The Conservation Of Momentum Environmental Sciences Essay

The Conservation Of Momentum Environmental Sciences Essay The conservation of momentum was shown in three types of collisions, elastic, inelastic and explosive. By getting mass and velocities for two carts during the collision the change in momentum and kinetic energy was found. In an elastic collision of equal massess ΔP = Pf-Pi =-8.595 and ΔKE = KEf-Kei = -4.762. In an inelastic collision of equal massess ΔP = -12.989 and ΔKE = -43.14. In an explosive collision of equal massess ΔP = -448.038 and ΔKE = -118.211. This shows that conservation of momentum is conserved in elastic and inelastic equations due to their very low change in momentum; however kinetic energy is conserved in the elastic collision but not in the inelastic collision. In an explosive collision momentum is not conserved since the two objects start at rest with no momentum and gain momentum once moving opposite. Introduction Just like Newtons laws, the conservation of momentum is a fundamental principal in physics that is integral in daily life. However unlike Newtons laws, the conservation of momentum does not seem to be entirely intuitive. If a ball is thrown in the air some momentum seems to be loss to the air. This makes proving the conservation of momentum tricky and difficult to do in a real life setting. To measure the conservation of momentum in the lab, two carts will be used along a frictionless track. This allows calculation to be easier since the vectors will be moving along only one axis. This way positive direction can be movement to the right while negative direction can be movement to the left. One cart will have a plunger which is ejected by a spring that will convert its potential energy to kinetic energy of the cart. This will knock the other cart and its momentum will be transferred either partially or entirely. These velocities of the two carts will be measured by a graphing device. This is shown in diagram 1. Diagram 1. Momentum is produced by mass and velocity, in other words: p = mv. It is important to point out that momentum is not conserved on an object by object basis, however it is conserved for the isolated system. This is shown in the equation: Psystem = P1 + P2. Therefore if momentum is conserved then the initial momentum of the entire system should equal the final momentum of the entire system. Thus this can be shown in the equation where: Psystem, initial = Psystem, final M1 X V1i + M2 X V2i = M1 X V1f + M2 X V2f In the lab collisions will be shown to illustrate the conservation of momentum. In elastic collisions energy is always conserved. Unfortunately for this lab kinetic energy can be converted into heat so that energy is lost to viable measurements. If the energy is conserved, the collision is considered to be elastic, but if the energy is not conserved, then the collision is considered inelastic. Kinetic energy is energy associated with motion where an object with mass and moving with a certain velocity the equation is: KE = Â ½ m |v|2 This allows to find the loss or gain in energy of a system much like for momentum where the change in kinetic energy of a system is determined by the equation: ΔKESYS = KEsys,final KEsys,intial For the two collisions stated earlier if ΔKESYS is equal to zero the collision is considered elastic, however if ΔKESYS does not equal zero then the collision is considered inelastic. There is also another type of collision that will be determined in this lab called an explosive collision. This can be considered the opposite of an inelastic collision since the energy is not conserved because the kinetic energy is transformed for potential energy to kinetic energy. These three types of collisions will be measured in the lab under differing conditions and the change in momentum and kinetic energy of the system will be calculated. Procedure In the lab the momentum and kinetic energy will be calculated by measuring different velocities for the two carts at different masses. Two carts will be set along a frictionless track. As stated earlier this allows for easier calculations since it allows working only in one dimension. One of the carts used has a plunger while the other car is just a regular car. Both carts have different sides which will allow the emulation of the different collision types. For and elastic collision the plunger cart will be placed against the side of the ramp and then set off by a small piece of wood. It will the knock the other cart and emulate a elastic collision because the carts have magnets facing each other that will help conserve energy and momentum by having the opposite sides face each other. Having magnets of opposite charge face each other help keep the collision elastic since major contact between the two carts can convert kinetic energy into heat and will be lost. This will be done in three different ways, first having equal mass carts, second having the plunger cart heavier than the regular cart, and lastly by having the plunger cart lighter than the regular cart. The velocities for these carts will be measured for the different variable for six different trails and averaged. For the inelastic the set up will be identical except to emulate this collision the carts will have Velcro sides that will be facing each other and cause the carts to stick together once they hit each other. This will be done in three different ways similar to the elastic collision, first having equal mass carts, second having the plunger cart heavier than the regular cart, and lastly by having the plunger cart lighter than the regular cart. The velocities for these carts will be measured for the different variable for six different trails and averaged also. For the explosive collision the two carts will be sitting next to each other. The plunger car will have its plunger faced toward the adjacent regular car so when the button is pressed the will move away from each other in opposite directions. This will only be done in two different ways, one way having the carts equal in mass and one ways have one cart heavier than the other cart. The velocities for these carts will be measured for the different variable for six different trails and averaged as well. Results Table 1. Elastic Collision Data Elastic Equal Mass regular car (g) 506.2 plunger car (g) 503.3 v1 (m/2) v1f (m/s) v2f (m/s) Pi = m1vi1+ m2 vi2 Pf = m1vf1 + m2 vf2 Kei = .5m1vi1 + .v5m2vi2 Kef= .5m1vf1 + .v5m2vf2 0.5 0 0.483 251.65 244.4946 62.9125 59.04545 0.494 0 0.482 248.6302 243.9884 61.41166 58.8012 0.574 0 0.505 288.8942 255.631 82.91264 64.54683 0.422 0 0.405 212.3926 205.011 44.81484 41.51473 ΔP = Pf-Pi 0.482 0 0.496 242.5906 251.0752 58.46433 62.26665 -8.595433333 0.516 0 0.498 259.7028 252.0876 67.00332 62.76981 ΔKE = KEf-KEi average 250.6434 242.048 62.91988 58.15744 -4.762437183 Elastic Heavy Int. regular car (g) 506.2 plunger car (g) 1000.9 v1 (m/2) v1f (m/s) v2f (m/s) Pi = m1vi1+ m2 vi2 Pf = m1vf1 + m2 vf2 Kei = .5m1vi1 + .v5m2vi2 Kef= .5m1vf1 + .v5m2vf2 0.412 0 0.501 294.3059 237.5554 84.94838 63.52835 0.502 0 0.59 310.6885 245.6916 126.1154 88.10411 0.321 0 0.466 324.3081 244.3456 51.56687 54.96218 0.462 0 0.544 337.2292 242.4102 106.818 74.9014 ΔP = Pf-Pi 0.51 0 0.602 354.5463 242.5007 130.167 91.72445 -81.71491849 0.486 0 0.52 324.2156 242.5007 118.2043 68.43824 ΔKE = KEf-KEi average 324.2156 242.5007 102.97 73.60979 -29.36021623 Elastic Light Int. regular car (g) 1003.8 plunger car (g) 503.3 v1 (m/2) v1f (m/s) v2f (m/s) Pi = m1vi1+ m2 vi2 Pf = m1vf1 + m2 vf2 Kei = .5m1vi1 + .v5m2vi2 Kef= .5m1vf1 + .v5m2vf2 0.563 0 0.309 468.8014 310.1742 79.76525 47.92191 0.396 0 0.243 495.1158 243.9234 39.46275 29.63669 0.697 0 0.351 523.2297 352.3338 122.2538 61.83458 0.554 0 0.296 563.0325 297.1248 77.23541 43.97447 ΔP = Pf-Pi 0.596 0 0.343 610.7959 344.3034 89.39011 59.04803 -227.7090311 0.493 0 0.278 532.195 279.0564 61.16328 38.78884 ΔKE = KEf-KEi average 532.195 304.486 78.21177 46.86742 -31.34434946 For the elastic collision with equal masses the change in momentum and kinetic energy is every small. Where as in the other two methods the change in momentum is much larger since the masses where different then the change in kinetic energy. Table 2. Inelastic Collision Data Inelastic Equal Mass regular car (g) 506.2 plunger car (g) 503.3 v1 (m/2) v1f (m/s) v2f (m/s) Pi = m1vi1+ m2 vi2 Pf = m1vf1 + m2 vf2 Kei = .5m1vi1 + .v5m2vi2 Kef= .5m1vf1 + .v5m2vf2 0.622 0.292 0.297 313.0526 297.305 97.35936 43.78238 0.481 0.242 0.243 242.0873 244.8052 58.222 29.68293 0.619 0.289 0.289 311.5427 291.7455 96.42247 42.15722 0.602 0.276 0.274 302.9866 277.6096 91.19897 38.17143 ΔP = Pf-Pi 0.51 0.236 0.237 256.683 238.7482 65.45417 28.23227 -12.98885 0.502 0.248 0.249 252.6566 250.8622 63.41681 31.16993 ΔKE = KEf-KEi average 279.8348 266.846 78.67896 35.5327 -43.14626406 Inelastic Heavy Int. regular car (g) 506.2 plunger car (g) 1000.9 v1 (m/2) v1f (m/s) v2f (m/s) Pi Pi = m1vi1+ m2 vi2 Pf = m1vf1 + m2 vf2 Kei = .5m1vi1 + .v5m2vi2 0.495 0.322 0.321 319.6722 484.78 122.6228 77.96833 0.506 0.343 0.342 323.0093 516.4291 128.1332 88.48103 0.497 0.317 0.318 336.2746 478.2569 123.6157 75.8842 0.499 0.312 0.312 352.9982 470.2152 124.6126 73.35357 ΔP = Pf-Pi 0.323 0.211 0.208 367.6309 316.4795 52.21145 33.23065 115.4745216 0.486 0.31 0.308 339.917 466.1886 118.2043 72.10332 ΔKE = KEf-KEi average 339.917 455.3916 111.5667 70.17019 -41.39646683 Inelastic Light Int. regular car (g) 1003.8 plunger car (g) 503.3 v1 (m/2) v1f (m/s) v2f (m/s) Pi Pi = m1vi1+ m2 vi2 Pf = m1vf1 + m2 vf2 Kei = .5m1vi1 + .v5m2vi2 0.575 0.181 0.181 480.8526 272.7851 83.20178 24.68705 0.589 0.172 0.163 506.4235 250.187 87.30267 20.77979 0.555 0.179 0.183 534.182 273.7861 77.51449 24.87125 0.563 0.186 0.186 573.035 280.3206 79.76525 26.06982 ΔP = Pf-Pi 0.367 0.115 0.113 619.6586 171.3089 33.89449 9.736832 -289.887818 0.574 0.178 0.179 542.8304 269.2676 82.91264 24.05466 ΔKE = KEf-KEi average 542.8304 252.9426 74.09855 21.6999 -52.3986526 For the inelastic collision the change in kinetic energy is much larger then it was in elastic collision. This holds true for the other all three methods used. Table 3. Explosive Collision Data Explosive Equal regular car (g) 506.2 plunger car (g) 503.3 v1 (m/2) v1f (m/s) v2f (m/s) Pi = m1vi1+ m2 vi2 Pf = m1vf1 + m2 vf2 Kei = .5m1vi1 + .v5m2vi2 Kef= .5m1vf1 + .v5m2vf2 0 0.482 0.503 0 497.2092 0 122.4709 0 0.448 0.471 0 463.8986 0 106.6245 0 0.489 0.512 0 505.2881 0 126.4901 0 0.438 0.469 0 457.8532 0 103.9089 ΔP = Pf-Pi 0 0.478 0.492 0 489.6278 0 118.7447 488.0378833 0 0.506 0.513 0 514.3504 0 131.0292 ΔKE = KEf-KEi average 0 488.0379 0 118.2114 118.2113751 Explosive- Unequal regular car (g) 506.2 plunger car (g) 1000.9 v1 (m/2) v1f (m/s) v2f (m/s) Pi = m1vi1+ m2 vi2 Pf = m1vf1 + m2 vf2 Kei = .5m1vi1 + .v5m2vi2 Kef= .5m1vf1 + .v5m2vf2 0 0.297 0.615 0 608.5803 0 139.8729 0 0.34 0.618 0 653.1376 0 154.517 0 0.292 0.619 0 605.6006 0 139.6484 0 0.307 0.633 0 627.7009 0 148.5813 ΔP = Pf-Pi 0 0.276 0.574 0 566.8072 0 121.5127 599.3574667 0 0.24 0.581 0 534.3182 0 114.2626 ΔKE = KEf-KEi average 0 599.3575 0 136.3992 136.399151 For the explosive collision the change in momentum is much larger than in the other two collisions. There is no initial momentum for this collision since the two carts started together at rest. Conclusion From momentum and the kinetic energies calculated from the formulas the different trails were averaged to find the initial and final momentum and kinetic energy for each of the eight conditions. They the change in momentum of the system was calculated for the system by subtracting the final momentum minus the initial momentum. This was then done for kinetic energy to find the change in kinetic energy by subtracting final minus initial as well. This produced different values for the different conditions. For the elastic collision the momentum and kinetic energy are supposed to be conserved. As table 1 shows, the momentum and kinetic energy for the equal mass carts is very close to zero, much closer than for the other conditions. For the heavier plunger cart, the initial force had much more inertia and caused the lighter second car to move much further. This is opposite in the other conditions where the plunger cart was much light. It had a harder time moving the second heavier cart. The main difference for the change in momentum and kinetic energy for the two unequal mass cart conditions was due to the fact the final velocity for cart one was never measured properly. It was assumed that the velocity was zero when in fact the plunger cart moved slightly after the collision. The assumption was due to careless human error. For the inelastic collision kinetic energy is not conserved. This is evident very much in the results for the change in kinetic energy. There is a much larger value or this change then in the elastic counterpart since the carts stick together and move as one unit. This close interaction allows for the loss of energy as heat. As for the explosive collision, the change in momentum is by far the largest. Since the system start at rest it is entirely potential energy. When the collision happened the carts move apart and become kinetic energy. Since the final momentum is subtracted by an initial momentum of zero, it is obvious why the change is so large.

A Cross-Cultural Study of British, Indian, and Portuguese College Students Essay

Love Styles: A Cross-Cultural Study of British, Indian, and Portuguese College Students Love is a feeling known to many people. Sometimes, it becomes a goal for different individuals to find their love through different ways. Moreover, love is a feeling which helps individuals to be compassionate to other people. Hence, love is not a mysterious thing yet no one has really attempted to study such topic for its very vague concept. However, there is a study which presented varying aspects regarding love. The article Love Styles: A Cross-Cultural Study of British, Indian, and Portuguese College Students. Furthermore this article provides a very significant understanding towards the subject of love. Therefore, as stated by the author is categorized into six different types of love. The types of love which were mentioned are Eros, Ludus and Storge which are categorized as the primary styles of love. The secondary styles are Pragma, Mania and Agape. These are all composed of different categories which mostly provide the concept of love in the current society. Eros is known to be the romantic and passionate type of love. Ludus is known to be the game playing love. Storge is a love which is based on friendship. Pragma is meant to name the practical kind of love. Mania is the possessive or dependent kind of love while Agape is the unselfish kind of love. Thus, all of these kinds of love is dynamic that could interchange or develop through time (Neto, 2007). Hence, the author was able to create a study which does not focus on the concept of romantic love but also the varying kinds of love in the society. Hence, in order to understand the concept of love, the author view various culture such as the British, Portuguese and the Indians in order for to attain the accurate results of the data. In the conclusion, the author realized that the need to view love must go through a gender, cultural and psychological aspects for there are great differences in the ethnicity of varying individuals. The research question in the author’s mind is definitely leaning on the concept of love through utilizing the concept of Lee on various students from India, Britain and Portugal. Hence, this is the main basis utilized by Neto in order to derive to such study. In order to get the accurate data, the author used various students of different nationalities such as Indians, British and Portuguese that shall provide different view points of the study. The study was composed of 562 individuals who are not equally divided between male and female. The results stated that different cultures of people are important in predicting the kind of styles of love which they make. Furthermore, the cultural perspective mainly predicts what kind of love the participants wanted to experience. Moreover, the gender of the students who participated is significant in the study. Therefore, there are studies which show that Eros is important to both genders. Hence, in the study Indians is much leaning into three secondary type of love such as Pragma, Mania and Agape (Neto, 2007). Thus, the Portuguese and British participants are much active in the Ludus kind of love. In the whole study, I view that all the information provided is very interesting. The kind of love that is being practiced by different individuals from varying countries is truly affected by the culture and environment they have experienced. Furthermore, I find it very interesting that Indians tend to have a Mania style of love. Although I have learned their culture, I do not believe that Indians’ love style could lead into a very deep and possessive state. I have learned that although love is needed by everyone, the types of love could be different. Love is not only defined through an Eros relationship but also through different kinds of love such as Storge or Agape. Most people see love as a serious relationship that could end up in marriage. Friendly relationships and companionships are also love but with limitations. References Neto, F. (2007). â€Å"Love Styles: A Cross-Cultural Study of British, Indian, and Portuguese College Students. † Journal of Comparative Family Studies. Retrieved on May 13, 2010 from http://findarticles. com/p/articles/mi_hb3248/is_200704/ai_n32226385/? tag=content;col1

Human Biology Digestive and Urinary System Essay

Abstract The human body needs to maintain proper homeostasis to survive. There are several different organ systems in the human body. Two of those systems are the digestive system and the urinary system. Both systems remove waste from the body but in an entirely different way. They also maintain homeostasis within our blood stream. Both systems are critical for survival. This paper will describe how each of those systems works on its own and also how they work together to maintain homeostasis for the body.How the Urinary and Digestive Systems Work Together to Maintain Homeostasis The digestive system works to bring nutrients into the body with a series of hollow organs working together extending from the mouth to the anus. These hollow organs are the mouth, pharynx, esophagus, stomach, small intestine, large intestine, rectum, and anus. These hollow organs make up the gastrointestinal (GI) tract. The GI tract is aided by four accessory organs, the salivary glands, liver, gallbladder, and pancreas. Johnson) There are five basic processes of the digestive system. These processes are motility, secretion, digestion, absorption, and excretion. Motility is the mechanical processing and movement of food taken into the body. Chewing breaks food into smaller pieces, and two types of movement mix the contents of the lumen and propel it forward. The lumen is the inside lining of the GI tract. Secretion is the fluid, digestive enzymes, acid, alkali, bile, and mucus that are secreted into the GI tract at various places.Digestion is where the contents of the lumen are broken down mechanically and chemically into smaller and smaller particles, culminating in nutrient molecules. Absorption is what happens when the nutrient molecules pass across the GI tract and into the blood. Elimination is all of the undigested material is eliminated from the body through the anus. (Johnson) These GI tract and the four accessory organs of the digestive tract work together to complete the fiv e processes in order to bring nutrients into the body, and eliminate waste. The large intestine absorbs nutrients and eliminates waste.The small intestine absorbs nutrients and water. The water and nutrients absorbed from the intestines go into the blood stream . (Johnson) The urinary system consists of the kidneys, ureters, bladder, and urethra. The ureters, bladder, and urethra transport and store urine until it is eliminated from the body. The kidneys produce urine. Urine is the nitrogenous waste removed from the blood stream. It is essential to the body to remove this waste in order to maintain homeostasis. (Johnson) The kidneys perform all of the main functions in the urinary system.They regulate water levels in the body. When you take in water every day it is up to the kidneys to excrete the excess water or to conserve as much as possible. The kidneys have a great capacity to adjust water excretion as necessary. Normally with a minimum of a half liter per day to one liter per hour. (Johnson) The kidneys also regulate the nitrogenous waste and other solutes in our blood. One waste that is toxic to our cells is ammonia. Ammonia is detoxified quickly by the liver by being combined with carbon dioxide to create urea.Urea is the main waste product in urine. Sodium and chloride are both regulated by the kidneys as well. Sodium and chloride are both very important to determine the volume of extracellular fluids, like blood. This directly affects blood pressure. (Johnson) Other substances that the kidneys regulate are potassium, calcium, hydrogen, and creatinine. It is up to the kidneys to maintain homeostasis with each of these substances. Creatinine is a waste product that is produced during metabolism. This is one of the wastes that give urine a yellow color. Johnson) The digestive system removes nutrients and water from the food that we eat and drink and transports it to our blood stream. Any solid wastes that are unable to be digested are then eliminated fr om the GI tract. Our urinary system then removes any of the unwanted and unneeded substances and wastes from the blood stream and excretes them in urine. This is how homeostasis is maintained in our body with waste, so that there is no toxic build up in our body from certain substances. Homeostasis is critical to be maintained within the body for survival.Homeostasis is the body’s ability to maintain relatively stable internal conditions even though the outside world is continuously changing. (Marieb) All living things must maintain an internal environment compatible with life, and the range of chemical and physical conditions compatible with life is very narrow. (Johnson) In order for the digestive and urinary systems to work together to maintain homeostasis both systems must be functioning properly. First we consume food and water. Then our digestive system absorbs the nutrients and water into our blood stream.Once the nutrients and water is in our blood stream our urinary system the filters the blood and removes any additional waste and maintains the proper homeostasis with the nutrients we consume. If our digestive system is not functioning properly then we are not able to absorb the proper nutrients and fluids in order to allow our kidneys to filter our blood and maintain homeostasis. If our urinary system is not functioning properly than the nutrients and fluids we consume can cause a buildup in our blood stream of toxic wastes and fluids and throw off homeostasis.Either way it would be detrimental to our body for this to happen. Our blood pressure would be affected severely and this would cause issues with not only our cardiovascular system but also our respiratory system. It is critical to keep both systems healthy to ensure that our body’s have the proper water and nutrients to survive and that all waste is secreted from our systems. References Johnson, Michael D. Human Biology: Concepts and Current Issues-sixth edition. Pearson Educatio n. 2012. Print. Marieb, Elaine Nicpone. Essentials oh Human Anatomy and Physiology-tenth edition. Pearson Education. 2012. Print.

Citizen Kane Free Essay Example, 1000 words

The symbolic "walking out" of Susan out of Kanes life achieves tragic inference owing to the fact that Kane like the audience does not expect Susans decision. This is so as Susan has been someone who has always been presented as inferior to him. Kane in his naivetà ©, which constitutes his tragic flaw, supposes that he has done everything for Susan. One salient element of the genre form is the role achieved by the manipulation of the camera in portraying certain scenes key to the thematic concerns of the art piece. In this scene the camera shoots both Kane and Susan from an upward-facing angle. This angle thus juxtaposes them as equals which is in the plot context and concept a significant reversal theme central in tragedy genre. Size and height symbolize that Susan is no longer engulfed in the realms of an inferiority complex in his relation and regard to Kane. This illuminates the dramatic twist and the reversal theme considering the inferiority complexities of Susan in previous scenes reflected through her deportment around Kane. In the previous scenes Susan is regularly portrayed kneeling before Kane while Kane towers and postures over her with a condescending disposition. We will write a custom essay sample on Citizen Kane or any topic specifically for you Only $17.96 $11.86/pageorder now In the ideal tragedy according to Pullicino (opcit), claims Aristotle, the protagonist will mistakenly bring about his own downfall—not because he is sinful or morally weak, but because he does not know enough. On Kanes fate this aspect evokes pathos from the audience interpreting Kanes downfall as a result of the imbalance between his noble endeavors and his villainous approach which culminate form his character flaw. The breakfast table scene also remarkably stands out for its contribution to the making of the narrative as a tragedy. This scene is significant for the representation of the widening chasm developing between Kane and Emily. The elaborate body language, the editing aspect of the characters change of costumes as well as a seamless dialogue form a important set of conceptual objects and events that constitute the development of a tragic plot. The development of their marriage is outlined efficiently in six segments separated by whip pans, the camera and spoken narrative. In another scene, when Kane is being lectured by Leyland after the election the framing has been used to emphasise Kane’s power by placing the camera down low so as to capture the ceiling and Kane’s stature. These techniques demonstrate the dexterity of Welles and the film director in employing these tools enhance meaning dissemination.