Examine the way in which the culture

 

He was an important part of the town from then on. The story is set in South America in Chile. The town is a remote one. The culture is very individual and the town would be small. Because the town is small and remote the people in the town are very close. Everyone will know everyone’s business. People would trust and respect each other and stick up for each other, especially for the woman who has helped them out so much, the schoolteacher. The town may be small and remote but it does have a police force that the town had to hide the body from and its whores who distracted the police while the people drove away the body.

The important themes relate into the story because the town is so small. The town is small and therefore the people will all know each other and look after each other. Also they relate with each other because when the woman’s’ sons’ murderer turns up in the town and the woman cuts off his head the towns people help her get rid of the headless body. The town gets out of its own way and shields what the woman has done from the law. She is the respected matriarch in the town. The people will help her to dispose the body before the law finds it. They also help her cover the body and hide it from where anyone can find it.

The schoolteacher receives a lot of respect in her town because she is a woman who is educated and looking after herself. She has her own business and lives alone. In the time the story happened, women were still seen as less capable of having control, being prosperous and less respected than men. Because she has had a lot of success by herself she receives so much respect in her town. Her culture and expectations has affected her life in that way. She would not normally receive all that respect as an ordinary. The town had not forgotten about the death and loss of the boy.

As soon as the towns’ people had heard about the news of the murderer being back the people became lively and excited by it. People seemed to lose their fear, in line 135 ‘Women carried their kitchen chairs out to the sidewalk and sat down to enjoy the cool air’. Line 135 suggests that the people are breathing a sigh and sense of relief that the man has been found. When the men return form barring the body they found that the lights of the town were still on and no one had gone to sleep. The people were celebrating. A Stench of Kerosene The main themes in this story are: Loyalty; A feeling or attitude of devoted attachment and affection.

Often used in the plural:” My loyalties lie with my family” Love; A feeling of intense desire and attraction toward a person with whom one is disposed to make a pair; the emotion of sex and romance Other important themes in the story are expectations, culture and marriage. All of these themes affect the main characters lives. They mainly affect the character Guleri’s life. All of the factors affect her life, decisions and expectations. The story is set in India in the mid-twentieth century, in a Hindu district. The culture in India expects that the people get married at a young age and have children at a young age.

The parents of the male want their son to have children early because they want to keep the family name. The wife of their son has to live with them and not the husband live with the wife’s family. The story is about love as are most of the other stories we read. The stories veronica, country lovers and the school teachers guest are also about love. In the story the culture the people live in effects the way the people behave and their expectations. The important characters, whose lives are affected by their culture, are: Guleri, Manak’s wife, (girl) Manak, Guleri’s husband, (boy)

and Manak’s parents. In the story the girl Guleri and her husband Manak are not yet adults. They are both very young still. Guleri lives with Manak and his parents. Guleri only sees her parents and friends once a year and they live in another town quite a way away. The couple do love each other, but in India there are arranged marriages. When Guleri leaves the lovers are split up for a bit as in the story Veronica. Guleri does not want to miss the opportunity to go to her parents, the same as the boy in Veronica who does not want to miss the opportunity for a better life.

This shows us that she is leading a very independent life. She is not allowed to make her own decisions. Guleri loves to go see her parents for those few days every year. But this year when she is gone Manak’s mother has something planned. Guleri doesn’t appear to know about it at first. Manak obviously tried to tell Guleri about it because he was trying to persuade her to stay this year and not go to see her parents. He started saying in the story: ‘My mother… ‘ But stopped. He knew about what his mother wanted and wanted Guleri to stay so he would not have to go through with it.

If she stayed then his mother could not bring the ‘other’ wife in to Manak and they certainly would not be able to force Manak and Guleri apart then. Manak couldn’t find it in his heart to say to Guleri what his mother had decided for him. The couple are obligated to have children and that comes into conflict with their love for each other. Guleri knew what was going to happen. Her love and loyalty to Manak overcame her decision to live. At the fair she burnt herself. Manak however responded to the new bride. He didn’t emotionally but physically. He only did so because it was his culture and the way things worked in India.

Guleri wouldn’t give Manak a child yet so his mother paid for a new bride for her son to bring a child to carry on the family name. The couples’ lives and decisions had been affected by their culture in their society. Manak felt that he had to respond to the new bride. He didn’t want to but he did not protest against it because it would be disrespectful to his custom and family. Guleri did not want to accept that that was the custom she had to live with. She was in love with Manak and didn’t want to be second place. She was not willing to accept the culture. At the end of the story Manak said: ‘Take him away!

He stinks of kerosene. ‘ This could be symbolic to say that the child reminds him too much of Guleri, whose death had something to do with kerosene. He did not want the child near him because it was not from the woman he loved. But again he had to accept it as it was his culture. Women’s expectations in that time were:  To produce an heir for the family name, therefore a son not a daughter.  They are only objects/possessions. Not people with feelings, emotions or many rights. * If the woman cannot have children she is unfit for the parents son. So the parents would find their son a new or second wife.

Since the Soviet Union

In this research essay, I will examine the development of the state of Mongolia since 1990. While doing research for this paper, I have noticed that an underlying theme in many academic sources is the development of Mongolia in the past decade. Since the Soviet Union has withdrawn major monetary support from Mongolia in the late 1980s, the nation has had to restructure most of its government, and economy, which also caused a social change. In an attempt to revitalize the economy, the government of Mongolia has taken steps to improve the nation in several ways.

I will examine the development processes in the context of the following areas of concern: demographics, history, nationalism, current issues, economics, government, communication and education, and finally, infrastructure. Demographics Mongolia is a large (1. 5 million km5), land-locked country located between two giant countries: Russia and the People’s Republic of China. Mongolia’s present population of 2. 6 million people is growing at 1. 4 percent. The national language is Mongolian, but there are other languages spoken, which are Turkic, Russian, Chinese, and English.

Ulaanbaatar, with a population of 650,000, is Mongolia’s national capital. Other major cities include Darkhan (90,000) and Erdenet (65,000) (ADB 2000: 3). Located deep within the interior of eastern Asia far from any ocean, Mongolia has a marked continental climate, with long, very cold winters, and short, cool to extremely hot summers (temperature highs to 40 degrees Celsius). Its variety of scenery comprises upland steppes, semi-deserts, and deserts, although in the west and north, forested, high mountain ranges alternate with dry, lake-dotted basins.

Mongolia is highland country, with an average altitude of 1,585 metres above sea level (Rupen 1978: 15). Following independence from China in the 1920s, Mongolia became the second country after Russia to adopt communism in 1924. Mongolia remained closely tied to the Soviet Union until the end of the 1980s. It received support from the Soviet Union economically and with its military, and generally followed Soviet guidance politically and culturally. Mongolia now practices a democratic government with multiple parties.

It is currently attempting to rebuild national culture which it lost during the Soviet communist establishment. I will further discuss its history in the “History” section. Mongolia’s major exports include copper, livestock, animal products, cashmere, wool, hides, fluorspar, and other non-iron metals. Major imports are machinery and equipment, fuels, food products, industrial consumer goods, chemicals, building materials, sugar, and tea. Mongolia’s training partners are Russia, China, Japan, South Korea and the United States.

Mongolia’s assets include abundant natural resources, a well-educated population, and a strategic trade location. Ninety-seven percent of the population is literate, compared to the average literacy rate of fourty nine percent in South Asia, and fifty three percent in low-income countries world-wide (ADB 2002). History In 1196, a small nomadic tribe lead by Chinggis Khan rose to power in Mongolia. The Khan tribe triggered the start of a great Mongolian expansion from Mongolia to as far west as Palestine.

For 200 years Chinggis Khan and his successors ruled a large portion of Asia as well as parts of the Middle East thanks to their intelligent and well planned military strategy. The empire fell to the Chinese Qing dynasty in the eighteenth century. The Qing dynasty separated Mongolia into Outer and Inner Mongolia, where Outer Mongolia extended from the Gobi Desert to the Russian border, and Inner Mongolia touched the Chinese borders. As a result of its proximity to China, Inner Mongolia experienced much more Chinese influence than did the people of Outer Mongolia.

There was as influx of Chinese settlers into Inner Mongolia, where they slowly gained economic power through the Qing implementation of taxes and money lending to the Mongolians of the area (Rupen 1978: 13). Even to this day, Inner Mongolia is ruled by the Chinese and remains separate from Outer Mongolia (now called the State of Mongolia). Outer Mongolia was largely left on its own for a hundred years, where they peacefully practised Buddhism introduced by the Tibetans. Eventually, however, the Qing dynasty was able to corrupt Buddhism in Outer Mongolia, and gained power in the region.

It was only in 1911, when the Qing dynasty collapsed, that the Russians had a great amount of influence on the Mongolians. With the help of Russian forces, Outer Mongolia immediately claimed their independence from China in 1921, and in 1924, again with the help of Soviet forces, the Mongolian People’s Republic was proclaimed. Until the end of the Soviet era, Mongolia was very much under Soviet control. It relied heavily on Soviet economic support and modelled its administration and economic development on the Soviet pattern.

In the 1960s the copying of the Soviets reached its peak: an intensive program of industrialization commenced. The implementation of the industrialization program conflicted with many environmental values viewed amongst most Mongolian pastoralists. The industrialization of the nation affected the Mongolian nation and culture for many years after. The Buddhist religious establishment, for example, controlled most aspects of Mongolian society, at least in Outer Mongolia, prior to the 1930s.

It played a large part of religious and cultural identity of the Mongolians, and was interwoven into everyday life: it funded and ran the education, judiciary, and health care systems. The Soviets, however, saw the Buddhists as a hindrance to the Mongolian revolution lead by the communist Soviets because of their different belief and value systems, and as a result, almost anything related to Buddhism was abolished in the country by the Soviets (Gilberg ; Svantesson 1996: 13). In the early 1980s, the rise of Gorbachev as the new Soviet leader spurred a negative reaction amongst Mongolians toward his new policies.

Because of this, opposition demonstrations in 1989 led the ruling Mongolian People’s Revolutionary Party (MPRP) to call multi-party elections following which, in 1992, the communist state was dismantled, and the People’s Republic of Mongolia became the State of Mongolia (Altangerel, 2001: 4). Nationalism Mongolia’s history continues to play a large role in Mongolian identity, especially the legacy of Chinggis Khan and his successors. While other nations like Russia might think of the Khan dynasty as destroyers, Mongolians have always seen him as the centre and origin of the national history and the founder of their independent statehood.

It is not surprising then, to see Mongolian interest and pride in Chinggis Khan (Sabloff 2001: 102 ). Prior to Russian rule, the Mongolians were generally a peaceful population of nomadic pastoralists, who, many of them, practiced Buddhism. However, in the past few decades, by mechanically copying other nations (like the Soviet industrialization during the 1960s), Mongolia almost lost its cultural roots, customs, and traditions. In 1989 the Mongolia People’s Revolutionary Party (MPRP) recognized this problem and initiated a “Mongolian nationalist revival” (Bruun ; Odgaard 1996: 33).

The Frankfurt school

Ever since the introduction of mass media through print, media has surrounded us and has had significant impact on our political views, our economy and even shapes our culture and society. With media so intertwined with such important aspects of our lives, it is important to develop an understanding of how media effects the political, economic and cultural meaning of our lives through media studies. For this essay, I will explain two political aspects of media and discuss the relationship between the economic power of the media (ie.advertising) and how this plays a considerable role in shaping our culture.

I will also point out how media studies assist us to understand these aspects better throughout the essay. The importance of understanding the relationship between the media and parliamentary politics is crucial to our political understanding in general. The media aims to provide a variety of political views and opinions so that as voters, we have a well rounded knowledge of all parties and issues involved.

However, many of us are unaware of the developing relationships between media owners and politicians. As a result of this camaraderie, journalists are mere puppets as their boss dictates which way they want the story to slant1 to support their ally, and hence, a bias is created. Media studies not only allows us to explore these controversial relationships but to also become more sophisticated in understanding the hidden language of the journalist and politician. In comparison to the written word, visual media seems to have a powerful control over the audiences’ political opinions.

One might assume that good looks may be a contributing factor to a political win. However, Windschuttle argues that “good looks are largely in the eye of the beholder… someone who has charisma will become good looking”2. As an example, Windschuttle refers to Richard Nixon’s television appearances during the 1972 Presidential election3. According to western standards, Nixon was not considered a ‘good looking’ man. However, his charisma and mastery of television catapulted him to the top.

Similarly, Prime Minister Bob Hawke was also known for his charisma and transparent nature, which were both factors that contributed to his popularity4. The second approach to politics in media is related to the politics within the media. Media studies is fundamentally a political study into the “power structures”5 behind each form of media. By using Harold Lasswell’s model of communication, we ask “Who says what, in which channel, to whom and with what effect? “6. This type of questioning allows us to develop a healthy suspicion so that everything is not taken for face value.

The theory of political economy also suggests that the media is essentially a corporate enterprise7, run by powerful people who play a dominant role in what the media says and how they say it. The media coverage of the infamous S11 protests began months prior to the date with journalists creating an assumption that it would be a bloody affair. However, journalists never revealed that during these months of preparation, protesters had participated in workshops educating themselves on the benefits of non-violent protesting8.

Due to this lack of knowledge, thousands of police men and women were briefed with the theory that the protesters were going to be violent and brutal. As a result of this, the second day of the protests saw police from the paramilitary Force Response Unit aggressively seize upon 50 civilians non-violently protesting against large global corporations9. The police provided no forewarning to the group or directions to move before the assault. The attack was repeated 12 hours later on a group of approximately 100 civilians, resulting in injuries to about 70 citizens. 10

So why did the media prematurely portray the protesters as aggressive, even when they were unsure of the outcomes? The media tycoons who run the media were part of the World Economic Forum and had given specific instructions to portray the protesters as violent so that police numbers would be high to ultimately protect themselves. This example of manipulation from behind the scenes gives us a new perspective on the issue. We often find that the inclusion of this political perspective gives us the knowledge to be a more critical member of society, which can be quite rewarding. 11.

Scenario of a Human Service Client

Scenario of a Human Services Client Working in the field of human services is a unique, yet rewarding career. Human services workers engage with diverse populations and carry different titles to include counselors, case managers, milieu therapists, social workers, etc. Regardless of what title a human service worker may hold, communication between a client and professional is vital. The purpose of this paper is to introduce a client from a particular background and address his presenting issues.

It will review how I would interview my client to include specific communication techniques, ways I would establish a working relationship, discuss possible boundary issues, and recommendations or referrals to certain groups for this client. The scenario I chose to write about is an actual experience I’ve had with a client working in a residential treatment center for troubled youth. In this paper, I will refer to my client as Joe Smith. Joe is a 13 year old Hispanic male. His mother and father are still alive, but he currently lives with his mother.

Joe has 3 siblings, all brothers, one of whom has been in and out of jail and seems to be the biggest influence in Joe’s life right now. Joe’s father is a well known drug dealer and has also been in and out of jail, and is now trying to be a part of his son’s life. Joe’s mother has good intentions of ensuring her son receives the proper treatment, but is losing patience and repeatedly states she is running out of options to provide a stable home for Joe. Joe’s referral to the treatment was because he got caught at school selling drugs. There were also allegations of inappropriate contact with a female peer.

Upon Joe’s admittance to the program, his mental health assessment diagnosed him with Attention Deficit Hyperactive Disorder, Post Traumatic Stress Syndrome, and Oppositional Defiance Disorder. His placement into out-patient services set a goal to help reduce his anger outbursts and teach him life skills that would help him reintegrate back into public schools with more positive behaviors. He is in a track with Mentally Retarded/Developmentally Delayed population. His treatment team consists of a therapist, milieu therapist, case manager, social worker, and probation officer.

My first clinical interview with Joe was productive. I went over goals and objectives to pass through his phases. His responses were short and made very little eye contact. Joe was very quiet his first few weeks in program. He was compliant with staff and stayed on task with class work and actively engaged in group and individual therapy sessions. As he got comfortable with his peers and staff, he began showing his personality, which allowed me, as his milieu therapist, to begin helping him develop coping skills and teach him more appropriate behaviors when he would lose control of his emotions.

His anger outbursts became more frequent, and some resulted in physical altercations with other peers. He was found with drugs and rumors of him selling drugs to other clients became an issue. As previously mentioned, I believe communication between a client and human services worker is of vital importance in any individual’s treatment. According to the Center for Negotiation Studies website, there are three major styles of communication: Aggressive, Passive, and Assertive (Communication Skills, n. d. ).

My communication style with Joe is Assertive. I am an effective, active listener with Joe. He does not open up very easily, and so when he does I need to ensure he feels like I am listening to every word he says. I set limits and expectations with Joe, and I took me awhile to learn that this was more effective in his treatment than just letting him talk without knowing my expectations of him. I would state my observations to him, without being judgmental or labeling him in any way. Establishing an effective relationship with Joe took time.

I took on the model of “You do for me, and I’ll do for you”. Gaining his trust took time, and I understood that he has been let down his whole life with empty promises from family and friends. Keeping my word with this client is how I began establishing a close relationship. If I told Joe I would stop by and see how he was doing in 3rd period, I made sure I was there, and if I could not be there I made sure he was aware as to why I was not. I gained respect from this kid, and was able to teach him coping skills within just a few months.

We were able to understand each other’s verbal and non verbal communication styles, which made his treatment much more effective. My interviews with Joe after developing a relationship were a lot more productive. He was able to state his own goals and more importantly what he needed from me, as his milieu therapist, to achieve those goals. Working with Joe on a daily basis did contribute to some boundary issues. There were times when he would want to process when no real crisis was present. His attachment to me was obvious, and limits had to be set.

I, along with my supervisor and clinical director, did not believe it was a personal attachment, but had finally found someone he could trust to be there for him without judgment. After confronting him with this concern, he was able to accept the limitations and with positive encouragement began using his coping skills on his own. After about 9 months of having Joe in the out-patient program, I reached a stopping point where I could no longer help him. He had developed coping skills and knew when he needed to use them. He recognized his triggers and was able to verbalize his wants and needs to his parents.

Unfortunately, the environment he was in at home was unstable. Joe’s father was continuing to try and get him to deal drugs, and his brother’s negative influence was overwhelming. My referral for Joe into Treatment Foster Care was the next step for my client. Getting Joe out of his family life where all he knew was drugs and jail was the best outcome for him. He is with a family who provides stability and is helping him become a productive member of society. References CNS. (n. d. ). Communication Skills. Retrieved from http://www. au. af. mil/au/awc/awcgate/sba/comm_style. htm

The data-out

VHDL, is it easy, or is it not, is it step forward, or just another complication in achieving the final goal? Do we need another programming language, even if it is meant to put software and hardware together, or we are much better off with pure electronics, soldering the wires, inhaling the fumes? The following report might clarify one’s mind about different approaches in designing digital circuits. A sequential synchronous circuit was designed using VHDL programming method, and then expanded based on the knowledge of the electronic components.

Program code has been debugged and simulated. The resulting waveforms confirmed the expected operation. On the other hand, some of the circuit components have been designed from scratch using digital design techniques. The procedure then has been compared to the programming method Even though, VHDL programming approach may be seen easier and faster, it should be noted that to actually be able not just to program, but to program well, one must have an appropriate ‘vision’ of the problem to be able to implement it in the programming code. And that is not a very transferable skill.

Therefore it is concluded that although majority of designers will benefit from the usage of VHDL programming, some may encounter certain difficulties and ought to use less advanced techniques. Introduction VHDL (Very high speed integrated circuit Hardware Description Language) is a programming language designed for the purposes of description and of the behaviour of digital circuits and systems by function, data flow behaviour and structure. It is used to configure a programmable logic device with a custom logic design. The design of fault free synchronous sequential circuit was carried out.

Two ways of the implementation have been considered: schematic approach and VHDL programming approach. While some may consider schematic approach more logical and hands-on method of designing, an attempt has been made to implement the specifications in the VHDL programming language first. Having debugged the code, separate sequential circuit components have been designed and put together to result in the working schematic implementation. Objectives The designed synchronous sequential circuit will have to take two unsigned 4-bit numbers at the serial input and will output the greater of the two numbers on serial output.

The circuit will include asynchronous Reset input, Clock input, and two synchronous inputs: Start and Data-in. The two synchronous outputs will be: Ready and Data-out. Fig. 1 shows the timing diagram of the expected behaviour of the circuit. The following is a sequence of the circuit events: 1. Upon receiving of the Reset signal (not shown on the diagram), circuit settles in the known “0” state; 2. After one clock event of the Start signal, circuit starts to accept the input data from the Data-in. 3. When all 8 bits are received, the two 4-bit numbers are compared.

4. Once compared, circuit outputs the Ready signal for one clock cycle. 5. After the Ready signal, the largest 4-bit number is outputted serially through the Data-out. Method Before designing any circuit, it is a good idea to picture the major components it is going to be constructed of, and their interconnections. The overall external view of the system to be designed is shown in Fig. 2. Some of the components are combinational devices themselves. Considering the above figure and the timing diagram specified, the following ASM chart has been built.

To further extend the understanding of the circuit functionality, the diagram of the components interconnected together is demonstrated below, and afterwards an explanation will be given as to the sequence of events happening within the circuit. Now that the components diagram is derived, a working sequence of the circuit can be explained.  Reset is an asynchronous input clearing the whole circuit independently of the clock and putting gates and flip-flops into known “0” state.  Once the Start becomes “1” for 1 clock cycle, the 4-bit counter initiates and starts to count.

Starting from the next clock cycle it generates Enable 1 “1” signal to activate shift register.  After 8 clock cycles, counter outputs “0” on the Enable 1 to stop shift register inputting data. At this point all the 8 bits have been propagated along the shift register and are stored in the flip-flops. On the next clock cycle, Enable 2 is activated by the counter to enable comparator Once comparator has taken the two 4-bit numbers, they are compared and when the largest is found, the Ready signal is transmitted for 1 clock cycle.  At last, the largest 4-bit number is outputted once, LSB first.

Van Vlack

A hard sphere model of the unit cell of the FCC crystal structure is shown below: where aFCC represents the length of each of the edges of the cubic unit cell (since the unit cell is cubic, all of these edges are equal in length), and RFCC represents the radius of an atom within the unit cell (since all the atoms within a unit cell of an elemental solid are made up of the same element, they all possess the same radius). From this diagram, the number of atoms in the unit cell of the FCC crystal structure can be calculated. There are eight corner atoms in the above unit cell, and each of these corner atoms is shared among eight unit cells.

Therefore, each unit cell contains one-eighth of each of these corner atoms. There are also six face centered atoms in the above unit cell, and each of these face centered atoms is shared among two unit cells. Therefore, each unit cell contains one-half of each of these face centered atoms. Mathematically, therefore, the number of atoms in the unit cell of the FCC crystal structure can be determined in the following manner: From the above hard sphere model of the FCC unit cell, we can see that the atoms of the unit cell touch each other along the diagonals of each of the faces. This diagonal has a length of 4*RFCC.

Therefore, we can use the pythagorean theorem to relate the edge length of the cube, aFCC, and the atomic radius, RFCC. We can then rearrange the above equation to solve for aFCC: The atomic radius of platinum is given in the question as 0. 1387 nm. Similarly, the atomic radius for copper is given as 0. 1278 nm. We can substitute these values into the above equation for aFCC to determine the edge lengths of the unit cells of platinum and copper: with significant figures applied with significant figures applied where aPlatinum is the edge length of the unit cell of platinum, and aCopper is the edge length of the unit cell of copper.

Since the three edges of a cubic unit cell are all equal in length, the volume of a cubic unit cell can be calculated by the following formula: where acubic_unit_cell is the edge length of a cubic unit cell. Substituting our values for edge lengths of the unit cells of platinum and copper into the above formula, we can obtain the volumes of the unit cells of platinum and copper: with significant figures applied with significant figures applied where (Volume/unit_cell) Platinum is the volume of a unit cell of platinum and (Volume/unit_cell) Copper is the volume of a unit cell of copper.

Using the conversion factor of 1 nm = 10-9 m, we can convert the volumes of the unit cells of platinum and copper to meters: with significant figures applied with significant figures applied As shown in question #1, the mass of a unit cell of a material can be calculated by the following formula: The density of a material can be defined by the following formula: Therefore, the density of a material can be defined in terms of the above two equations: The (Mass/mole) of a substance is defined as its atomic weight or molar mass.

According to Appendix B of Van Vlack, the mass/mole (or molar mass) of platinum is given as 195. 1 g/mole. The mass/mole (or molar mass) of copper is given as 63. 54 g/mole. As stated previously, for atoms, Avogadro’s number is equal to 6. 022×1023 atoms/mole. Using these values in the above formula for density, the densities of platinum and copper can be determined. with significant figures applied with significant figures applied According to the question, copper forms a substitutional solid solution for concentrations of up to approximately 6 wt% copper at room temperature.

Therefore, under the assumption that the alloy was formed at room temperature, an alloy containing 5 wt% copper and 95 wt% platinum will form a substitutional solid solution, with copper acting as the solute and platinum acting as the solvent. According to equation 4. 10a on page 72 of Callister, where ? ave is the density of a binary alloy (an alloy composed of two elements), C1 is the concentration of the first element in the alloy (in wt%) C2 is the concentration of the second element in the alloy (in wt%), ? 1 is the density of the first element in the alloy, and ?

2 is the density of the second element in the alloy. Let us define platinum to be the first element in our alloy, and copper to be the second element in our alloy. Since the platinum-copper alloy in this question is defined as containing 5 wt% copper and 95 wt% platinum, this implies that CCopper = 5 and CPlatinum = 95. Using these values, and the densities of copper and platinum, we can apply the above formula for the density of a binary alloy to determine the density of the platinum-copper alloy: with significant figures applied where densityalloy is the density of our copper-platinum alloy.

According to equation 4. 11a on page 73 of Callister, where Aave is the atomic weight (molar mass) of a binary alloy (an alloy composed of two elements), C1 is the concentration of the first element in the alloy (in wt%) C2 is the concentration of the second element in the alloy (in wt%), A1 is the atomic weight (molar mass) of the first element in the alloy, and A2 is the atomic weight (molar mass) of the second element in the alloy. Note that according to page 73 of Callister, the above equation for atomic weight of a binary alloy does not always yield exact results.

This is largely due to the fact that the above equation is derived based on the assumption that the alloy volume is exactly equal to the sum of the volumes of each of the elements that compose the alloy. For the majority of alloys, this assumption is typically false, but the assumption does not create significant errors in most practical situations. Please refer to page 73 of Callister for more details. Using our values for the molar masses and concentrations (in wt%) of copper and platinum, we can apply the above formula for atomic weight of a binary alloy to determine the atomic weight of our copper-platinum alloy.with significant figures applied where Aalloy is the atomic weight (molar mass) of our copper-platinum alloy. We have previously defined the density of a material in the following manner:

We can then rearrange this formula to solve for the volume of a unit cell. We have previously calculated the values for the atomic weight (molar mass) of our copper-platinum alloy, Aalloy, and the density of our alloy, ? alloy. According to an email message I received from Adam Stefanski, “a solution of two elements with the same structure will have the same structure as the individual elements”.

Therefore, since both copper and platinum have the FCC crystal structure, the copper-platinum alloy will also have the FCC crystal structure. As has previously been shown, the FCC crystal structure has 4 atoms per unit cell. Additionally, we know from our previous definition that Avogadro’s number is equal to 6. 022×1023 atoms/mole. Applying these values to the above formula for the volume of a unit cell of our copper-platinum alloy: with significant figures applied I have used the previously given formula for the volume of a cubic unit cell:

Rearranging this formula to solve for acubic_unit_cell, the length of each of the edges of the cubic unit cell, yields: Using the volume of the unit cell of our copper-platinum alloy, we can use this formula to solve for the edge length of the cubic unit cell of our alloy. with significant figures applied 3). According to Appendix B in Van Vlack, lithium has a BCC crystal structure, while aluminum has a FCC crystal structure. In question #2, the number of atoms in the unit cell of the FCC crystal structure were calculated:

Additionally, the edge length of the unit cell of the FCC crystal structure, aFCC, was related to the atomic radius of an atom in the FCC crystal structure, RFCC, by the following equation: A hard sphere model of the unit cell of the BCC crystal structure is shown below: where aBCC represents the length of each of the edges of the cubic unit cell (since the unit cell is cubic, all of these edges are equal in length), and RBCC represents the radius of an atom within the unit cell (since all the atoms with a unit cell of an elemental solid are made up of the same element, they all possess the same radius).

From this diagram, the number of atoms in the unit cell of the BCC crystal structure can be calculated. There are eight corner atoms in the above unit cell, and each of these corner atoms is shared among eight unit cells. Therefore, each unit cell contains one-eighth of each of these corner atoms. There is also one body centered atom in the above unit cell, and this atom is completely enclosed within the unit cell. Therefore, each unit cell contains the entirety of this body centered atom.

Mathematically, therefore, the number of atoms in the unit cell of the BCC crystal structure can be determined in the following manner: From the above hard sphere model of the BCC unit cell, we can see that the body and corner atoms of the unit cell touch each other along the cube diagonals. This diagonal has a length of 4*RBCC. Therefore, we can use the pythagorean theorem to relate the edge length of the cube, aBCC, and the atomic radius, RBCC. We can then rearrange the above equation to solve for aBCC: According to Appendix B in Van Vlack, the atomic radius of lithium is 0.1519 nm.

BCO unit cell

The three different crystallographic planes shown are for a unit cell of a hypothetical metal. I have completed this question based on the assumption that each of the planes is rectangular. Therefore, the angles between each of edges of the planes must equal 90o. Each of the plane dimensions must completely be enclosed by the boundaries of the unit cell of the metal. Even though there are infinite parallel planes having Miller indices equivalent to those of the given planes, these are not enclosed by the unit cell of our metal.

Thus, these equivalent planes do not need to be considered to determine the geometry, crystal system and crystal structure of our unit cell. Although we are given the edge lengths of the three crystallographic planes, this gives no information about the orientation of those edges. Therefore, since each plane is rectangular, there are two possible orientations for each plane. The correct orientation must allow the dimensions of all three planes to be correlated with one another. An initial diagram of the three planes of the unit cell is shown below:

We can see from the above figure that one edge of the (001) plane must be equal in length to one edge of the (101) plane. There is only one possible orientation of the two planes that can produce this condition – the alignment of the 0. 40 nm edge of the (001) plane to the 0. 40 nm edge of the (101) plane. The (001) plane and (101) planes have been orientated so that their 0. 40 nm edges are coincident in the following diagram of the unit cell: The diagonals of each of these planes can be calculated using the pythagorean theorem. Regardless of the orientation of any of the planes, their diagonals will always be of the same length.

The diagonal of the (001) plane is of particular interest, because it is aligned with one of the edges of the (110) plane. This implies that the length of the diagonal of the (001) plane must be equal to the length of one of the edges of the (110) plane. The length of the diagonal of a given plane can be calculated by the pythagorean theorem in the following manner: where lengthplane_diagonal is the length of the diagonal of a given plane, lengthplane_edge_1 is the length of one of the edges of that plane, and lengthplane_edge_2 is the length of the other edge of that plane. Rearranging this formula,

For the (001) plane, length(001)_plane_edge_1 = 0. 30 nm, and length(001)_plane_edge_2 = 0. 40 nm. We can therefore apply the above formula to calculate the length of the diagonal of the (001) plane. with significant figures applied The diagonal of the (001) plane is coincident with one of the edges of the (110) plane, so their lengths must be equal. Therefore, the edge of the (110) plane that is coincident with the diagonal of the (001) plane must have a length of 0. 50 nm. There is only one possible orientation of the (110) plane that will allow this situation. This orientation has been established in the following diagram of the unit cell.

I have used the symbols ? , ? , and ? , as defined on page 38 of Callister, to denote the three interaxial angles. Symbols ? 1 and ? 2 have also been used to denote two angles of interest. As a final verification that the above diagram of the unit cell is correct, we can compare the diagonals of the (110) and (101) planes. It can be seen from the above diagram that these diagonals should be coincident in our unit cell. In order for the two diagonals to be coincident, they must have the same length. We can verify that the diagonals possess the same length by applying the pythagorean theorem to determine the lengths of the diagonals.

with significant figures applied with significant figures applied Using the appropriate number of significant figures, the diagonals of the (110) plane and (101) plane indeed have the same length. We must now determine the interaxial angles of the unit cell. This can be accomplished by considering the intersections between the given planes. Miller indices are defined in the cartesian coordinate system. The x-axis, y-axis and z-axis of the cartesian coordinate system are orthogonal by definition. The (001) plane is defined to be parallel to the x-y plane. Therefore, as shown in the above diagram of the unit cell, the 0.

30 nm edge of the (001) plane must lie parallel to the x-axis, and the 0. 40 nm edge of the (001) plane must lie parallel to the y-axis. Therefore, by shifting the 0. 40 nm edge, we can see that it equivalently lies along the y-axis. By shifting the 0. 30 nm edge, we can see that it equivalently lies along the x-axis. The interaxial angle between the x-axis and the y-axis, which is denoted by the symbol ? , must be defined by the angle between the 0. 30 nm edge of the (001) plane and the 0. 40 nm edge of the (001) plane. Since I have assumed that the three given planes are rectangular, ? must therefore be 90o.

We can also determine the interaxial angle between the y-axis and the z-axis, which is denoted by the symbol ? , by considering the (101) and (110) planes of the unit cell. From the above diagram of the unit cell, we can see that the 0. 35 nm edge of the (110) plane lies parallel to the z-axis. Therefore, by shifting this edge, we can see that it equivalently lies along the z-axis. From the above diagram of the unit cell, we can see that the 0. 40 nm edge of the (101) plane lies parallel to the y-axis. Therefore, by shifting this edge, we can see that it equivalently lies along the y-axis.

Since the z-axis and y-axis in the cartesian coordinate system are orthogonal by definition, ? must therefore be 90o. We can also calculate the interaxial angle between the x-axis and the z-axis, which is denoted by the symbol ? , by considering 0. 46 nm edge of the (101) plane along with the 0. 35 nm and 0. 30 nm edges of the unit cell. These three sides form a triangle, as shown in the above diagram of the unit cell. Since the lengths of the sides of the triangle are known, we can use the cosine law to determine the angles of the triangle.

The cosine law is defined (The Cosine Law, 2003) as:where side1, side2 and side3 represent the three sides of the triangle in question, lside1, lside2, and lside3 represent lengths of these three sides, respectively, and ? represents the angle between side2 and side3. This formula can be rearranged to solve for ?. The angles ? 1, ? 2 and ? , as defined in the above diagram of the unit cell, can now all be determined. with significant figures applied with significant figures applied with significant figures applied Thus, using the appropriate number of significant figures, we currently have determined that ? = 90o, ? = 90o and ? = 90o.

These values for ? , ? , ? , ? 1 and ?2 can then be substituted into our above diagram for the unit cell. b). From the above diagram of the unit cell of our hypothetical metal, we can see that all three edges of the unit cell have a different length. According to table 3. 2 on page 39 of Callister, the only crystal system which satisfies the conditions of all three edges having different lengths and all interaxial angles being equal to 90o is the orthorhombic crystal system. Therefore, our unit cell must belong to the orthorhombic crystal system. c). For each of the given planes, the locations of the atoms residing within those planes has been provided.

Those planes have been oriented so that the locations of the atoms in our unit cell are now fixed. In the above diagram of the unit cell of our hypothetical metal, we can observe the positions of each of the atoms within our unit cell. There are atoms at each of the corners of the unit cell. Additionally, the centre of the (101) plane and the centre of the (110) plane both lie at the centre of the unit cell of our hypothetical metal. Therefore, the atom at the centre of the (101) plane is coincident with the atom at the centre of the (110) plane.

This produces an atom at the centre of the unit cell. Therefore, our unit cell must have a body centered crystal structure. The unit cell also belongs to the orthorhombic crystal system. The crystal structure of the unit cell would therefore be called body centered orthorhombic, or BCO. d). From the above diagram of our BCO unit cell, the number of atoms in this unit cell can be calculated. There are eight corner atoms in the above unit cell, and each of these corner atoms is shared among eight unit cells. Therefore, each unit cell contains one-eighth of each of these corner atoms.

There is also one body centered atom in the above unit cell, and this atom is completely enclosed within the unit cell. Therefore, each unit cell contains the entirety of this body centered atom. Mathematically, therefore, the number of atoms in this unit cell can be determined in the following manner: The mass of a unit cell of a material can be calculated by the following formula: The density of a material can be defined by the following formula: Therefore, the density of a material can be defined in terms of the above two equations: We can rearrange this formula to solve for (Mass/mole) in the following manner:

The (Mass/mole) of a substance is defined as its atomic weight. Therefore, the above formula can be rewritten: Since the three edges of our unit cell all possess different lengths, (do I have to make a change for monoclinic) the volume of this unit cell be calculated by the following formula: where lengthunit_cell_edge_1, lengthunit_cell_edge_2, and lengthunit_cell_edge_3 represent the three different edge lengths of our unit cell of the hypothetical metal. Since the values for the edge lengths of our unit cell have been determined previously, we can obtain the volume of our unit cell: with significant figures applied.

Using the conversion factors of 1 nm = 10-9 m and 1 cm = 10-2 m , we can convert the volume of the unit cells to meters: with significant figures applied The number of atoms in one mole of an element (or molecules in one mole of a compound) is defined by Avogadro’s number. For atoms, Avogadro’s number is equal to 6. 022×1023 atoms/mole. The density of the metal is given in the question to be 8. 95 g/cm3. Using these values in the above formula for atomic weight, the atomic weight of the metal can be determined. with significant figures applied 2). According to the question, copper and platinum both have the FCC crystal structure.

The relationship between wire length

 

This shows me that if I double the wire length I will double the resistance (see construction lines on graph). From this graph I can now make a near prefect estimate to the resistance if I have the wire length. I can accurately state this because all of my points are very close to my line of best supporting my conclusion further. However only between 10cm and 80cm. I would be able to make a very educated guess beyond this range I have no evidence to prove that the two factors will remain directly proportional. This evidence fully supports my prediction because as you can see the data is directly proportional.

The reason why the resistance increases as I increase the length is because Resistance (Ohms) = p(rho)length Area of the wire. Therefor if I increase the length of the wire (numerator) then the resistance increases. -Results 2 – Length = 15 cm Wire width = variable TEST 1 TEST 2 TEST 3 Wire width (mm) Amps (A) Volts (V) Resistance (? ) Amps (A) Volts (V) Resistance (? ) Amps (A) Volts (V) Resistance (? ) Average resistance ( The highlighted results above are done so because all the resistances are quite different. Because there is not two results similar. Therefor it is impossible to say which of the results is anomalous. From the above table I managed to calculate the below graph: The above graph shows the relationship between the wire width and the wires resistance.

As you can see there is definitely no directly proportional results. To try and manipulate the data to show directly proportional results I could alter the width of the wire to area of the wire. I can do this by halving the wire width or diameter to create the radius. Then I can use the formula ? ri?? Having done the above I was left with the following results: TEST 1 TEST 2 TEST 3 Wire Area (mmi?? ) Amps (A) Volts (V) Resistance (? ) Amps (A) Volts (V) Resistance (? ) Amps (A) Volts (V) Resistance (? )

Average resistance From the results shown on the previous page I can then produce the following graph: The above graph still does not give me directly proportional data. To manipulate the data further I am going to say that adversely proportional. This means that the wire area is directly proportional to 1/area.

When I manipulated the data in this fashion I created the following graph: As you can see I now have a scatter with a straight-line trend line through the origin. You can see that from the ri?? formula that again there is a near perfect positive relationship. It is probable that ri?? formula would be 1 if it where not for the slight unreliability’s with the experiment. The reason that the resistance decreases as the width increases is because Resistance (Ohms) = p(rho) length Area of the wire. Therefor if I increase the area of the wire (denominator) then the resistance decreases. It is like any division sum.

If you increase the second number then you will get a smaller answer. -Summary: When altering the length of the wire – length of wire is directly proportional to the resistance When altering the thickness of the wire – area of the wire is inversly proportional to the resistance. -Evaluation In general I thought that this experiment went very well. I thought that in general I collected good reliable results that made it easier for me to see any anomolous results. This was made easier by the repition of results as when there wwere two similar results and one wild one it was obvious that the wild result was anomoulus.

However if I had only had two sets of results then it would have been difficult to point out anomolous results. Because alll of my anomolous results where easy to spot they have been highlighted and removed from any average equations. There could be several reasons why a few of my results where anomolous. Probobly the main reason would be that the wire had heated up and therefor increased the resitance. However another possibility is that if you weren’t carefull the crocodile clips could rest in a thicker wire. Because there is a less resistance in a thicker wire the electricity would pass through that wire and give anomolous results also.

In general the results where reliable. One of the points in this experiment that could be argued is that all the wires heated up. If all the wires heated up proprtionally then I would still appear to get accurate results. However there is always a possibility that if the wires where kept at a controlled temperature then the resulta may have been different. Therefor to improve the reliability I could keep the wires at a controlled temperature. Also another thing that could slightly affect the results is the tightness of the wire. For example if one wire was very taught when I measured out 15cm on the board the taught wire would be 15cm long.

Howeverif there was a loose wire there might be a dip in the middle this would ever so slightly increase the length of the wire causing a greater resistance. To avoid this I should have tightened all the wires brfore commencement of the experiment. If I wanted to carry on the experiment then I could try longer and thicker wire to extend my graph. I could also compare the resistance in different wires made from different materials. I could also add time into the equation. From this I could then work out how the resistance of a wire changes as it heats up over time.

The purpose of this experiment was to demonstrate

The purpose of this experiment was to demonstrate the uncertainty of experimental measurements. The free-fall times of two metal balls of varying weight were measured. This data was used to calculate the best estimate of these measurements and its consistency. Then, the acceleration due to gravity was determined using the data obtained. Procedure The heavier metal ball was used first. It was placed in the release mechanism of the electronic free-fall timer apparatus.

The receptor plate was placed right below the release mechanism and the release mechanism was placed on the stand so that the measured distance from the top of the receptor pad to the bottom of the ball was the specified length given, which was 90 centimeters. Once the timer was set to zero, the ball was released by loosening the thumb screw to start the timer; the timer ended when the ball hit the pad. The fall time was recorded in seconds in a table and was repeated for twenty trials.

The average time of the twenty trials was calculated using the following equation where xN is the fall time and N is the number of trials. Then , the standard deviation was determined using the equation , where x is fall time. The experimental value of the free-fall time of the heavy metal ball was, which is the best estimate i?? uncertainty. Another table was made to determine the number of trials that occurred for each value of time. A graph resembling a Gaussian distribution was plotted using this table where the number of trials occurred was a function of time value.

The acceleration due to gravity in cm/s2 was then calculated using the formula , where t is the average time and d is the specified distance. This process was repeated using the lighter steel ball using five trials only. The experimental value of the acceleration due to gravity was obtained by calculating the class average of g and ?. The percent error was calculated using the literature value of the acceleration due to gravity, which is 980 cm/s2, and the experimental value ( . A graph was plotted using the class results where the distance, y, was a function of tavg, x.

The experimental value was determined by multiplying the slope of the graph by two. . The percent error was calculated using experimental value attained from the graph as well. Data Table 1: Heavy Steal Ball Fall Time Trial # Tis.

The x(t) graph of the falling puck is a parabolic, concave-up curve and the points of the v(t) graph of the falling puck can be best described as a linear curve. For the position versus time graph, the points reveal a well-defined parabolic curve. However, the velocity versus time graph, it was more difficult to distinguish what curve it followed; instead of velocity increasing the same amount per second, the velocity fluctuated a few times, but generally followed an increasing fashion. The equation of the line of best fit (regression) was obtained.

The parameters obtained by fitting the program was significant because the acceleration of the puck on the horizontal plane (ax) is the slope of the x(t) graph, which was 0. 355 m/s2. Using this value, the experimental value of gravity was obtained, which was, 10. 6 m/s2. The graph of a(t) should look like a straight horizontal line, where One may expect to get better agreement for larger, rather than smaller values of ? because the larger the angle, the more it resembles a motion of a free-falling object, which falls perpendicularly from the horizontal.

Conclusion The motion of a puck sliding down an inclined plane, without air resistance, resembles a free-falling object and should have had a 9. 80 m/s2 acceleration due to gravity. However, there were possibly two dominant sources of error in this experiment: the table was not completely level, which may have effected the motion and the velocity of the puck, and there was a lot of glare from the sunlight in the video, which made it difficult to see the dot in the middle of the puck.

Therefore, there was random error due to the inaccuracy of the cursor tracking of the image of the middle of the puck. These were the probable causes of error in the experimental results. It can be concluded that the force acting upon the puck undergoing a two-dimensional projectile motion is gravity and its motion can be described in two different components independently, its horizontal and vertical motions.

The objective of this laboratory was to measure

 

I then found the actual speed of sound by using the equation V speed of sound = 331. 4 + 0. 6Tc m/s. The experimental speed of sound was 344. 064 m/s and the actual speed of sound was 345. 8 m/s. By knowing these values I could then find the percent error. My calculations are above in the calculations section. The calculations show that the percent error was 0. 502 %. I found this percent error to be acceptable and due to random error. The independent variables in this experiment would be temperature and composition of the air. The dependent variable would be the speed of sound.

Depending on the composition of the medium through which the sound wave is travelling through, which in this case is air (78% nitrogen, 21% oxygen, and 1% other gases), the speed of sound would vary. The more elastic a medium, the faster sound travels. The less elastic a medium, the slower sound travels. Because sound is traveling through air (gases) in this case, the altitude affects the speed of sound because of the air pressure. At lower altitudes sound travels faster because of higher air pressure. At higher temperatures sound travels faster because as explained by the equipartition theorem of thermodynamics.

This means that each molecule has 1/2kT of energy per degree of freedom, and this is related to its kinetic energy through the usual 1/2mv 2 rule (Stony Brook University Staff, 2010). As stated earlier, the experimental results were very close to the actual values. The reason for the . 0502% error was due to both problems with an idealized equation and random error. The best way to decrease uncertainty would be to repeat the experiment many times over. The results could then be averaged. Standard deviation could then also be determined. The biggest drawback to the experimental procedure was the measurement of the first harmonic.

If there were a measurement of distance integrated into the actual tube, it would have been easier to get a more precise measurement. Interpretation of Results: The results of the experiment show that I was able to calculate the speed of sound at which sound travelled through the air very accurately. I know this because the percent error was very small. I believe I did an adequate job of keeping uncertainty to a minimum in this experiment. I think if I had conducted this experiment at a controlled temperature, elevation, and with more precise instruments I would have found more accurate values for the speed of sound.

The theory behind this exercise was that the velocity of sound was equal to the frequency multiplied by the wavelength. My results agree with this theory because I found the experimental value of the speed of sound using the process described in the results section and it was extremely close to the accepted actual value that was calculated in the calculations section. My experimental results show that the speed of sound was 344. 064 m/s and the actual speed of sound was 345. 8 m/s. This is a percent difference of 0. 502%.

I was surprised that I could accurately calculate the speed of sound so easily and the results forced me to change this preconceived notion. I also had originally thought that sound travelled faster in objects that were more dense. I was forced to change this belief, not based on my results, but by researching the topic. I found that this was not true. Elasticity, the composition of the air, temperature, and pressure are factors that determine how fast sound travels. I always thought that sound travelled in water at a faster speed than through air because of the density. I guess this was not true.

I also thought sound would travel faster at colder temperatures, due to the density of the air, and I found that this was also not true (to an extent). I was surprised to learn that sound travelled faster at higher temperatures in air. All in all, I am confident with my results and glad that I researched the topic and found additional information that corrected my preconceived notions. Error Sources and Why: Errors were caused by random error in this laboratory. It was difficult to get a good measurement of the first harmonic. As stated earlier, this could be corrected by introgration of a measurement device onto the outside of the tube.

I also think that the equation used to find the actual speed of sound was idealized and could contribute to a small and varying amount of error at different temperatures. Additionally, I believe that my altitude and the amount of humidity in the air contributed to the error. I had no way of knowing my altitude. I also had no way of knowing what the humidity was. If I had instruments to measure these factors, it would have helped. Then, sources of error could be eliminated by knowing what the altitude and humidity values were when the equation used to find the actual speed of sound was made.

I could then construct a new equation and find a more precise actual value for the speed of sound travelling through air by using the values I measured.

References: Stony Brook University Staff. 12 February 2010. Lecture 36: First Law of Thermodynamics. Stony Brook Physics 141/142. Web. Retrieved 29 April 2011. http://www. ic. sunysb. edu/Class/phy141md/doku. php? id=phy141:lectures:36 Mathpages Staff, n. d. The Speed of Sound. Mathpages. com. Web. Retrieved 29 April 2011. http://www. mathpages. com/home/kmath109/kmath109. htm Jeschofnig, Peter, Ph. D. 2009. Determining the Speed of Sound. LabPaq Manual PK-S. pp. 115-117.