Future of Project Management Team Performance Essay

Future of Project Management Team Performance - Essay Example Presently the HP organization is engaged in streamlining process. With the increase in collaborative activities around the world, the works of companies are outsourced to various service providers in a bid to lower the cost (Mullaney, Stubbings and Clarke, 2013). Until now, most of the HP team members had to work  in  one place and route their communication through  landlines  and direct meetings with colleagues.  With advanced research and development of  technology in the Internet and others, it is easy to contact any person and share the requisite information.   The requirement of the technology is limited to the smart phone or any wired device for connection with  the Internet.   As HP is a largest technology company  around  the world and requires  a constant  inflow of knowledge workers, who are experts in current technologies employed  at  the company and  have the capability of assimilation of the continuous innovation. With the continuous research and development in  communication  technology, employees of HP are able to work and  communicate  to  their colleagues and office people with various electronic devices. So, it is possible for them to work from home or any other premises. This facility has facilitated the engagement of talented workforce  that is not locally available.  With  technological improvement, HP has started implementation of flexibility in work in various offices.   With continuous development and improvement in knowledge,  it is imperative that, the project management team should be trained in contemporary styles regularly.  This is also true about various types of certifications that are needed to keep the edge in the changing market. HP continues to implement the policy regarding the training  to  their employees with contemporary knowledge and certifications (Stanleigh, 2015).   It is a fact that with time, everything changes including trends in the workplace. With ongoing research in  the

Three Phase System Outline

Three Phase System Outline Single phase systems are defined by having an AC source with only one voltage waveform. Figure 1 is a simple AC circuit. Single-phase power distribution is widely used especially in rural areas, because the cost of a single-phase distribution network is low. Figure 1:- Single phase system schematic diagram Today most of the electrical power generated in the world is three-phase. Three-phase power was first conceived by Nikola Tesla. Three-phase power was the most efficient way that electricity could be produced, transmitted, and consumed. A three-phase generator has three separate but identical windings that are 1200 electrical apart from one another. 5.2 Three Phase Circuit Three-phase voltage systems are composed of three sinusoidal voltages of equal magnitude, equal frequency and separated by 120 degrees, as shown in Figure 2. It is one voltage cycle of a 3 phase system. It is labeled 0 to 360 ° (2 Ï€ radians) along the time axis. The plotted lines show the variation of instantaneous voltage (or current) over time. This power wave cycle will repeat usually  50 (50Hz), 60 (60Hz), or 400 (400Hz)  times per second, depending on the power system  frequency (Hz). The colors of the lines are in the  American Color Code for 3-phase wiring. It is black=VL1  red=VL2blue=VL3. Figure 2:- Three phase waveforms Three phase systems may or may not have a neutral wire. The neutral wire allows 3 phase systems to use a higher voltage while still supporting lower voltage 1 phase appliances. In  high voltage 3 phase distribution  situations it is common not to have a neutral wire as the loads can simply be connected between phases (phase-phase connection). 5.2.1 Advantage over Single Phase system Three phase system is better to single phase system. The reason for the advantage over single phase system is given below. The horsepower rating of three-phase motors and the KVA (kilo-volt-amp) rating of three-phase transformers is about 150% greater than for single-phase motors or transformers with a similar frame size. Figure 3:- Single-phase power falls to zero three times each cycle. Figure 4:- Three-phase power never falls to zero. The power delivered by a single-phase system pulsates, as shown in Figure 3. The power falls to zero three times during each cycle. The power delivered by a three-phase circuit pulsates also, but it never falls to zero, as shown in Figure 4. In a three-phase system, the power delivered to the load is the same at any instant. This produces superior operating characteristics for three-phase motors. In a balanced three-phase system, the conductors need be only about 75% the size of conductors for a single-phase two-wire system of the same KVA rating. This helps offset the cost of supplying the third conductor required by three-phase systems. If a magnetic field is rotate through the conductors of a stationary coil then a single phase alternating voltage can be produced. This explanation is shown in Figure 5. Figure 5:- A single-phase voltage. Since alternate polarities of the magnetic field cut through the conductors of the stationary coil, the induced voltage will change polarity at the same speed as the rotation of the magnetic field. The alternator shown in Figure 5 is single phase because it produces only one AC voltage. Figure 6:- The voltages of a three-phase system are 120 ° out of phase with each other. If three separate coils are spaced 120 ° apart, as shown in Figure 6, three voltages 120 ° out of phase with each other will be produced when the magnetic field cuts through the coils. This is the manner in which a three-phase voltage is produced. 5.2.2 Classification Three-phase supply voltages and load systems have two basic configurations: a). wye or star connection and b). delta connection. 5.3 Star and Delta connection The Wye is a 4-wire system. Wye configurations typically include a neutral line (N) connected to the common point (3 phase plus neutral for a total of four wires), as shown in Figure 7. Figure 7:- A wye connections is formed by joining one end of each of the windings together. The Delta, as shown in Figure 8, is a 3-wire system which is primarily used to provide power for three-phase motor loads. The system is normally ungrounded and has only one three-phase voltage available. The lack of a system ground makes it difficult to protect for ground faults. Often, a ground detection scheme, employing ground lamps, is used to provide an indication or alarm in the event of a system ground. The Delta System is sometimes corner grounded to protect for ground faults on the other two phases. Figure 8:- Three-phase delta connection 5.4 Phasor diagrams 5.4.1 Star connection The voltage measured across a single winding or phase is known as the phase voltage, as shown in Figure 9. The voltage measured between the lines is known as the line-to-line voltage or simply as the line voltage. The currents flowing in the phases are called phase currents and currents flowing in the lines are called line currents. Figure 9:- Line and phase voltages are different in a wye connection. The parallelogram method of vector addition for the voltages in a wye-connected three-phase system is shown in Figure 10. Figure 10 shows how the line voltage may be obtained using the normal parallelogram addition. Figure 10:- Phasor diagram of Star connection Voltage However, the line voltage is not equal to the phase voltage. The line voltage V1-2 is equal to the phasor difference of VA and VB. The line voltage V2-3 is equal to the phasor difference of VB and VC. The line voltage V3-1 is equal to the phasor difference of VC and VA. The line voltages are defined as: V1-2 = VA VB, V2-3 = VB-VC, and V3-1 = VC-VA. Here V1-2, V2-3, V3-1 are the line voltage (VLine) and VA, VB, VC are the phase voltage (VPhase) of Wye connection. VA, VB, VC are the reverse phase voltage of VA,VB, VC. The two phasors VA and VB are 600 apart. V1-2 = VLine = VA VB = [VPhase (-VPhase)] cos(600/2) = 2 VPhase cos300 = √3 VPhase The two phasors VB and VC are 600 apart. V2-3 = VLine = VB-VC = √3 VPhase The two phasors VC and VA are 600 apart. V3-1 = VLine = VC-VA = √3 VPhase  Ã…“ V1-2 = V2-3 = V3-1 = line voltage = VLine =√3 VPhase Current On a Wye system or star connected supply, the phase unbalance current is carried by the neutral. On a Wye system, the line current (current in the line) (ILine) is equal to the phase current (current in a phase) (IPhase) i.e. ILine = IPhase Power Total power P = 3 Power in each phase = 3 VPhase IPhase cosÃŽ ¦ = 3 (VLine/√3) ILine cosÃŽ ¦ [for Wye connection] = √3 VLine ILine cosÃŽ ¦ Where VLine and ILine are the line voltage and the line current of a star connected supply. The term cosÃŽ ¦ is called power factor of the circuit and its value is given by; cosÃŽ ¦ = R/Z Where R and Z are the resistance and impedance of a circuit. 5.4.2 DELTA CONNECTIONS In Figure 11, voltmeters have been connected across the lines and across the phase. Ammeters have been connected in the line and in the phase. Figure 11:- Voltage and current relationships in a delta connection The delta connection is similar to a parallel connection because there is always more than one path for current flow. Since these currents are 120 ° out of phase with each other, vector addition must be used when finding the sum of the currents, as shown in Figure 12. Figure 12:- Phasor Diagram of Delta connection Voltage In the delta connection, the three voltages are equal in magnitude but displaced 1200 from one another. In the delta connection, line voltage (VLine) and phase voltage (Vphase) are the same. VLine = Vphase Current In the delta connection, the line current and phase current are different. The line current is the vector sum of two individual phase currents. The line current I1 is equal to the phasor difference of IA and IC. The line current I2 is equal to the phasor difference of IB and IA. The line current I3 is equal to the phasor difference of IC and IB. The line currents are defined as: I1 = IA IC, I2 = IB IA and I3 = IC IB. Here I1, I2, I3 are the line current (ILine) and IA, IB, IC are the phase current (IPhase) of Wye connection. IA, IB, IC are the reverse phase current of IA, IB, IC. The two phasors IA and IC are 600 apart. I1 = ILine = IA IC = [IPhase (-IPhase)] cos(600/2) = 2 IPhase cos300 = √3 IPhase The two phasors IB and IA are 600 apart. I2 = ILine = IB IA = √3 IPhase The two phasors IC and IB are 600 apart. I3 = ILine = IC IB = √3 IPhase  Ã…“ I1 = I2 = I3 = ILine = line current = √3 IPhase However, the line current of a delta connection is higher than the phase current by a factor of the square root of 3 (1.732). Power Total power P = 3 Power in each phase = 3 VPhase IPhase cosÃŽ ¦ = 3 VLine- (ILine/√3) cosÃŽ ¦ [for delta connection] = √3 VLine ILine cosÃŽ ¦ Where VLine, ILine and cosÃŽ ¦ are the line voltage, the line current and power factor of a delta connected supply. 5.5 Relationship between line and phase quantities 5.5.1 Star connection On a Wye system, the line current is equal to the phase current i.e. ILine = IPhase Where ILine and IPhase are the line current and phase current of Wye connection. In a wye connected system, the line voltage is higher than the phase voltage by a factor of the square root of 3 (1.732). Two formulas used to compute the voltage in a wye connected system are: VLine = √3 VPhase = 1.732 VPhase  Ã…“ VPhase = VLine / 1.732 Where VLine and VPhase are the line voltage and phase voltage of Wye connection. 5.5.2 Delta connection In the delta connection, line voltage and phase voltage are the same. VLine = Vphase Where VLine and VPhase are the line voltage and phase voltage of delta connection. Formulas for determining the current in a delta connection are: Where ILine and IPhase are the line current and phase current of delta connection. 5.6 Power measurement by two watt meters method In two wattmeters method, current coils of the two wattmeters are connected in any two terminals of Wye system, as shown in Figure 13. The algebraic sum of two wattmeters gives the total power consumed whether the load is balanced or not i.e. Total power = W1 + W2 Figure 13:- Wye connected load Figure 14:- Phasor Diagram The power factor angle of load impedance being ÃŽ ¦ lag. The currents will lag behind their respective phase voltages by ÃŽ ¦ as shown in Fig. 14. Current through current coil of W1 = IA. Potential difference across potential coil of W1, V1-2 = VA VB. The phase angle between V1-2 and IA is (300 + ÃŽ ¦).  Ã…“ W1 = V1-2 IA cos(300 + ÃŽ ¦) Current through current coil of W2 = IB. Potential difference across potential coil of W2, V2-3 = VB-VC. The phase angle between V2-3 and IB is (300 ÃŽ ¦).  Ã…“ W2 = V2-3 IB cos(300 ÃŽ ¦) Here load is balanced, V1-2 = V2-3 = VLine = line voltage and IA = IB = ILine = line current.  Ã…“ W1 = VLine ILine cos(300 + ÃŽ ¦)  Ã…“ W2 = VLine ILine cos(300 ÃŽ ¦)  Ã…“ W1 + W2 = VLine ILine [cos(300 + ÃŽ ¦) + cos(300 ÃŽ ¦)] = VLine ILine(2cos300cosÃŽ ¦) = √3VLine ILine cosÃŽ ¦  Ã…“ W2 W1 = VLine ILine [cos(300 ÃŽ ¦) cos(300 + ÃŽ ¦)] = VLine ILine(2sin300sinÃŽ ¦) = VLine ILine sinÃŽ ¦ tanÃŽ ¦ = [√3 (W2 W1)] / (W1 + W2) Thus from the two wattmeter method, we can find ÃŽ ¦. PROBLEM 1. Three coils, each having a resistance of 20- and an inductive reactance of 15-, are connected in star to a 400V, 3-phase, 50Hz supply. Calculate (i) the line current (ii) power factor and (iii) power supplied. Solution:- VPhase = VLine / 1.732 = 400/1.732 = 231V ZPhase = √(202 + 152) = 25- (i) IPhase = VPhase/ ZPhase = 231/25 = 9.24A = ILine (ii) Power factor = cosÃŽ ¦ = RPhase/ ZPhase = 20/25 = 0.8 lag (iii) P = √3VLine ILine cosÃŽ ¦ = √3 400 9.24 0.8 = 5121W 2. A balanced star-connected load of impedance (6 + j8)- per phase is connected to a 3-phase, 230V, 50Hz supply. Find the line current and power absorbed by each phase. Solution:- ZPhase = √(62 + 82) = 10- VPhase = VLine / 1.732 = 230/1.732 = 133V Power factor = cosÃŽ ¦ = RPhase/ ZPhase = 6/10 = 0.6 lag IPhase = VPhase/ ZPhase = 133/10 = 13.3A = ILine P = √3VLine ILine cosÃŽ ¦ = √3 230 13.3 0.6 = 1061W 3. Three similar coils, connected in star, take a total power of 1.5kW at a power factor of 0.2 lagging from 3-phase, 400V, 50Hz supply. Calculate the resistance and inductance of each coil. Solution:- VPhase = VLine / 1.732 = 400/1.732 = 231V P = √3VLine ILine cosÃŽ ¦  Ã…“ ILine = P / (√3VLine cosÃŽ ¦) = 1500 / (1.732 400 0.2) = 10.83A = IPhase ZPhase= VPhase/ IPhase = 231 / 10.83 = 21.33- RPhase = ZPhase cosÃŽ ¦ = 21.33 0.2 = 4.27- XPhase = √(21.332 4.272) = 20.9- LPhase = XPhase/ 2Ï€f =20.9 / (2Ï€ 50) = 0.0665H 4. The load to a 3-phase supply comprises three similar coils connected in star. The line currents are 25A and kVA and kW inputs are 20 and 11 respectively. Find (i) the phase and line voltages (ii) the kVAR input and (iii) resistance and reactance of each coil. Solution:- VPhase = Apparent power / (3 IPh) = (20-103) / (3 25) = 267V VLine= √3 VPhase=1.732-267 = 462V Input kVAR = √ (kVA2 kW2) = √ (202 112) = 16.7kVAR Power factor = cosÃŽ ¦ = kW/kVA = 11/20 ZPhase= VPhase/ IPhase = 267 / 25 = 10.68- RPhase = ZPhase cosÃŽ ¦ = 10.68 11/20 = 5.87- XPhase = √(10.682 5.872) = 8.92- 5. A balanced 3-phase, delta-connected load has per phase impedance of (25+j40)-. If 400V, 3-phase supply is connected to this load, find (i) phase current (ii) line current (iii) power supplied to the load. Solution:- ZPhase = √(252 + 402) = 47.17- IPhase= VPhase/ ZPhase = 400 / 47.17 = 8.48- ILine= √3 IPhase=1.732-8.48 = 14.7A Power factor = cosÃŽ ¦ = RPhase/ ZPhase = 25/47.17 = 0.53 lag P = √3VLine ILine cosÃŽ ¦ = √3 400 14.7- 0.53 = 5397.76W 6. A balanced 3-phase load consists of three coils, each of resistance 6-, and inductive reactance of 8-. Determine the line current and power absorbed when the coils are delta-connected across 400V, 3-phase supply. Solution:- ZPhase = √(62 + 82) = 10- cosÃŽ ¦ = RPhase/ ZPhase = 6/10 = 0.6 lag VPhase = VLine = 400V IPhase= VPhase/ ZPhase = 400 / 10 = 40A ILine= √3 IPhase=1.732-40 = 69.28A P = √3VLine ILine cosÃŽ ¦ = √3 400 69.28 0.6 = 28799W 7. Two-wattmeter method is used to measure the power absorbed by a 3-phase induction motor. The wattmeter readings are 12.5kW and -4.8kW. Find (i) the power absorbed by the machine (ii) load power factor (iii) reactive power taken by the load. Solution:- W2 = 12.5kW ; W1 = -4.8kW Power absorbed = W2 + W1 = 12.5 + (-4.8) = 7.7kW tanÃŽ ¦ = [√3 (W2 W1)] / (W1 + W2) = (12.5+4.8) / 7.7 = 3.89 ÃŽ ¦ = tan-13.89 = 75.60 Power factor = cosÃŽ ¦ = cos75.60 = 0.2487lag Reactive power = √3 (W2-W1) = √3 (12.5 + 4.8) = 29.96kVAR P O I N T S TO REMEMBER 1. The voltages of a three-phase system are 120 ° out of phase with each other. 2. The two types of three-phase connections are wye and delta. 3. Wye connections are characterized by the fact that one terminal of each device is connected together. 4. In a wye connection, the phase voltage is less than the line voltage by a factor of 1.732. The phase current and line current are the same. 5. In a delta connection, the phase voltage is the same as the line voltage. The phase current is less than the line current by a factor of 1.732. IMPORTANT FORMULAE 1. On a wye system, the relation between line and phase current is: ILine = IPhase 2. On a wye system, the line voltages are defined as: V1-2 = VA VB, V2-3 = VB-VC, and V3-1 = VC-VA. 3. In the delta connection, the relation between line and phase voltage is: VLine = Vphase 4. In the delta connection, the line currents are defined as: I1 = IA IC, I2 = IB IA and I3 = IC IB 5. On a wye system, the relation between line and phase voltage is: VPhase = VLine / 1.732 6. In the delta connection, the relation between line and phase current is: OBJECTIVE QUESTIONS 1. In a two phase generator, the electrical displacement between the two phases or winding is: (a) 1200 (b) 900 (c) 1800 (d) none of these 2. The advantage of star-connected supply system is that: (a) line current is equal to phase current (b) two voltages can be used (c) phase sequence can be easily changed (d) it is a simple arranged 3. In a balanced star-connected system, line voltage are ahead of their respective phase voltages. (a) 300 (b) 600 (c) 1200 (d) none of these 4. In a star connected system, the relationship between the line voltage VL and phase voltage VPh is: (a) VL = VPh (b) VL = VPh / √3 (c) VL = √3VPh (d) none of these 5. The algebraic sum of instantaneous phase voltages in a three-phase circuit is equal to: (a) zero (b) line voltage (c) phase voltage (d) none of these 6. If one line conductor of a 3-phase line is cut, the load is then supplied by: (a) single phase voltage (b) two phase voltage (c) three phase voltage (d) none of these 7. The resistance between any two terminals of a balanced star-connected load is 12-. The resistance of each phase is: (a) 12- (b) 24- (c) 6- (d) none of these 8. A 3-phase load is balanced if all the three phases have the same (a) impedance (b) power factor (c) impedance and power factor (d) none of these REVIEW QUESTIONS 1. How many degrees out of phase with each other are the voltages of a three-phase system? 2. What are the two main types of three-phase connections? 3. A wye-connected load has a voltage of 480 V applied to it. What is the voltage drop across each phase? 4. A wye-connected load has a phase current of 25 A. How much current is flowing through the lines supplying the load? 5. A delta connection has a voltage of 560 V connected to it. How much voltage is dropped across each phase? 6. A delta connection has 30 A of current flowing through each phase winding. How much current is flowing through each of the lines supplying power to the load? 7. A three-phase resistive load has a phase voltage of 240 V and a phase current of 18 A. What is the power of this load? 8. If the load in question 7 is connected in a wye, what would be the line voltage and line current supplying the load? 9. An alternator with a line voltage of 2400 V supplies a delta-connected load. The line current supplied to the load is 40 A. Assume the load is a balanced three-phase load, what is the impedance of each phase? 10. If the load is pure resistive, what is the power of the circuit in question 9? PRACTICE PROBLEMS 1. Three similar coils are star connected to a 3-phase, 400V, and 50Hz supply. If the inductance and resistance of each coil are 38.2mH and 16- respectively, determine (i) line current (ii) power factor (iii) power consumed. 2. Three 50- resistors are connected in star across 400V, 3-phase supply. (i) Find phase current, line current and power taken from the main. (ii) What would be the above value if one of the resistors were disconnected? 3. Calculate the active and reactive components of current in each phase of a star-connected 10,000 volts, 3-phase generator supplying 5,000kW at a lagging power factor 0.8. Find the new output if the current is maintained at the same value but the power factor is raised to 0.9 lagging. 4. Three 20 µF capacitors are star-connected across 420V, 50Hz, 3-phase, three wire supplies. (i) Calculate the current in each line. (ii) If one of the capacitors is short-circuited, calculate the line currents. (iii) If one of the capacitors is open-circuited, calculate the line currents and potential difference across each of the other two capacitors. 5. If the phase voltage of a 3-phase star connected alternator be 231V, what will be the line voltages (i) when the phases are correctly connected (ii) when the connections of one of the phases are reversed? 6. Calculate the phase and line currents in a balanced delta connected load taking 75kw at a power factor 0.8 from a 3-phase 440V supply. 7. Three identical resistances, each of 18-, are connected in delta across 400V, 3-phase supply. What value of resistance in each leg of balanced star connected load would take the same line current? 8. Three similar resistors are connected in star across a 415V, 3-phase supply. The line current is 10A. Calculate (i) the value of each resistance (ii) the line voltage required to give the same line current if the resistors were delta-connected. 9. Two wattmeters are used to measure power in a 3-phase balanced load. The wattmeter readings are 8.2kW and 7.2kW. Calculate (i) total power (ii) power factor and (iii) total reactive power. 10. A balanced 3-phase load takes 10kW at a power factor of 0.9 lagging. Calculate the readings on each of the two wattmeters connected to read the input power. 11. Three identical coils, each having a resistance of 20- and a reactance of 20- are connected in (i) star (ii) delta across 440v, 3-phase lines. Calculate for each method of connection the line current and readings on each of the two wattmeters connected to measure the power.

Gay and Lesbian Liberation Essay example -- Homosexuality

Generations ago, the United States was a country of the male wardrobe. Today's movements for the rights of lesbian, gay, bisexual and transgender (LGBT) community are leveraging the existence of more globalized and open systems. Besides, the promotion of the lesbian, gay, bisexual and transgender population have been acknowledged through smart partnerships within conventional, political and economic scene, while the males and heterosexuals are still pervasive. This paper will focus on the oppression of the LGBT population denying their human rights. However, at the same token showing their struggle to ultimately be accepted in society to some degree will be addressed as well. The gay and Lesbian liberation movement presented the later involvement of the bisexual and transgendered groups. This was considered a significant social force that introduced the reality of this culture during the last several decades. Fortunately, the individuals who had been profoundly isolated from traditional sources gained some support. The sources entail family members, friends and communities. Unfortunately, the LGBT population encountered harsh treatment by their oppressors, as many people were against their way of life. Many of these LGBT individuals were terrorized by other individuals, which also resulted in death. A controversial issues regarding the LGBT population, particularly gays and lesbian has been their privilege to be married. Homosexuality was decriminalized in most states in the Americas. However, homosexuality remains criminally punishable in Guyana, to Belize and part of Caribbean islands (in Jamaica, in Dominica in Saint Lucia, in Barbados, in St. Vi ncent and the Grenadines in Grenada, in Trinidad and Tobago to Antigua a... ...tudies from 1978–2000. Scandinavian Journal of Psychology vol. 43 pp. 335–351 Bawer, Bruce (1993) A Place at the Table: The Gay Individual in American Society. New York: Touchstone Books Boswell, J. (1994). Same-sex unions in pre-modern Europe. New York: Villard. Dexter, P. (2003). Countering the counterfeit: A case for traditional marriage. . Miller, A. V. (2000). â€Å"Our own voices: a directory of lesbian and gay periodicals, 1890s-2000s†, Canadian Gay and Lesbian Archives: http:/ / www.clga.ca One, Inc. v. Olesen, 355 U.S. 371 Rasmussen, M. and Kenway, J. (2004), â€Å"Queering the youthful cyberflaneur Globalizing identities, consuming queers: Issues in education and globalization†, Journal of Gay and Lesbian Issues in Education vol. 2 no. (1). pp. 47–63. Streitmatter, R. (1995). â€Å"Unspeakable: The rise of the gay and lesbian press in America†, Boston: Faber & Faber

The Grapes of Wrath

The movie â€Å"The Grapes of Wrath† I watched recently is the classic adaptation of John Steinbeck’s novel written and published in 1939 only one year before the movie was released. When the movie production was approved the director John Ford and his crew were able to accomplish a major task and finish the project without getting involved in the conflict with labor unions what was very important since many members of the crew belonged to them. The script which is based on Steinbeck’s book adopted a very specific dialect, the language spoken by the villagers living in rural Oklahoma so the authenticity impression was preserved. The movie presents the times of the Great Depression in the American Mid West and West coasts. This was a tough era of American history that got the historically accurate documentation and John Ford’s movie shows us perfectly the experience of many people those days – farmers and their families, drifters and strikers. For many Americans who watched the movie in the theaters in the big cities it was not easy to find out how hard and stressful it was for the people living in the country to be forced out of the lands they were working on. Here the director introduces us to the family of poor farmers from Oklahoma Dust Bowl and their struggling during the tough times of the economic crisis. While watching the movie I observed the contrast between the representatives of the lower class and the wealthy landowners who take an advantage of the unfortunate circumstances. From the events presented I’ve learned that the economic crisis and the advancing technology which was used to farm the land were responsible for drastic and extreme changes which were imposed on thousands of people against their own will. All the farmers’ families which for many generations made the living out of their farms in the short period of time received the notifications demanding to move out and abandon the properties which they were residing at. Obviously it was not their choice and the decision was made on their behalf without giving them the alternative opportunities. The main characters in the movie are the members of Joad’s family who are not only dealing with the problems inside of their family but also with the basis of their existence which is endangered. Those poor people understand very well that when fighting all the odds the only solution is to stay together and rely on each other and when facing the unpleasant reality they decide to remain as a family while entering the new chapter of their lives. In order to find a solution to their problems Joad’s family decides to embark on the journey to California. They have high hopes and strong beliefs that if they will get there, their situation will change for better, they will be able to find jobs and as a result will settle in their new home. On the way to their destination they experience many unexpected events which they have to deal with as they appear. They loose two members of their family and while staying at the camps for the similar migrants they learn about the sad reality which was awaiting them. It appeared that all the information they have been provided before they left Oklahoma were not exactly true and the land owners were trying to earn profits by using the unfair labor practices. Being fully aware of the desperate conditions the owners of the large farms offer low wages jobs and exploit all those hired to work on their lands. The new comers were not welcomed and treated with very suspicious manners. The local residents who were afraid of losing their own source of income were acting very unfriendly and openly demonstrated their opposition. People like Joad’s family while forced to stay in the camps and awaiting the potential employment were treated by local authorities like trouble makers. All those who tried to stand up for their rights were considered difficult and wrongly accused of trouble causing. Rich and wealthy class representatives were simply misusing the power of the local authorities in order to protect their own interest and increase the potential profits. Whoever was brave enough to disagree with those kinds of methods was considered as a threat and forced to leave or had to face the mistreatment and abuse. While majority of migrating families were subjected to the hardship and constant struggling, the smaller number of this group were lucky enough to find the temporary accommodations provided by the Federal authorities as a form of social assistance during the economic depression. Once they were able to reach such facilities they were offered decent and reasonable housing with the basic sanitary installations. The life in those places was organized by the rules which campers could set up so all the residents were treated with respect and there was no tolerance for any form of abuse. However some of the local land owners did not like the situation and were trying to sabotage the camp. Quite often those government managed facilities were subjected to the organized attempts of purposely initiated disturbances which will give the arguments to the local authorities to act while â€Å"preserving law and order†. It should be noticed that even tough it was not the director’s intention the conditions showed and the form of management in those camps succeeded in building up the reputation of the Government and helped in recognizing efforts directed towards continuous improvement during the crisis. The democratic methods which were giving all the residents the right to protect their status were successfully maintaining the stabilized life while staying within the borders of the camp. Even the police authorities were not allowed to enter without the warrant. Many residents quickly realized how fragile and valuable it is for them to continue preventing the established order in the camp so they could organize the system of the comities which were overseeing all the key aspects of every day living conditions. Under those circumstances surviving the difficult times until the employment opportunities appear was much more acceptable and helped to build the confidence in the better future. The movie definitely proves that people while facing major difficulties have much higher chances of prevailing if they help each other and stay together instead of resisting and fighting those problems just on their own. Even the strongest and most determined attempts to resist and fighting the reality by one single person will not last long and at the end will not be considered successful. The Joad’s family proved to themselves that together they can manage to survive and use the difficulties which they went through as the learning experience. By using the combination of wisdom and wise assumptions together with good will and honest intentions they were able to help each other in the process of the decisions making. As we all now go through the inconveniences and carry the burden of struggling economy, while watching the movie we can relate in some certain way to the presented story. We are fully aware that unfortunately seventy years later there are many Americans who do have to find answers to resolve their problems and react to the drastic changes which the economy imposed on them. As Casy, the former preacher perfectly summarizes his view when performing the funeral rites by saying: â€Å"I wouldn’t pray just for an old man that’s dead, ‘cause he’s all right. If I was to pray, I’d pray for folks that’s alive and don’t know which way to turn†. The fact that the American Academy of Motion Pictures nominated John Ford’s movie was one of the reasons why even the President of the USA Franklin D. Roosevelt was present there and gave the speech which consisted of comments regarding the movie unique character. For someone like me whose knowledge of the United States modern history is not so deep I found the movie as being very educating and presenting Americans’ lives during the Great Depression from a different perspective. Comparing to the recent production, the â€Å"Cinderella Man† I found quite few similarities and issues which were particularly important however the director John Ford is not trying to create a fairy tale but instead is exposing the dark side in which both the rules of economy and human nature compete with each other in the most important fight as the only possible way of surviving without loosing the human values such a dignity and pride. We can say that the main characters, their interactions and behavior even if originated from the natural instincts of human beings slowly but steady helped them to built up the resistance and eventually made them stronger than they were ever before. Just like Russell Crowe in the â€Å"Cinderella Man† was given a second chance and did not waste it but succeeded in his journey to the victory on the boxing ring, Joad’s family under the spiritual leadership of Tom’s mother proudly emerged out of all difficulties and remained faithful about awaiting them much better future. In my opinion the history of this country taught us a very valuable lesson which proved that even the Great Depression was not able to destroy the strong American spirit and discourage people from using their skills and joined efforts to help each other in the times of need. People need to believe that the tough times will end and they will be able to enjoy living again. Nothing motivates people in need to take an action and engage in many long term efforts better than having hope and faith that the better future is in their hands, that nothing else can help them better than their own determination.

Physiological Basis of Human Behavior Essay

Heredity is the passing of traits to offspring from its parents or ancestor. This is the process by which an offspring cell or organism acquires or becomes predisposed to the characteristics of its parent cell or organism. Through heredity, variations exhibited by individuals can accumulate and cause some species to evolve. The study of heredity in biology is called genetics, which includes the field of epigenetics. Both hereditary and hormones affect human behavior but in different ways. It is a combination of these two factors which results in â€Å"normal† behavior. Heredity and hormones work together to influence behavior. Heredity includes genetics, behavioral genetics, and evolutionary psychology; while hormones are produced by the endocrine system, in concert with the nervous system. Hormones switch on behaviors. Looking at the relationships between hormones and heredity provides a complete picture of the effects of heredity on human behavior. Genetics is the study of how traits are passed from one generation to the next through genes, which are found on chromosomes. They are a small part of DNA, and they direct particular traits of groups of traits. They are the essence of a human being, their make-up, and they are unmodifiable. They are carried by tiny threadlike bodies called chromosomes. They vary in size and shape and they come in pairs. Genes are responsible for the development of the nervous and endocrine systems; therefore, genes can influence the chance of a certain behavior occurring in a certain set of circumstances. They are indirect to behavior, while hormones are direct. Genes are most often considered in a physiological manner, as in whether one is predisposed to being tall or short, or being blonde haired or blue eyed. Behavior genetics are the other side of genetics, more specifically how genes can be considered to pass on psychological traits rather than physiological ones. Each species has a constant number of chromosomes that never varies, weight, height, skin color and many other traits are produced by the interaction of several genes. This process is called polygenic inheritance, and each of the genes contributes individually to the overall effect. As the chromosomes carries the genes, so does the bloodstream, hormones.