Amino Acid Definition and Examples

Amino Acid Definition and Examples Amino acids are important in biology, biochemistry, and medicine. Learn about the chemical composition of the amino acids, their functions, abbreviations, and properties: Key Takeaways: Amino Acids An amino acid is an organic compound characterized by having a carboxyl group, amino group, and side chain attached to a central carbon atom.Amino acids are used as precursors for other molecules in the body. Linking amino acids forms polypeptides. Polypeptides may be modified and combined to form proteins.The genetic code is basically a code for proteins made within cells. DNA is translated into RNA. Three bases (combinations of adenine, uracil, guanine, and cytosine) code for an amino acid. There is more than one code for most amino acids.Amino acids are made in the ribosomes of eukaryotic cells.Some amino acids may not be made by an organism. These essential amino acids must be present in the organisms diet.In addition to making amino acids from the genetic code and obtaining them from the diet, other metabolic processes convert molecules into amino acids. Amino Acid Definition An amino acid is a type of organic acid that contains a carboxyl  functional group (-COOH) and an amine functional group (-NH2) as well as a side chain (designated as R) that is specific to the individual amino acid. Amino acids are considered to be the building blocks of polypeptides and proteins. The elements found in all amino acids are carbon, hydrogen, oxygen, and nitrogen. Amino acids may contain other elements on their side chains. Shorthand notation for amino acids may be either a three-letter abbreviation or a single letter. For example, valine may be indicated by V or val; histidine is H or his. Amino acids may function on their own, but more commonly act as monomers to form larger molecules. Linking a few amino acids forms peptides. A chain of many amino acids is called a polypeptide. Polypeptides may become proteins. The process of producing proteins based on an RNA template is called translation. Translation occurs in ribosomes of cells. There are 22 amino acids involved in protein production. These amino acids are considered to be proteinogenic. In addition to the proteinogenic amino acids, there are some amino acids that are not found in any protein. An example is the neurotransmitter gamma-aminobutyric acid. Typically, nonproteinogenic amino acids function in amino acid metabolism. The translation of the genetic code involves 20 amino acids, which are called canonical amino acids or standard amino acids. For each amino acid, a series of three mRNA residues acts as a codon during translation (the genetic code). The other two amino acids found in proteins are pyrrolysine and selenocysteine. These two amino acids are specially coded, usually by an mRNA codon that otherwise functions as a stop codon. Common Misspellings: ammino acid Examples: lysine, glycine, tryptophan Functions of Amino Acids Because they are used to build proteins, most of the human body consists of amino acids. Their abundance is second only to water. Amino acids are used to build a variety of molecules and are used in neurotransmitter and lipid transport. Amino Acid Chirality Amino acids are capable of chirality, where the functional groups may be on either side of a C-C bond. In the natural world, most amino acids are the L-isomers. There are a few instances of D-isomers. An example is the polypeptide gramicidin, which consists of a mixture of D- and L-isomers. One and Three Letter Abbreviations The amino acids most commonly memorized and encountered in biochemistry are: Glycine, Gly, GValine, Val, VLeucine, Leu, LIsoeucine, Leu, LProline, Pro, PThreonine, Thr, TCysteine, Cys, C  Methionine, Met, MPhenylalanine, Phe, FTyrosine, Tyr, Y  Tryptophan, Trp, W  Arginine, Arg, RAspartate, Asp, DGlutamate, Glu, EAparagine, Asn, NGlutamine, Gln, QAparagine, Asn, N Properties of the Amino Acids The characteristics of the amino acids depend on the composition of their R side chain. Using the single-letter abbreviations: Polar or Hydrophilic: N, Q, S, T, K, R, H, D, ENon-Polar or Hydrophobic: A, V, L, I, P, Y, F, M, CContain Sulfur: C, MHydrogen Bonding: C, W, N, Q, S, T, Y, K, R, H, D, EIonizable: D, E, H, C, Y, K, RCyclic: PAromatic: F, W, Y (H also, but doesnt display much UV absorption)Aliphatic: G, A, V, L, I, PForms a Disulfide Bond: CAcidic (Positively Charged at Neutral pH): D, EBasic (Negatively Charged at Neutral pH): K, R

Wreak and Pique Revisited

Wreak and Pique Revisited Wreak and Pique Revisited Wreak and Pique Revisited By Maeve Maddox A plaintive email from a reader has prompted this post on these two misused and abused rhyming verbs: A new civil trialis poised to wreck havoc on the 100-year-old institutions reputation. Shouldnt that be wreak? And shouldnt My interest was peaked be My interest was piqued†? I see that everywhere it seems. Though peaked might be an okay substitute- it sort of means something similar. 1. Yes, the phrase should be â€Å"to wreak havoc.† 2. No, peaked is not an okay substitute for piqued. In modern usage, wreak [REEK] is a transitive verb usually followed by a limited number of object words that include vengeance, havoc, and damage. Storms are the most common wreakers. The past tense form is wreaked [REEKT]. Here are some examples of wreak being used correctly: Tropical storm Arthur expected to wreak havoc on East Coast Storms wreaking havoc across UK Northeasters also wreaked damage in 1991 and 1992. January Jones Discusses Wreaking Vengeance in the Sundance Film ‘Sweetwater’ The word pique [PEEK], as both noun and verb, has more than one meaning. The verb’s most common use is in the sense of stimulate or arouse. The past form is piqued [PEEKT]. Here are some examples in which the verb is spelled correctly: The request piqued my interest and I began what has become a continuing search for documentaries relating to the Comanches. Foreign cricket players hope to pique Lebanese interest New Study Provides Insight into How Piquing Curiosity Changes Our Brains It’s not surprising when entertainment site comments and self-published novels contain errors like these: I still have the feeling that Stavros is alive and the two of them will connect and reek havoc on Pt. Charles. It’s my understanding that you have been using him to wreck vengeance on the descendants of the clergy, and soldiers of New France because of some perceived wrong doing [sic]. I thought [Grimm] was ok. Ill probably keep watching, but the pilot didnt peak my interest right from the start. As one does expect news sources and professional publications to use words correctly, the following errors are less tolerable: Gov. Martin OMalley declared a state of emergency one day before a winter  storm  is  expected to wreck havoc  in Maryland- Baltimore Post Examiner. If  they  come from violent and abusive homes, children learn to be violentwill grow up to  wreck vengeance  on themselves and those around  them.- Social justice site. Four houses destroyed by fire and lightning as the weekends thunderstorms wrecked havoc across Britain- Daily Mail. Extremely high rain soaked [sic] winds wrecked havoc by downing trees and disrupting schools and traffic in the Bay Area- ABC News. All the teachers are engaging and do their best to peak the interest of the student.- Site advertising private school in Washington DC. Though we were enjoying a near perfect day in Oakland, hearing the name Birmingham not only peaked his interest but also placed him back on the Jim Crow bus system in Alabama.- Huffington Post columnist. Misspelling pique is perhaps more understandable than misspelling wreak because peak, peek, and pique are all pronounced the same. Pronunciation offers no excuse for mixing up wreak [REEK] and wreck [REK], however. Bottom line: Speakers who care about the language don’t require excuses for misspelling words they use in daily speech. They learn the differences. Related posts: Wreck, Wreak, and Other [rek/reek] Words Reeking and Wreaking Please, Let Your Interest Be Piqued Want to improve your English in five minutes a day? Get a subscription and start receiving our writing tips and exercises daily! Keep learning! Browse the Misused Words category, check our popular posts, or choose a related post below:50 Idioms About TalkingDo you "orient" yourself, or "orientate" yourself?1,462 Basic Plot Types

The Importance of Digital Security Essay Example | Topics and Well Written Essays - 3500 words

The Importance of Digital Security - Essay Example The most essential need of every single organization is the digital security. Most importantly, the effective internet security has become a dire need for any kind of organization, small, medium or large which use the information technology and web based services to carry out their work in an easy and effective manner. As these organizations depend upon the internet, the implementation of internet security and monitoring of networks inside the organization has increased dramatically. The risks of security have increased to a great extent after the launch of broadband internet. Now-a-days, the home users and professionals, both are using this kind of internet connection. The different companies, be it private, public, non-government organizations, they all are depending upon the internet for the exchange of information. Internet also serves as the major means of communication between different channels. The risks of leakage of information and hacking of security have increased with th e dependency of organizations on the internet. KINDS OF THREATS Based on the goals and purposes of the attacks on any digital system, the threats can be categorized as STRIDE. It is an acronym which categorizes different types of threats. STRIDE stands for: Spoofing It gains access to a system by using a false identity. It can be done by using stolen user credentials or using a false IP. Tampering As data flows between two computers, it can be altered in an unauthorized manner. Repudiation These attacks are difficult to prove as the users deny that they performed any specific action. Information Disclosure When private data is unwontedly exposed, it is called information disclosure. Denial of Service It is the process that makes a system or application unavailable. Elevation of Privilege It happens when personnel with limited privileges takes up the identity of a privileged one, and performs certain privileged actions (McClure & Kurtz, 2009). Network Threats Routers, switches and fi rewalls make up the infrastructure of network. They are the gatekeepers that guard the system and applications from intrusions and attacks. The networks threats are as follows: i. Information gathering ii. Sniffing iii. Spoofing iv. Session Hijacking v. Denial of Service Information Gathering The attackers or hackers first scan the ports. After the identification of the ports, they detect the types of devices, operating system and versions of

Porche Essay Example | Topics and Well Written Essays - 750 words - 1

Porche - Essay Example Ferdinand had the right mixture of the brains required to successfully run a car company and the passion required to drive the growth of Porsche. In the Prince Henry trials in 1910, the innovatively designed car model which was driven by Ferdinand himself won the most prestigious award in the sports category (Boatcallie, Chase, Salehi, Skrisovsky and Volio 2). In 1931, Ferdinand opened up his own company under the name of Porsche in Stuttgart, Germany for performing the activities of engineering and consultation. The company soon received contracts from the car manufacturers like the Wanderer, Auto Union and NSU for supply and design of cars. Porsche had the vision to anticipate the demands of the present and link it to the changing trends in future. With innovative designs and influencing the partners to manufacture car model that would accepted worldwide, Porsche delivered the model of the Beetle which was taken up by the VW Group. The Beetle designed by Porsche underwent mass prod uction and had not lost its popularity for the next 75 years. Ferdinand who was the founder of Porsche was a race-driver by passion and actively participated in the racing events. Ferdinand was also an engineer by profession. The founder of the company was able to mix his passion with the entrepreneurial activities as an engineer and applied his ideas and innovations in the field of manufacturing newly designed and creative car models. After the death of the founder, his son took over the charge of Porsche and started to expand the trading activities through sustenance of efficiently manufactured innovative car models. Porsche became famous for its racing cars and emerged victorious in the various rallies that it participated. The 911 model was one of the famous racing models built by Porsche (Porsche1 1). The 911 model was traded in the 70s, 80s and the 90s. In 1996, the

Introducing a New Food in Australia Assignment Example | Topics and Well Written Essays - 2000 words

Introducing a New Food in Australia - Assignment Example The Australian National Food Safety Standards has the role of labeling food standards to be introduce. In Australian Capital Territory (ACT) the agency responsible for ensuring the safety of food is the Health Protection Service (HPS) of ACT Health. The HPS under its Food Sampling Working Group (FSWG) looks after the development, overall implementation and co-ordination of the Food Survey Program (Program). These government agencies monitor the introduction of new food as well as supply to ensure that it is safe and pose no risk to the consumers. The new food should comply with standards for microbiological contaminants, pesticide residue limits and chemical contamination. Introduction of new food in Australia has to meet all the food surveillance data from public health units in Australia. This data includes the results of compliance testing, and specialty targeted surveys. If the new product is canned food then it comes under the Canned Food Information Service Inc (CFIS Inc) for the promotion and review of the product. The CFIS aims to convince consumers about the foods contained, and to dispel misconceptions and so generate increased purchases. The nutrition programme of CFIS is aimed to create the awareness of the influences of public opinion. After that the authority issue a license only after carrying out a comprehensive risk assessment process so that Australian environment and human health and safety would not be at risk. The national regulatory scheme does not look into the marketing issues of new food. The main objective of the authority is to provide an unambiguous recognition, under Commonwealth law. Nutrition labelling Mandatory nutrition labelling is necessary for the launch of any new food product as it has significant impacts on health in the community. The method of nutrition labelling divided into two steps, first, it identify the risk factors of diet-related disease and study their impact on health systems expenditure and the value of life. Secondly, it estimates the level of reduction in risk factors. Risk factors can be identified through diet-related diseases, which are associated with three risk factors, namely, obesity, hypertension and high blood cholesterol. It is no denying fact that nutrition information greatly influences consumer choice. According to an American study of the impact of the introduction of mandatory nutrition labelling the consumers principally respond to negative nutrition information. Nutrition labelling really boosts the consumer behaviour as it reduces the risk factors. In Australia food products that carry nutrition labels is considered as healthy. A nutritional analysis programme is vital for any new food product to know the size and complexity of the product and a careful assessment of the resources, skills, courage and discipline required to progress the task to completion (Scheelings 1987). The programme is more about data evaluation, which identifies the critical elements of quality assurance. The Nutrition Committee of the National Health and Medical Research Council (NHMRC) has developed work programme for the current revision of the Australian food tables. The work programme is composed of four components: the analytical programme; the processing and validation of food composition data and preparation for publication; the establishment of the Australian Nutrient Data Bank to store and process data; and the development of the

Fluids In Rigid Body Motion Engineering Essay

Fluids In Rigid Body Motion Engineering Essay 11-38C A moving body of fluid can be treated as a rigid body when there are no shear stresses (i.e., no motion between fluid layers relative to each other) in the fluid body. 11-39C A glass of water is considered. The water pressure at the bottom surface will be the same since the acceleration for all four cases is zero. 11-40C The pressure at the bottom surface is constant when the glass is stationary. For a glass moving on a horizontal plane with constant acceleration, water will collect at the back but the water depth will remain constant at the center. Therefore, the pressure at the midpoint will be the same for both glasses. But the bottom pressure will be low at the front relative to the stationary glass, and high at the back (again relative to the stationary glass). Note that the pressure in all cases is the hydrostatic pressure, which is directly proportional to the fluid height. 11-41C When a vertical cylindrical container partially filled with water is rotated about its axis and rigid body motion is established, the fluid level will drop at the center and rise towards the edges. Noting that hydrostatic pressure is proportional to fluid depth, the pressure at the mid point will drop and the pressure at the edges of the bottom surface will rise due to rotation. 11-42 A water tank is being towed by a truck on a level road, and the angle the free surface makes with the horizontal is measured. The acceleration of the truck is to be determined. ax  Ã‚ ± = 15 ° Water tank Assumptions 1 The road is horizontal so that acceleration has no vertical component (az = 0). 2 Effects of splashing, breaking, driving over bumps, and climbing hills are assumed to be secondary, and are not considered. 3 The acceleration remains constant. Analysis We take the x-axis to be the direction of motion, the z-axis to be the upward vertical direction. The tangent of the angle the free surface makes with the horizontal is Solving for ax and substituting, Discussion Note that the analysis is valid for any fluid with constant density since we used no information that pertains to fluid properties in the solution. 11-43 Two water tanks filled with water, one stationary and the other moving upwards at constant acceleration. The tank with the higher pressure at the bottom is to be determined. Tank A 8 m Water az = 5 m/s2 Tank B 2 m Water g z 0  · 2  · 1  · 2  · 1 Assumptions 1 The acceleration remains constant. 2 Water is an incompressible substance. Properties We take the density of water to be 1000 kg/m3. Analysis The pressure difference between two points 1 and 2 in an incompressible fluid is given by or since ax = 0. Taking point 2 at the free surface and point 1 at the tank bottom, we have and and thus Tank A: We have az = 0, and thus the pressure at the bottom is Tank B: We have az = +5 m/s2, and thus the pressure at the bottom is Therefore, tank A has a higher pressure at the bottom. Discussion We can also solve this problem quickly by examining the relation . Acceleration for tank B is about 1.5 times that of Tank A (14.81 vs 9.81 m/s2), but the fluid depth for tank A is 4 times that of tank B (8 m vs 2 m). Therefore, the tank with the larger acceleration-fluid height product (tank A in this case) will have a higher pressure at the bottom. 11-44 A water tank is being towed on an uphill road at constant acceleration. The angle the free surface of water makes with the horizontal is to be determined, and the solution is to be repeated for the downhill motion case. z x az ax  Ã‚ ¡ = 20 ° - Ã‚ ± Downhill motion Uphill motion z x ax Free surface az Water tank  Ã‚ ¡ = 20 °  Ã‚ ± Horizontal Assumptions 1 Effects of splashing, breaking, driving over bumps, and climbing hills are assumed to be secondary, and are not considered. 2 The acceleration remains constant. Analysis We take the x- and z-axes as shown in the figure. From geometrical considerations, the horizontal and vertical components of acceleration are The tangent of the angle the free surface makes with the horizontal is  ®  Ã‚ ± = 22.2 ° When the direction of motion is reversed, both ax and az are in negative x- and z-direction, respectively, and thus become negative quantities, Then the tangent of the angle the free surface makes with the horizontal becomes  ®  Ã‚ ± = 30.1 ° Discussion Note that the analysis is valid for any fluid with constant density, not just water, since we used no information that pertains to water in the solution. 11-45E A vertical cylindrical tank open to the atmosphere is rotated about the centerline. The angular velocity at which the bottom of the tank will first be exposed, and the maximum water height at this moment are to be determined.  Ã‚ · 2 ft z r 0 Assumptions 1 The increase in the rotational speed is very slow so that the liquid in the container always acts as a rigid body. 2 Water is an incompressible fluid. Analysis Taking the center of the bottom surface of the rotating vertical cylinder as the origin (r = 0, z = 0), the equation for the free surface of the liquid is given as where h0 = 1 ft is the original height of the liquid before rotation. Just before dry spot appear at the center of bottom surface, the height of the liquid at the center equals zero, and thus zs(0) = 0. Solving the equation above for  Ã‚ · and substituting, Noting that one complete revolution corresponds to 2 Ã‚ ° radians, the rotational speed of the container can also be expressed in terms of revolutions per minute (rpm) as Therefore, the rotational speed of this container should be limited to 108 rpm to avoid any dry spots at the bottom surface of the tank. The maximum vertical height of the liquid occurs a the edges of the tank (r = R = 1 ft), and it is Discussion Note that the analysis is valid for any liquid since the result is independent of density or any other fluid property. 11-46 A cylindrical tank is being transported on a level road at constant acceleration. The allowable water height to avoid spill of water during acceleration is to be determined D=40 cm ax = 4 m/s2  Ã‚ ± htank =60 cm  Ã¢â‚¬Å¾z Water tank Assumptions 1 The road is horizontal during acceleration so that acceleration has no vertical component (az = 0). 2 Effects of splashing, breaking, driving over bumps, and climbing hills are assumed to be secondary, and are not considered. 3 The acceleration remains constant. Analysis We take the x-axis to be the direction of motion, the z-axis to be the upward vertical direction, and the origin to be the midpoint of the tank bottom. The tangent of the angle the free surface makes with the horizontal is (and thus  Ã‚ ± = 22.2 °) The maximum vertical rise of the free surface occurs at the back of the tank, and the vertical midplane experiences no rise or drop during acceleration. Then the maximum vertical rise at the back of the tank relative to the midplane is Therefore, the maximum initial water height in the tank to avoid spilling is Discussion Note that the analysis is valid for any fluid with constant density, not just water, since we used no information that pertains to water in the solution. 11-47 A vertical cylindrical container partially filled with a liquid is rotated at constant speed. The drop in the liquid level at the center of the cylinder is to be determined. z r  Ã‚ · zs R = 20 cm Free surface ho = 60 cm g Assumptions 1 The increase in the rotational speed is very slow so that the liquid in the container always acts as a rigid body. 2 The bottom surface of the container remains covered with liquid during rotation (no dry spots). Analysis Taking the center of the bottom surface of the rotating vertical cylinder as the origin (r = 0, z = 0), the equation for the free surface of the liquid is given as where h0 = 0.6 m is the original height of the liquid before rotation, and Then the vertical height of the liquid at the center of the container where r = 0 becomes Therefore, the drop in the liquid level at the center of the cylinder is Discussion Note that the analysis is valid for any liquid since the result is independent of density or any other fluid property. Also, our assumption of no dry spots is validated since z0(0) is positive. 11-48 The motion of a fish tank in the cabin of an elevator is considered. The pressure at the bottom of the tank when the elevator is stationary, moving up with a specified acceleration, and moving down with a specified acceleration is to be determined. Fish Tank  · 2 az = 3 m/s2 h = 40 cm g z Water  · 1 0 Assumptions 1 The acceleration remains constant. 2 Water is an incompressible substance. Properties We take the density of water to be 1000 kg/m3. Analysis The pressure difference between two points 1 and 2 in an incompressible fluid is given by or since ax = 0. Taking point 2 at the free surface and point 1 at the tank bottom, we have and and thus (a) Tank stationary: We have az = 0, and thus the gage pressure at the tank bottom is (b) Tank moving up: We have az = +3 m/s2, and thus the gage pressure at the tank bottom is (c) Tank moving down: We have az = -3 m/s2, and thus the gage pressure at the tank bottom is Discussion Note that the pressure at the tank bottom while moving up in an elevator is almost twice that while moving down, and thus the tank is under much greater stress during upward acceleration. 11-49 vertical cylindrical milk tank is rotated at constant speed, and the pressure at the center of the bottom surface is measured. The pressure at the edge of the bottom surface is to be determined. z r  Ã‚ · zs R = 1.50 m Free surface g 0 ho Assumptions 1 The increase in the rotational speed is very slow so that the liquid in the container always acts as a rigid body. 2 Milk is an incompressible substance. Properties The density of the milk is given to be 1030 kg/m3. Analysis Taking the center of the bottom surface of the rotating vertical cylinder as the origin (r = 0, z = 0), the equation for the free surface of the liquid is given as where R = 1.5 m is the radius, and The fluid rise at the edge relative to the center of the tank is The pressure difference corresponding to this fluid height difference is Then the pressure at the edge of the bottom surface becomes Discussion Note that the pressure is 14% higher at the edge relative to the center of the tank, and there is a fluid level difference of nearly 2 m between the edge and center of the tank, and these large differences should be considered when designing rotating fluid tanks. 11-50 Milk is transported in a completely filled horizontal cylindrical tank accelerating at a specified rate. The maximum pressure difference in the tanker is to be determined. Æ’-EES ax = 3 m/s2  · 1 z x 0 g  · 2 Assumptions 1 The acceleration remains constant. 2 Milk is an incompressible substance. Properties The density of the milk is given to be 1020 kg/m3. Analysis We take the x- and z- axes as shown. The horizontal acceleration is in the negative x direction, and thus ax is negative. Also, there is no acceleration in the vertical direction, and thus az = 0. The pressure difference between two points 1 and 2 in an incompressible fluid in linear rigid body motion is given by  ® The first term is due to acceleration in the horizontal direction and the resulting compression effect towards the back of the tanker, while the second term is simply the hydrostatic pressure that increases with depth. Therefore, we reason that the lowest pressure in the tank will occur at point 1 (upper front corner), and the higher pressure at point 2 (the lower rear corner). Therefore, the maximum pressure difference in the tank is since x1 = 0, x2 = 7 m, z1 = 3 m, and z2 = 0. Discussion Note that the variation of pressure along a horizontal line is due to acceleration in the horizontal direction while the variation of pressure in the vertical direction is due to the effects of gravity and acceleration in the vertical direction (which is zero in this case). 11-51 Milk is transported in a completely filled horizontal cylindrical tank decelerating at a specified rate. The maximum pressure difference in the tanker is to be determined. Æ’-EES z x  · 2  · 1 g ax = 3 m/s2 Assumptions 1 The acceleration remains constant. 2 Milk is an incompressible substance. Properties The density of the milk is given to be 1020 kg/m3. Analysis We take the x- and z- axes as shown. The horizontal deceleration is in the x direction, and thus ax is positive. Also, there is no acceleration in the vertical direction, and thus az = 0. The pressure difference between two points 1 and 2 in an incompressible fluid in linear rigid body motion is given by  ® The first term is due to deceleration in the horizontal direction and the resulting compression effect towards the front of the tanker, while the second term is simply the hydrostatic pressure that increases with depth. Therefore, we reason that the lowest pressure in the tank will occur at point 1 (upper front corner), and the higher pressure at point 2 (the lower rear corner). Therefore, the maximum pressure difference in the tank is since x1 = 7 m, x2 = 0, z1 = 3 m, and z2 = 0. Discussion Note that the variation of pressure along a horizontal line is due to acceleration in the horizontal direction while the variation of pressure in the vertical direction is due to the effects of gravity and acceleration in the vertical direction (which is zero in this case). 11-52 A vertical U-tube partially filled with alcohol is rotated at a specified rate about one of its arms. The elevation difference between the fluid levels in the two arms is to be determined. z r 0 h0 = 20 cm R = 25 cm Assumptions 1 Alcohol is an incompressible fluid. Analysis Taking the base of the left arm of the U-tube as the origin (r = 0, z = 0), the equation for the free surface of the liquid is given as where h0 = 0.20 m is the original height of the liquid before rotation, and  Ã‚ · = 4.2 rad/s. The fluid rise at the right arm relative to the fluid level in the left arm (the center of rotation) is Discussion Note that the analysis is valid for any liquid since the result is independent of density or any other fluid property. 11-53 A vertical cylindrical tank is completely filled with gasoline, and the tank is rotated about its vertical axis at a specified rate. The pressures difference between the centers of the bottom and top surfaces, and the pressures difference between the center and the edge of the bottom surface are to be determined. Æ’-EES h = 3 m D = 1.20 m z r 0 Assumptions 1 The increase in the rotational speed is very slow so that the liquid in the container always acts as a rigid body. 2 Gasoline is an incompressible substance. Properties The density of the gasoline is given to be 740 kg/m3. Analysis The pressure difference between two points 1 and 2 in an incompressible fluid rotating in rigid body motion is given by where R = 0.60 m is the radius, and (a) Taking points 1 and 2 to be the centers of the bottom and top surfaces, respectively, we have and . Then, (b) Taking points 1 and 2 to be the center and edge of the bottom surface, respectively, we have , , and . Then, Discussion Note that the rotation of the tank does not affect the pressure difference along the axis of the tank. But the pressure difference between the edge and the center of the bottom surface (or any other horizontal plane) is due entirely to the rotation of the tank. 11-54 Problem 11-53 is reconsidered. The effect of rotational speed on the pressure difference between the center and the edge of the bottom surface of the cylinder as the rotational speed varies from 0 to 500 rpm in increments of 50 rpm is to be investigated. g=9.81 m/s2 rho=740 kg/m3 R=0.6 m h=3 m omega=2*pi*n_dot/60 rad/s DeltaP_axis=rho*g*h/1000 kPa DeltaP_bottom=rho*omega^2*R^2/2000 kPa Rotation rate , rpm Angular speed  Ã‚ ·, rad/s  Ã¢â‚¬Å¾Pcenter-edge kPa 0 50 100 150 200 250 300 350 400 450 500 0.0 5.2 10.5 15.7 20.9 26.2 31.4 36.7 41.9 47.1 52.4 0.0 3.7 14.6 32.9 58.4 91.3 131.5 178.9 233.7 295.8 365.2 11-55E A water tank partially filled with water is being towed by a truck on a level road. The maximum acceleration (or deceleration) of the truck to avoid spilling is to be determined. ax  Ã¢â‚¬Å¾h = 2 ft  Ã‚ ± Water tank hw = 6 ft z x 0 L=20 ft Assumptions 1 The road is horizontal so that acceleration has no vertical component (az = 0). 2 Effects of splashing, breaking, driving over bumps, and climbing hills are assumed to be secondary, and are not considered. 3 The acceleration remains constant. Analysis We take the x-axis to be the direction of motion, the z-axis to be the upward vertical direction. The shape of the free surface just before spilling is shown in figure. The tangent of the angle the free surface makes with the horizontal is given by  ® where az = 0 and, from geometric considerations, tan Ã‚ ± is Substituting, The solution can be repeated for deceleration by replacing ax by ax. We obtain ax = -6.44 m/s2. Discussion Note that the analysis is valid for any fluid with constant density since we used no information that pertains to fluid properties in the solution. 11-56E A water tank partially filled with water is being towed by a truck on a level road. The maximum acceleration (or deceleration) of the truck to avoid spilling is to be determined. ax  Ã¢â‚¬Å¾h = 0.5 ft  Ã‚ ± Water tank hw = 3 ft z x 0 L= 8 ft Assumptions 1 The road is horizontal so that deceleration has no vertical component (az = 0). 2 Effects of splashing and driving over bumps are assumed to be secondary, and are not considered. 3 The deceleration remains constant. Analysis We take the x-axis to be the direction of motion, the z-axis to be the upward vertical direction. The shape of the free surface just before spilling is shown in figure. The tangent of the angle the free surface makes with the horizontal is given by  ® where az = 0 and, from geometric considerations, tan Ã‚ ± is Substituting, Discussion Note that the analysis is valid for any fluid with constant density since we used no information that pertains to fluid properties in the solution. 11-57 Water is transported in a completely filled horizontal cylindrical tanker accelerating at a specified rate. The pressure difference between the front and back ends of the tank along a horizontal line when the truck accelerates and decelerates at specified rates. Æ’-EES ax = 3 m/s2 z x 0 1  · 2 g  · Assumptions 1 The acceleration remains constant. 2 Water is an incompressible substance. Properties We take the density of the water to be 1000 kg/m3. Analysis (a) We take the x- and z- axes as shown. The horizontal acceleration is in the negative x direction, and thus ax is negative. Also, there is no acceleration in the vertical direction, and thus az = 0. The pressure difference between two points 1 and 2 in an incompressible fluid in linear rigid body motion is given by  ® since z2 z1 = 0 along a horizontal line. Therefore, the pressure difference between the front and back of the tank is due to acceleration in the horizontal direction and the resulting compression effect towards the back of the tank. Then the pressure difference along a horizontal line becomes since x1 = 0 and x2 = 7 m. (b) The pressure difference during deceleration is determined the way, but ax = 4 m/s2 in this case, Discussion Note that the pressure is higher at the back end of the tank during acceleration, but at the front end during deceleration (during breaking, for example) as expected. Review Problems 11-58 The density of a wood log is to be measured by tying lead weights to it until both the log and the weights are completely submerged, and then weighing them separately in air. The average density of a given log is to be determined by this approach. Properties The density of lead weights is given to be 11,300 kg/m3. We take the density of water to be 1000 kg/m3. Analysis The weight of a body is equal to the buoyant force when the body is floating in a fluid while being completely submerged in it (a consequence of vertical force balance from static equilibrium). In this case the average density of the body must be equal to the density of the fluid since Lead, 34 kg Log, 1540 N FB Water Therefore, where Substituting, the volume and density of the log are determined to be Discussion Note that the log must be completely submerged for this analysis to be valid. Ideally, the lead weights must also be completely submerged, but this is not very critical because of the small volume of the lead weights. 11-59 A rectangular gate that leans against the floor with an angle of 45 ° with the horizontal is to be opened from its lower edge by applying a normal force at its center. The minimum force F required to open the water gate is to be determined. Assumptions 1 The atmospheric pressure acts on both sides of the gate, and thus it can be ignored in calculations for convenience. 2 Friction at the hinge is negligible. Properties We take the density of water to be 1000 kg/m3 throughout. Analysis The length of the gate and the distance of the upper edge of the gate (point B) from the free surface in the plane of the gate are FR F 45 ° B 0.5 m 3 m A The average pressure on a surface is the pressure at the centroid (midpoint) of the surface, and multiplying it by the plate area gives the resultant hydrostatic on the surface, The distance of the pressure center from the free surface of water along the plane of the gate is The distance of the pressure center from the hinge at point B is Taking the moment about point B and setting it equal to zero gives Solving for F and substituting, the required force is determined to be Discussion The applied force is inversely proportional to the distance of the point of application from the hinge, and the required force can be reduced by applying the force at a lower point on the gate. 11-60 A rectangular gate that leans against the floor with an angle of 45 ° with the horizontal is to be opened from its lower edge by applying a normal force at its center. The minimum force F required to open the water gate is to be determined. Assumptions 1 The atmospheric pressure acts on both sides of the gate, and thus it can be ignored in calculations for convenience. 2 Friction at the hinge is negligible. Properties We take the density of water to be 1000 kg/m3 throughout. FR F 45 ° B 1.2 m 3 m AAnalysis The length of the gate and the distance of the upper edge of the gate (point B) from the free surface in the plane of the gate are The average pressure on a surface is the pressure at the centroid (midpoint) of the surface, and multiplying it by the plate area gives the resultant hydrostatic on the surface, The distance of the pressure center from the free surface of water along the plane of the gate is The distance of the pressure center from the hinge at point B is Taking the moment about point B and setting it equal to zero gives Solving for F and substituting, the required force is determined to be Discussion The applied force is inversely proportional to the distance of the point of application from the hinge, and the required force can be reduced by applying the force at a lower point on the gate. 11-61 A rectangular gate hinged about a horizontal axis along its upper edge is restrained by a fixed ridge at point B. The force exerted to the plate by the ridge is to be determined. Assumptions The atmospheric pressure acts on both sides of the gate, and thus it can be ignored in calculations for convenience. FR 3 m A 2 m ypProperties We take the density of water to be 1000 kg/m3 throughout. Analysis The average pressure on a surface is the pressure at the centroid (midpoint) of the surface, and multiplying it by the plate area gives the resultant hydrostatic force on the gate, The vertical distance of the pressure center from the free surface of water is 11-62 A rectangular gate hinged about a horizontal axis along its upper edge is restrained by a fixed ridge at point B. The force exerted to the plate by the ridge is to be determined. Assumptions The atmospheric pressure acts on both sides of the gate, and thus it can be ignored in calculations for convenience. FR 3 m yP h = 2 m A Properties We take the density of water to be 1000 kg/m3 throughout. Analysis The average pressure on a surface is the pressure at the centroid (midpoint) of the surface, and multiplying it by the wetted plate area gives the resultant hydrostatic force on the gate, The vertical distance of the pressure center from the free surface of water is 11-63E A semicircular tunnel is to be built under a lake. The total hydrostatic force acting on the roof of the tunnel is to be determined. Assumptions The atmospheric pressure acts on both sides of the tunnel, and thus it can be ignored in calculations for convenience. Properties We take the density of water to be 62.4 lbm/ft3 throughout. Analysis We consider the free body diagram of the liquid block enclosed by the circular surface of the tunnel and its vertical (on both sides) and horizontal projections. The hydrostatic forces acting on the vertical and horizontal plane surfaces as well as the weight of the liquid block are determined as follows: Horizontal force on vertical surface (each side): Fy W Fx Fx Vertical force on horizontal surface (downward): R = 15 ft Weight of fluid block on each side within the control volume (downward): Therefore, the net downward vertical force is This is also the net force acting on the tunnel since the horizontal forces acting on the right and left side of the tunnel cancel each other since they are equal ad opposite. 11-64 A hemispherical dome on a level surface filled with water is to be lifted by attaching a long tube to the top and filling it with water. The required height of water in the tube to lift the dome is to be determined. Assumptions 1 The atmospheric pressure acts on both sides of the dome, and thus it can be ignored in calculations for convenience. 2 The weight of the tube and the water in it is negligible. Properties We take the density of water to be 1000 kg/m3 throughout. Analysis We take the dome and the water in it as the system.à ¢Ã¢â€š ¬Ã¢â‚¬Å¡ When the dome is about to rise, the reaction force between the dome and the ground becomes zero. Then the free body diagram of this system involves the weights of the dome and the water, balanced by the hydrostatic pressure force from below. Setting these forces equal to each other gives FV R = 3 m h W Solving for h gives Substituting, Therefore, this dome can be lifted by attaching a tube which is 2.02 m long. Discussion This problem can also be solved without finding FR by finding the lines of action of the horizontal hydrostatic force and the weight. 11-65 The water in a reservoir is restrained by a triangular wall. The total force (hydrostatic + atmospheric) acting on the inner surface of the wall and the horizontal component of this force are to be determined. FR h = 25 m ypAssumptions 1 The atmospheric pressure acts on both sides of the gate, and thus it can be ignored in calculations for convenience. 2 Friction at the hinge is negligible. Properties We take the density of water to be 1000 kg/m3 throughout. Analysis The length of the wall surface underwater is The average pressure on a surface is the pressure at the centroid (midpoint) of the surface, and multiplying it by the plate area gives the resultant hydrostatic force on the surface, Noting that the distance of the pressure center from the free surface of water along the wall surface is The magnitude of the horizontal component of the hydrostatic force is simply FRsin  Ã‚ ±, Discussion The atmospheric pressure is usually ignored in the analysis for convenience since it acts on both sides of the walls. 11-66 A U-tube that contains water in right arm and another liquid in the left is rotated about an axis closer to the left arm. For a known rotation rate at which the liquid levels in both arms are the same, the density of the fluid in the left arm is to be determined. 1*  ·  · 1 Fluid Water

Creative Story: The Chronic Swamp Murders :: essays research papers

Creative Story: The Chronic Swamp Murders   Ã‚  Ã‚  Ã‚  Ã‚  One day while Joe and Jill Hemp were walking through Chronic swamp they came across a trail of blood in the water. They followed the trail until it stopped at a dead body. The body was of a man who was wearing a camouflage outfit. They immediately ran back to their house, which was not far from the murder site and called the police. Their house was located right on the edge of the swamp.   Ã‚  Ã‚  Ã‚  Ã‚  When the police got there they roped off the whole area so they could start their investigation. At first it looked as if it was definitely a murder, but after a few days of investigation the police concluded that the man was probably a hunter who had either fallen out of a tree or just tripped and broken his neck. A broken neck was definitely what killed the man. The only problem with this hypothesis was that it left a few unanswered questions. If the man was a hunter where did his gun or bow go? How often did you find a dead hunter just lying in the middle of a swamp? Even with these questions police told the Hemps that it was an accident and they were in no danger.   Ã‚  Ã‚  Ã‚  Ã‚  The Hemp's life went on with no interruptions until about two weeks after the hunter was found. Another body had been found in the swamp. This time the body was a male whom had a business suit on. The police came back and investigated this death. After about a week they concluded again that it was a broken neck that had killed the victim. There were no signs of a struggle so the investigators said that it was some type of freak accident. They also told the Hemps to stay out of the swamp.   Ã‚  Ã‚  Ã‚  Ã‚  The Hemps never went back into the swamp again, but one night they were awakened by a loud pounding noise on the front door. When Mr. Hemp got up to see what it was, all he saw was something large running into the swamp. He then made sure that all the doors were locked and he got his shotgun out of the closet. He waited in his dark living room for about an hour and then went back to his bed. He didn't tell his wife what had happened so she wouldn't be scared   Ã‚  Ã‚  Ã‚  Ã‚  The next day when Joe was coming home from work he noticed the door was wide open. When he got closer he noticed that the door frame was broken and the Creative Story: The Chronic Swamp Murders :: essays research papers Creative Story: The Chronic Swamp Murders   Ã‚  Ã‚  Ã‚  Ã‚  One day while Joe and Jill Hemp were walking through Chronic swamp they came across a trail of blood in the water. They followed the trail until it stopped at a dead body. The body was of a man who was wearing a camouflage outfit. They immediately ran back to their house, which was not far from the murder site and called the police. Their house was located right on the edge of the swamp.   Ã‚  Ã‚  Ã‚  Ã‚  When the police got there they roped off the whole area so they could start their investigation. At first it looked as if it was definitely a murder, but after a few days of investigation the police concluded that the man was probably a hunter who had either fallen out of a tree or just tripped and broken his neck. A broken neck was definitely what killed the man. The only problem with this hypothesis was that it left a few unanswered questions. If the man was a hunter where did his gun or bow go? How often did you find a dead hunter just lying in the middle of a swamp? Even with these questions police told the Hemps that it was an accident and they were in no danger.   Ã‚  Ã‚  Ã‚  Ã‚  The Hemp's life went on with no interruptions until about two weeks after the hunter was found. Another body had been found in the swamp. This time the body was a male whom had a business suit on. The police came back and investigated this death. After about a week they concluded again that it was a broken neck that had killed the victim. There were no signs of a struggle so the investigators said that it was some type of freak accident. They also told the Hemps to stay out of the swamp.   Ã‚  Ã‚  Ã‚  Ã‚  The Hemps never went back into the swamp again, but one night they were awakened by a loud pounding noise on the front door. When Mr. Hemp got up to see what it was, all he saw was something large running into the swamp. He then made sure that all the doors were locked and he got his shotgun out of the closet. He waited in his dark living room for about an hour and then went back to his bed. He didn't tell his wife what had happened so she wouldn't be scared   Ã‚  Ã‚  Ã‚  Ã‚  The next day when Joe was coming home from work he noticed the door was wide open. When he got closer he noticed that the door frame was broken and the