | Day | Start Time | Programme | TVR | 1 | 4 | 02/06/2012 | Sat | 21:00 | KNORR SOUPY DID LITTLE MASTER 2 | 3.95 | 2 | 13 | 29/05/2012 | Tue | 22:30 | PUNARVIVAH | 2.48 | 3 | 20 | 29/05/2012 | Tue | 21:00 | PAVITRA RISHTA | 2.18 | 4 | 24 | 28/05/2012 | Mon | 22:00 | MRS KAUSHIK KI PAANCH BAHUEIN | 1.88 | 5 | 25 | 01/06/2012 | Fri | 21:30 | PHIR SUBAH HOGI | 1.88 | 6 | 28 | 30/05/2012 | Wed | 19:30 | SAPNE SUHANE LADAKPAN KE | 1.74 | 7 | 31 | 30/05/2012 | Wed | 20:00 | HITLER DIDI | 1.57 | 8 | 36 | 29/05/2012 | Tue | 20:30 | YAHAN MEIN GHAR GHAR KHELI | 1.3 | 9 | 50 | 27/05/2012 | Sun | 12:05 | HFF PLAYERS | 0.95 | 10 | 52 | 28/05/2012 | Mon | 23:00 | AFSAR BITIYA | 0.94 | 11 | 56 | 02/06/2012 | Sat | 17:29 | HFF VIVAH | 0.85 | 12 | 92 | 27/05/2012 | Sun | 20:29 | KNORR SOUPY DID LITTLE MASTER 2 LIMELIGH | 0.66 | Target Group :...
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...Detailed Lesson Plan (Grade 7) 1. Objectives After providing the necessary materials, each student; 1.1 investigates the different types of solutions: -unsaturated -saturated -supersaturated 1.2 performs an actual activity about solubility. 1.3 values the common solutions that can be found at home and can be used in daily living. 1. Learning Tasks 2.1 Topic: Solutions 2.2 Concept: The unsaturated solution has a less amount of solute to be dissolved. The saturated solution can hold no more solute to be dissolved. The supersaturated solution cannot hold more solute. 2.3 Materials 2.3.1 Textbook/Other Reference -Science Grade 7: Matter (K-12 Curriculum), pages 1-16 - http://www.infoplease.com/encyclopedia/science/solution-heat-solution.html 2.3.2 Instructional Materials - video clips, materials enumerated for the experiment and activity sheet. 2. Methodology 3.1 Daily Activities 3.1.1 Prayer/ Greetings Christ Jesus Whom we open our eyes, may you be there; When we open our ears, may you be there; When we open our mouths, may you be there; When we open our diaries, may you be there. Help us to see with your eyes; Help us to hear with your ears; Help us to speak your truth in love; Help us to make time for you… for others… for ourselves. Amen. 3.1.2 Checking of Attendance 3.1.3 Checking of Assignment 3.2 Preparatory Activities 3.2.1 Review Teacher: Class...
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...Osmosis Introduction: Tissues are made up of cells that have the same structure and function. In this experiment osmosis will be measured in a piece of tissue. Water potential can be determined by the changes in dimensions of the tissues when it is placed in solutions of different concentrations. Research Question: What is the effect of sucrose solution on potato cells Hypothesis: Osmosis will increase when the solutions are less concentrated with sucrose Variables: Dependant: Osmosis Independent: Length Materials: 12 large test tubes or specimen tubes with bungs, marker pen, potato chip cutter, scalpel, fine forceps, 3 watch glasses or petri dishes, ruler, 6 labeled beakers one containing 50cm^3 of distilled water and other four containing 50cm^3 of 0.2 mol dm^-3, 0.4 mol dm^-3, 0.6 mol dm^-3 and 0.8 mol dm^-3. Method: 1- Using the potato chip cutter, cut 24 chips from the potato. Cut the chips into 5cm lengths using the scalpel. Be as accurate as possible. Place 4 chips into each of six, labeled test tubes, one test tube for each of the different sucrose solutions. Pour in enough of each respective solution to cover the potato tissue. Put a bung in each one of the test tubes and label this series “potato”. 2- Repeat the method using the apple fruit or turnip root, putting four chips of tissue 5cm long into each of the second series of test tubes. Label these tubes “apple” or “turnip”. 3- Leave the tubes for at least an hour 4- After at least...
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...FUNDAMENTALS OF SURFACE MODES: ¥ COLLOIDS ¥ means ÒglueÓ in Greek ¥ was coined in 1861 by Thomas Graham. ¥ ¥ ¥ ¥ ¥ WHAT IS A COLLOID? usually consists of two phases; one continuous phase in which the other phase is dispersed. Size of particles: larger than the size of molecules and small enough for the dispersed phase to stay suspended for a longer period of time. ¥ No strict boundaries for the size limits. OFFICIAL DEFINITION ¥ In 1903 Wolfgang Ostwald formulated the official definition of a colloid: ¥ a system containing entities having at least one length scale in between 1nm and 1µm. ¥ For smaller particles there is no distinct boundaries between the phases and the system is considered a solution; ¥ for larger entities the particles will fall to the bottom due to the gravitational force, and the phases are separated. 1 2 3 MESOSCOPIC PHYSICS ¥ The particle size is in the so-called mesoscopic range in between the macroscopic and microscopic limits. LARGE INTERFACIAL AREA ¥ One very important quality of the colloids is the large interfacial area between the dispersed and the continuous phases. WHAT EFFECTS HAS THIS? ¥ This means that interface effects and hence the electromagnetic surface modes, are very important for the properties of the colloids. ¥ It costs energy to create this much surface and the particles would clump together if this isnÕt prevented. ¥ Usually the particles are charged and hence repel each other. 5 6 4 Four states...
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...encouraged to ask question/s that is related to this module and that you don’t know the answer to and want it to be answered. The question/s can be answered by anyone in class. The instructor will allow sometime for other students to answer the question/s before contributing. You can post your questions in MODULE 2 forum under the Discussion Forums. Complete Assignment #2. The assignments are posted in the ASSIGNMENTS area of the website. Even though the homework assignments are not to be turned in and graded, you are strongly encouraged to do them to understand the material and to prepare for quizzes and tests. A complete solution to this assignment will be posted on the course website under ASSIGNMENTS on Thursday (Sept 4th). You are encouraged to review the solutions and compare your work to the solution, ensuring that you understand the reasons the solutions appear as they do. You are also encouraged...
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... * Has no secretary The most important problem would be catching the flight for the meeting in Chicago since it’s an emergency and work related. Solutions: Getting work done for the day on the air plane on the way to the meeting. Call the secretary to find out at least some of the info for the meeting. Item#2 Problems: * Internal politicking in the company * Finding someone to replace F.T. Dickenson * Dealing with eliminating his overtime hours The two most important problems would the politicking, because someone else may feel the same way and decide to leave the company too. Also, finding someone to replace him and do his unfinished work in such a short period of time. Solution: Search for a new employee ASAP. Item#3 Problems: * Workers threatening to walk out over a co-worker * 10 votes to dismiss Foreman Edward George The workers are the most important asset to the company, so their interest should be first. Second, you have to figure out what to do about the votes to dismiss Ed George. Solutions: Talk to Ed George about his problems with his co-workers. Hold a meeting and get even more workers involved in the voting process, the take another vote. Item#4 Problems: * Overloading which can result in interruption of electrical power. * Not being reachable for Southern Power Solution: Contact Southern Power ASAP. See if they could come out and help with the problem. Item#5 Problem: * Balancing your work and his until Wednesday ...
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...Name Date Class COLLIGATIVE PROPERTIES OF SOLUTIONS Section Review Objectives • Identify the three colligative properties of solutions • Describe why the vapor pressure, freezing point, and boiling point of a solution differ from those properties of the pure solvent. Vocabulary • colligative properties • freezing-point depression • boiling-point elevation Part A Completion Use this completion exercise to check your understanding of the concepts and terms that are introduced in this section. Each blank can be completed with a term, short phrase, or number. In a solution, the effects of a nonvolatile _______ on the properties of the solvent are called _______. They include _______ point and vapor pressure _______, and boiling point _______. In each case, the magnitude of the effect is _______ proportional to the number of solute molecules or ions present in the _______. Colligative properties are a function of the number of solute _______ in solution. For example, one mole of sodium chloride produces _______ as many particles in solution as one mole of sucrose and, thus, will depress the freezing point of water _______ as much. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. Part B True-False Classify each of these statements as always true, AT; sometimes true, ST; or never true, NT. 11. When added to 1000 g of water, 2 moles of a solute will increase the boiling point by...
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... it increases the water movement out of the plant by a process called transpiration. When there is a shortage of water, the guard cells well get smaller and close the stomata, and the transpiration process will move slower. The osmolarity would be tested with solute concentrations ranging from 0.0 M to 0.6 M. The concentrations increased 0.1 M each time and the solute used was sucrose. The goal of this experiment was to determine which concentration had the least effect on the potato after being incubated. This information helped us estimate the osmolarity of the potato tuber tissue. Our group hypothesis is that the osmolarity of the potato will have the concentration with the 0.1 M solution. This hypothesis is based off the predicted outcome that smaller molarity concentrations will have the least effect on the potato. The more solute added to a solution decreases the concentration of water in most cases (Kosinski). The decrease in water concentration would then lead to a lower weight of the potato tuber once it has finished incubating. Methods & Materials: For experiment A you will need the following items: 1 large potato tuber Forceps petri dish razor blade DI water metric ruler 7 250 mL beakers/ or disposable cups balance that weighs to the nearest 0.01g sucrose...
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...The data solute concentration of the egg was found by taking the average percent change in mass for each percent solute concentration used in the lab and putting the data into the graph(Mass vs. Solute Concentration Over 3 Days). According to the data from the class, the solute concentration inside an egg is 25.1%. The egg would be in an isotonic solution at 25.1% solute concentration. This is when the concentration outside the egg is equal with the concentration of water inside the egg. The point of equilibrium is the point at which the trendline crosses the x-axis. At this point, the egg would no longer gain or lose mass. If the concentrations are the same inside and outside the egg, diffusion will not be able to take place. All averages...
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...Analysis of Carcinus maenas ingestion rates Materials and methods Part 1 - Control experiment Two mussels (mytilus edulis) were opened and the mussels inside cut into 4 pieces, producing 8 pieces of roughly equal size in total. A balance was then tared with a weighing boat placed on it. Each piece of mussel flesh was placed individually and in turn into the weighing boat to be weighed. The pieces were then placed into a plastic bucket half filled with filtered seawater. The weight, time of weighing and time of immersion into the bucket were all noted down for each piece of mussel. The position of each piece of flesh inside the bucket was also taken down so as not to mix them up. One of the eight pieces was left in the bucket for the duration of the experiment until being removed and weighed at the end and was written down as “the last”. Another piece was removed frequently throughout the experiment and was given the name “the one”. The last 6 pieces were each removed once and weighed at regular intervals to provide a range of submersion times. The experiment was run for roughly 20 minutes. The method for removing and weighing the pieces of mussel was kept consistent throughout the experiment. Each piece was removed, blotted clear of any excess water and weighed. If the piece of mussel was “the one”, it was returned to the bucket of seawater, if not, it was kept aside and moist for the crab feeding (Part 2). Once all of the pieces were removed and weighed (multiple times for “the...
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... The purpose of the experiment was to see how abrasion works and how it affects rocks and minerals with the use of sugar cubes. I measured the mass of all 5 sugar cubes together then placed it in a jar with a lid on and shook it 20 times, as I was shaking the jar my lab partner recorded the mass that was taking with the number of times I shake the jar. After which we drew the shape of the sugar cubes when it was poured out on a paper, then we calculated the percent change. This procedure was repeated 5 times and with each trial the shaking time was increased by 20. After all the calculations was done, we graphed the data. After each trial the size and mass of the sugar cubes reduced, this was due to the pressure put on the jar as we were shaking it. Supposing little pressure was put on the jar, the outcome would have been different and would have shown little abrasion effects on minerals and rock so therefore this made me conclude that, for abrasion to have a greater impact on minerals and rocks the pressure of the water or wind must be strong to push the minerals against each other. Another factor that contributes to how abrasion works could be the amount of space the minerals and rocks have because during the experiment, the sugar cubes pumped into each other and the walls of the jar and since there wasn’t enough space for movement the sugar cubes weathered and reduce in size with the mass decreasing which I believe wouldn’t occur if there were to be enough space for the sugar...
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...Polyimides are a synthetic polymeric resin of a class resistant to high temperatures, wear, and corrosion, used primarily as a coating or film on a substrate substance. Polyimides are also polymers that usually consist of aromatic rings coupled by imide linkages-that is, linkages in which two carbonyl groups are attached to the same nitrogen atom. Polyimides cover the whole range of high performance polymers. They are a sophisticated family of materials which have applications in highly technical end use fields from aerospace to microelectronics. In this paper I will give an introduction to polyimides, and then I will talk about the formation of Polyimides, and finally the effects of solvents in Polyimides. Polyimides possess very unique key properties such as thermoxidative stability, high modulus, high mechanical strength, excellent electrical properties, and superior chemical resistance. Because of these merits, general difficulty in processing polyimides and their high demand cost did not deter advanced exploration of new compositions and new processing methods which points at value-added niche markets in advanced technology applications. A very important consideration in the successful synthesis of Polyimides is a design/creation of proper composition and proper decisions of synthesis methods. Ultimately, the latter is determined by the chemical and physical properties of monomers and polymers and intermediates. When a diamine and a dianhydride are added into a dipolar...
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...The experience of the egg drop project was one that will help you realize that not every way that you try will work, and just because you fail once doesn't mean you can't try again. I believe my egg broke possibly because when I was testing mine I was working with a larger egg. So when it came to the drop it wasn't supported by the walls as much as it should have been. I believe also because the straws on the bottom didn't really reach the egg since it was a smaller egg than what I was practicing with. I feel if I had also dropped it a couple more times or had made it a tiny bit smaller it might have concealed the egg better , and hopefully wouldn't have cracked than.I believe also that when it came to the test it didn't land straight on...
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...dissolved in H2S c. CO2 dissolved in ammonia d. Mg(NO3)2 dissolved in water 2. At sea level, the partial pressure of O2 is 0.21 atm. However, the Dead Sea is not at sea level -- it’s at a much lower altitude, where the atmospheric pressure is 806 mmHg and the temperature is 20˚C. Use Henry’s Law and Table 13.2 to calculate the molar concentration of O2 in the Dead Sea. 3. Suppose we make a solution by dissolving 42.5 g of I2 in 1.50 mol of CCl4. a. What is the molality of this solution? b. Suppose we wanted to use our 42.5 g of I2 to make a 0.500 m solution. What volume of CCl4 would we need? The density of CCl4 is 1.589 g/mL. 4. We dissolve 200.0 g of NaCl in 684.5 g of water and find that the density of the solution is 1.15 g/mL. Calculate: a. The % by mass of NaCl b. The mole fraction of NaCl c. The molarity of NaCl d. The molality of NaCl 5. Calculate the number of moles of solute in each solution. a. A solution of Br2 dissolved in 50.0 g of...
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