NEC Internet of Things Security Globally Interlinked to The Internet Project

University of Phoenix University Writing Skills for Criminal Justice Response

University of Phoenix University Writing Skills for Criminal Justice Response.

I’m working on a criminal justice discussion question and need support to help me learn.

Respond to the following in a minimum of 175 words: One of the skills that is overlooked but very important in policing is writing. The article “Writing Skills: Hiring and Training Police Officers” provides some of the ways an officer will use his or her writing skills.Choose at least two qualities listed in Table 2: Importance of Qualities to Officers’ Writing Skills.Explain how learning these skills are important for a career in policing or any other career. Why might it be better to learn those skills now, while earning your degree, than later on the job?(The 2 Qualities listed in table 2 are Accuracy of information and Completeness of information)
University of Phoenix University Writing Skills for Criminal Justice Response

Ashworth College Week 2 Aerobic versus Anaerobic Discussion

best essay writers Ashworth College Week 2 Aerobic versus Anaerobic Discussion.

Your initial discussion thread is due on Day 3 (Thursday) and you have until Day 7 (Monday) to respond to your classmates. Your grade will reflect both the quality of your initial post and the depth of your responses. Refer to the Discussion Forum Grading Rubric under the Settings icon above for guidance on how your discussion will be evaluated. Aerobic Versus Anaerobic: What is the Difference? [CLOs: 2, 3, 5] [WLOs: 1, 3]After reading Chapter 5 and 6 in the course text and viewing Physical Training Strategies: Preparing for a Purpose: Energy System in the Body, select one of the following: a triathlete, a football player, a gymnast, and one phase of their sport. For example, if you choose the triathlete you can choose the cycling phase of their activity, or if you choose the football player, you could choose the sprint phase of their activity, or if you choose the gymnast, you may choose the backflip phase of their activity.As your athlete performs the chosen activity, discuss whether rapid or slow glycolysis is the most effective means of energy transfer?What physiological factors contributed to your analysis (e.g. hydrogen release, lactate formation, glucose catabolism, etc.)?Explain the benefits of lactate for optimal performance of the chosen activity.Your research and claims must be supported by your course text and a minimum of two additional scholarly sources. Use proper APA formatting for in-text citations and references as outlined in the Ashford Writing Center.Guided Response: Review some of your classmates’ posts. Identify at least two classmates who chose a different athlete than you chose and analyze the appropriateness of the type of glycolysis they discussed. Then, explain if you agree with the benefits of lactate identified for their chosen activity. Support your reasoning for each response to your classmates with at least one scholarly source.mine Discussion 1My chosen individual is a footballer and the phase of activity is during sprinting. The phase is important because energy usage is crucial and can be quantified or related to the physiological processes that are in place.During the sprint phase, rapid glycolysis is the most effective way of producing energy to help the footballer meet the energy requirements of the body (Katch, McArdle & Katch, 2015). Necessarily, the footballer requires enough energy to meet the needs of the body during the sprint phase. Rapid glycolysis yields more energy in the form of ATP that is used in the body for supporting the body during the rigorous exercise.The physiological factors that contributed to the analysis are lactate formation whereby during the sprinting, lactate is formed hence calling for more production of energy (Scott & Fountaine, 2013). Additionally, excess hydrogen release during exercise is a significant issue that makes the athlete need more energy in the form of ATP hence rapid glycolysis can help in promoting more energy release (Scott & Fountaine, 2013). Finally, glucose catabolism is a factor during the exercise and calls for a rapid glycolysis process for fastening the process of producing ATP.Lactate has a great benefit in a footballer for sprinting since it is always taken to the liver and converted to glucose and glycogen (Hall 2010). With lactate, more glucose is produced, and therefore the need for rapid glycolysis is accomplished which makes things better and perfect. Essentially, the focus and emphasis are on how things work and the level of precision associated with the same. ReferencesHall, K.D. (2010). Predicting metabolic adaptation, body weight change, and energy intake in humans. Am J PhysiolEndocrinolMetab, 298(3):E449-66.doi: 10.1152/ajpendo.00559.2009Katch, V., McArdle, W., & Katch, F. (2015). Essentials of exercise physiology. (5th ed.). Retrieved from https://www.vitalsource.com/Scott, C.B., Fountaine, C. (2013). Estimating the energy costs of intermittent exercise. J Hum Kinet, 38:107-13. doi: 10.2478/hukin-2013-0050Carmen GreeneJan 13, 2021 at 6:07 PMAerobic Versus Anaerobic – The TriathleteRapid or Slow Glycolysis? A triathlete is an endurance athlete who competes in running, cycling, and swimming. Focusing on the running portion of the event, the athlete typically has to run 26.2 miles, and because of this, the most effective means of energy transfer is slow aerobic glycolysis. Rapid anaerobic glycolysis only releases about 5% of the potential energy of a glucose molecule. This is useful for sports that require short, tough spurts of power of roughly 90-second durations but not so much for the endurance runner. Endurance running requires much more, so the remaining energy needed needs to be pulled from slow aerobic glycolysis (Katch et al., 2015). A Russian study done on the best way to train endurance runners found that those trained aerobically were better at adapting to long term endurance work. In contrast, those trained with anaerobic and mixed types of energy metabolism adapted better to speed-power work over endurance (Bakayev & Bolotin, 2020).Physiological factors contributing to my analysis Rapid glycolysis is anaerobic, the end product is lactate, and the process makes ATP much faster, but the amount is limited and not enough for a long endurance race. Slow glycolysis is aerobic. The end product is pyruvate, so further carbohydrate breakdown is needed and a much slower process, but it makes a lot more ATP, which is necessary to keep the endurance runner going (Katch et al., 2015). The benefits of lactate for optimal performance I always thought of lactate as a negative thing causing the side pain when I worked out too hard. It happens when the workout exceeds our circulation’s ability to meet our 02 requirements (Katch et al., 2015). It is our body’s way of maintaining homeostasis, and our text showed me that it accumulates with intense workouts but is a valuable energy source. Hydrogen attached to lactate is used to form more energy or ATP, and lactate is used to replenish glycogen reserves that were depleted during the intense workout. A 2020 study on blood lactate in shock noted that during exercise, the circulating blood lactate metabolism contributed more than twice as much energy as the blood glucose (Levitt et al., 2020). References:Bakayev, V., & Bolotin, A. (2020). Differentiated Training Model for Marathon Runners on Building Tempo and Speed Endurance Based On the Types of Energy Metabolism. Sport Mont, 18(3), 31–34.Katch, V., McArdle, W., & Katch, F. (2015). Essentials of exercise physiology. (5th ed.). Retrieved from https://www.vitalsource.com/Levitt, D. G., Levitt, J. E., & Levitt, M. D. (2020). Quantitative Assessment of Blood Lactate in Shock: Measure of Hypoxia or Beneficial Energy Source. BioMed Research International, 1–24. https://doi-org.proxy- library.ashford.edu/10.1155/2020/2608318
Ashworth College Week 2 Aerobic versus Anaerobic Discussion

MGT 312 Saudi Electronic University Decision Making Process Case Study Discussion

MGT 312 Saudi Electronic University Decision Making Process Case Study Discussion.

Course Learning Outcomes-Covered Demonstrate a solid understanding of decision making process for complex issues pertaining to business environment both internally and externally. (1.2) Explain critical thinking and cognitive psychology as it pertains to analyze and synthesize information for problem solving and decision making. (2.7) Identify and analyze different perspectives on understanding problems for different situations. (3.1) Assignment Instructions: Log in to Saudi Digital Library (SDL) via University’s websiteOn first page of SDL, choose “English Databases”From the list find and click on EBSCO database.In the search bar of EBSCO find the following article: Title: “The elements of good judgement- How to improve your decision-making” Author: Sir Andrew Likierman, Professor, London Business School Date of Publication:January–February 2020   Published: Harvard Business Review Assignment Questions:(Marks 05) Read the attached article titled as “The elements of good judgement- How to improve your decision-making” by Sir Andrew Likierman, published in Harvard Business Review, and answer the following Questions: [5 Marks] Summarize the article and explain the main issues discussed in the article. (700 words) What do you think about the article in relations to what you have learnt in the course about improving decision-making and problem solving skills?Use additional reference to support you argument. (500 words)
MGT 312 Saudi Electronic University Decision Making Process Case Study Discussion

A Study On Hookes Law Mechanics Essay

HOOKE’S LAW: Hooke’s lawof elasticity is an approximation that states that the extension of a spring is in direct proportion with the load added to it as long as this load does not exceed the elastic limit. Materials for which Hooke’s law is a useful approximation are known as linear elasticor “Hookean” materials. If a metal is lightly stressed, a temporary deformation permitted by an elastic displacement of atoms in space takes place. Removal of stress results in a gradual return of metal to its original shape. Mathematically, Hooke’s law states that Where, xis the displacement of the end of the spring from its equilibrium position; Fis the restoring force exerted by the material; and kis theforce constant(orspring constant). DIAGRAMATICALLY:- When no weight is applied to the spring, the strain is zero, And, we can measure its length,. and when we apply a force F to the spring It stretches And it extends length,x, that is, the strain, caused by the stress is F = mg. Also, In terms of mechanics hooks state that:- “For an elastic material stress applied on a body is directly proportional to the strain produced” That is, σ α e Or σ = E e Where, σ is the stress applied e is the strain developed E is the YOUNGS MODULUS OF ELASTICITY Now STRESS it is the force causing the deformation. It is measured in units of force per unit area of cross-section (N.m-2) denoted byσ(“sigma”). That is σ= F/A Units of stress are Pascal… Strain is the deformation that takes place in the body. It is the ratio of the increase in length,DLto the original length (L), Represented by symbolε(“epsilon”) or e. That is e=DL/L It is dimensionless. And according to hook’s law: σ = E e Or, E = σ/e Putting values of stress and strain in above equation we get:- E = FÃ-L/AÃ-DL Young’s modulus of elasticity (E) is defined as the ratio of unit stress to unit strain . GENERALIZED HOOK’S LAW: The generalized Hooke’s Law can be used to predict the deformations caused in a given material by an arbitrary combination of stresses. The linear relationship between stress and strain applies for The generalized Hooke’s Law also reveals that strain can exist without stress. For example, if the member is experiencing a load in the y-direction (which in turn causes a stress in the y-direction), the Hooke’s Law shows that strain in the x-direction does not equal to zero. This is because as material is being pulled outward by the y-plane, the material in the x-plane moves inward to fill in the space once occupied, just like an elastic band becomes thinner as you try to pull it apart. In this situation, the x-plane does not have any external force acting on them but they experience a change in length. Therefore, it is valid to say that strain exist without stress in the x-plane. STRESS-STRAIN CURVE:- Thestress-straincurve is a graphical representation of the relationship betweenstress, derived from measuring the load applied on the sample, andstrain, derived from measuring thedeformationof the sample, i.e. elongation, compression, or distortion. The nature of the curve varies from material to material. ELASTIC LIMIT: The elastic limit is where the graph departs from a straight line. If we go past it, the spring won’t go back to its original length. When we remove the force, we’re left with apermanent extension. Below the elastic limit, we say that the spring is showing “elastic behaviour”: the extension is proportional to the force, and it’ll go back to it’s original length when we remove the force. Beyond the elastic limit, we say that it shows “plastic behaviour”. This means that when a force is applied to deform the shape, it stays deformed when the force is removed. YIELD POINT: Theyield strengthoryield pointof a materialis defined in engineering and material science as the stress at which a material begins to deform plastically . Prior to the yield point the material will deform elastically and will return to its original shape when the applied stress is removed. Once the yield point is passed some fraction of the deformation will be permanent and non-reversible. True elastic limit: The lowest stress at whichdislocationsmove. This definition is rarely used, since dislocations move at very low stresses, and detecting such movement is very difficult. Proportionality limit: Up to this amount of stress, stress is proportional to strain hookes law so the stress-strain graph is a straight line, and the gradient will be equal to the elastic modulus of the material. Elastic limit (yield strength): Beyond the elastic limit, permanent deformation will occur. The lowest stress at which permanent deformation can be measured. This requires a manual load-unload procedure, and the accuracy is critically dependent on equipment and operator skill. For elastomers such as rubber the elastic limit is much larger than the proportionality limit. Also, precise strain measurements have shown that plastic strain begins at low stresses. Offset yield point (proof stress) This is the most widely used strength measure of metals, and is found from the stress-strain curve. A plastic strain of 0.2% is usually used to define the offset yield stress, although other values may be used depending on the material and the application. The offset value is given as a subscript, e.g., Rp0.2=310 MPa. In some materials there is essentially no linear region and so a certain value of strain is defined instead. Although somewhat arbitrary, this method does allow for a consistent comparison of materials. Upper yield point and lower yield point Some metals, such as mild steel reach an upper yield point before dropping rapidly to a lower yield point. The material response is linear up until the upper yield point, but the lower yield point is used in structural engineering as a conservative value. If a metal is only stressed to the upper yield point, and beyond rubber band can develop. NUMERICALS:- Q1) When a 13.2-kg mass is placed on top of a vertical spring, the spring compresses 5.93 cm. Find the force constant of the spring. Solution: Mass = 13.2 kg Weight = 13.2Ã-9.8 = 129 Compression (x) = 5.93 = 0.0593 m From Hooke’s Law: F = kx The force on the spring is the weight of the object, i.e.(F) = 129 N Putting values of force and compression in above equation; 129 = (0.0593) Ã- k Or, k = 2181 N/m Answer Q2) A 3340 N ball is supported vertically by a 2m diameter steel cable assuming cable has a length of 10m, determine stress and strain in the cable. Young’s modulus for steel is 200N/sq.m. Solution: Force (F) = 3340N Diameter = 2m Radius (r) = 1m Length of cable = 10m Young’s modulus (E) = 200N/sq.m Now we know, Stress (σ) = F/A Area = = 3.14Ã-1Ã-1 = 3.14 So, σ = 3340/3.14 σ = 1063.69N/m.sq Also, strain (e) = σ/E Putting values… e = 1063.69/200 e =5.3184 Answer Q3) If a spring has a spring constant of 400 N/m, how much work is required to compress the spring 25.0 m from its undisturbed position? Solution: spring constant (K) = 400 N/m compression (x) = 25m we know, force required for compression:- F = kx i.e. F = 400×25 = 10000N and work done = force x compression w = F x X w = 10000 x 25 w = 25,000 Joules Answer Q4) On a of steel rod of length 15m and diameter 6m a force of 60N is applied. Calculate the extension and new length of the rod. Young’s modulus of steel is 250N/m.sq. Solution: : Force (F) = 60 N Diameter = 6m So, Radius (r) = 3m Length (L) = 15 m Young’s modulus (E) = 250N/m.sq. Now, Area (A) A = 3.14 x 3 x 3 A = 28.26 sq.m Also, , E = FÃ-L/AÃ-DL Or, DL = FÃ-L/AÃ-E DL = 60Ã-15/28.26Ã-250 DL = 0.127m SO, new length = 15 0.127 L’ = 15.127m ANSWER REFERENCES:- 1) www.physicsworld.com 2) www.wikipedia.org 3) www.123iitjee.com 4) www.physicsforum.com 5) www.matter.org.content/HookesLaw 6) www.webphysics.davidson.edu/hook 7) www.scienceworld.com