Finance Manager in Home and Personal Care Products Essay - 37

Finance Manager in Home and Personal Care Products - Essay Example The researcher states that while investigating the financial activities and the operations of the newly acquired company, she found out that the company had issues with appropriate accounting practices that were geared towards inflating sales and earnings of workers. Most conspicuously, the company was involved in the very expensive commercial aimed at marketing and creating awareness of the company. The expenses were exorbitantly high amounting to millions of dollar. Secondly, the company had a number of assets without a clear list of their nature. However, on the deeper examination of these assets, she found out that, the controversial and expensive commercials were among the assets. To deal with the issues, Lisa as a finance manager had to follow the right procedure in solving an ethical issue as presented by IMA. To do so, she went on to Anderson, who was her immediate supervisor in the newly acquired company. Before doing the visit, she prepared herself adequately well with the plan of valuations of the parent company. The plan was supposed to inform Anderson on the requirement of the mergers and acquired the company by the mother company’s policy on a valuation of her assets and financial control at large. However, to her surprise, Anderson did not pay much attention to the plan that was fundamental in harmonizing the organization with the operation of the parent company. The issue at hand juts needed the intervention of the chief financial officer of the newly acquired company Anderson, to be resolved by providing clear information on the assets that the company held. This is because they are the major stakeholders of the stalemate that has been identified by Lisa. He rubbished the need of Lisa to gain more understanding of the other accrued liabilities. Anderson turns her down and insists that it is none of her business. Anderson is quick to assert that Lisa does not understand and comprehend the business, and the journal entries are standard pra ctice. He claims that if Lisa understood the nature of uncertainty of the prestige fragrance business, she would be able to go in line with what the company wants.

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

The Civil War of Rome :: Ancient Rome Roman History

The Civil War of Rome The Civil War in the eyes of most people is not glorious, but rather one of the worst crimes you could possibly commit when the state is all-important. Only under the most extreme circumstances should one be allowed to (in the eyes of the people that is) begin a Civil War with just cause. Caesar took this into consideration, but too many things were going wrong in Rome for him not to begin the war. The first of many problems was the collapse of the Triumvirate. The Triumvirate was one of the main parts of the government of Rome, with which there were three leaders, which at the time were Caesar, Pompey, and Crassus. This was never truly working all that great, but held itself together by the marriage of Caesar's daughter Julia, to Pompey, and the friendship Caesar and Crassus shared. But, all this came to an end when Crassus was killed in a battle against a Parthian army. Then, not too long afterwards, Julia was murdered by someone who had broken into her home. This, destroyed the bond between Caesar and Pompey, and made them drift apart. Caesar seeing all this taking place, attempted to restore the bond by proposing to Pompey's only daughter, but was not allowed to by Pompey. To only make matters worse, Rome was slowly slipping into total anarchy. The government was becoming corrupt with bribery. The elections were being stopped, and there wasn't a consul elected in 53 or 52 B.C. Most authority was lost, the streets became rioted, and unsafe. During this time, Pompey tried to annul the Law of Ten Tribunes without notifying Caesar. If this happen, it would of removed a lot of Caesar's power. Caesar saw what he was trying to do, and stopped it before this action took place, and now knew for sure that Pompey was no longer his ally, but instead an enemy. Pompey tried another devious act against Caesar, which this time worked. He had the senate pass a law that made Pompey and Caesar both give up troops, and send them to the East, where they were supposedly needed against the Parthians. This seemed fair, but it made Caesar lose two legions, one that was lent to him by Pompey in the Gallic Wars, and one of his own. Once they were positioned there, Pompey decided they were no longer needed, and sent them to Capua (a city in Rome) under his command.

Ship Breaking Industries of Bangladesh

Shipbreaking Activities in Bangladesh and collision of Marine Biodiversity Prabal Barua Associate Program Officer YPSA The marine environment of the coastal water is vital to mankind on a global as well as on local basis concerning energy. Man is becoming a dominant part of the ecosystem in many regions, due to his various uses of the marine environment. So the health of marine ecosystem is an important factor in man own existence.The Bay of Bengal which is a potential bode of marine life as well as for it’s vast coastal communities is now continually polluted by different types of pollutant through influx of land base and other sources and put an alarming signal of awareness about pollution in the sea. The coastal areas of Chittagong Support a complex trophic organization sustain a high biodiversity including some endemic species and are highly susceptible to interference from activities. Coastal ecosystem makes a sustainable livelihood particularly to coastal fishing communi ties.Ship breaking yards along the coast of Chittagong (Faujdarhat to Kumira) has become a paramount importance in the macro-and micro-economic context of poverty- stricken Bangladesh. Shipbreaking activities present both challenge and opportunity for coastal zone management in holistic manner. The history of ship breaking is as nearly old as shipbuilding. As we know that a ship is relatively a large vessel capable of operating in the deep ocean. The term ‘vessels’ applies to the vessels of over 5000 tons and that can navigate in open seas.In Bangladesh ship breaking is popularly known as ‘Beaching’. Ship breaking started as a business in Bangladesh in 1972. Prior to that, 2/3 ships were scrapped during Pakistan period. It started automatically when a 20000 DWT vessel was drive ashore by the devastating tidal bore of 1965. That was the first ship scrapped on the 2 Chittagong sea beach. At present, ship breaking is conducted in an area of about 10 km by 32 o ut of 110 ship Breaking yards from Bhatiari, Sonaichhari, kumira under the Sitakunda upazilla of Chittagong.The Department of Environment (DoE) has categorized the Ship Breaking Industry (SBI) as ‘Red’ in 1995(EIA guidelines for the Industries, 1997). The Environmental Impact assessment (EIA) is not conducted before the establishment of SBI. As there is no monitoring cell, the Shipyard owners are operating their business overwhelmingly as well as indiscriminately. They are less conscious about hazards, toxicity and environmental pollution whereas more conscious about their benefit. Wastes of the scrapped ships are discharged directly into its adjacent areas which are ultimately draining into the Bay of Bengal.These wastes especially oil and oily substances, PCBs, TBTs, PAHs etc. and different types of trace and heavy metals (Cd, Pb, and Hg) are being accumulated into the marine biota. As a result, marine fisheries diversity of the Chittagong coast that supports highly d iversified marine water fishes, mollusks and benthic organisms etc. is at the stake right at this moment. Moreover the Coastal inhabitants/fisher folks lead not only their livelihoods but also solely depend on the coastal resources for their protein source.The CPUE (Catch Per Unit effort) has drastically been reduced to more than half comparatively of a few decades ago. As a consequence, the coastal fisher folks are at the stake of their existence. They are either leaving their hereditary profession or migrating to other places and becoming ‘environmental refugees’. That is why their socio-economic status is below the poverty level. There are few studies was done to find out the linkage between Ship breaking activity (SBA) and the marine pollution, impact on fisheries biodiversity and livelihoods of the fishermen community.In those researches, investigators considered Bhatiary to Kumira as affected area and Sandwip Island as control area from the shipbreaking activity. The eastern side of Sandwip has been considered as control site because these are diagonally opposite and off the SBYs and the water and soil qualities are apparently free from pollutants as revealed from the earlier studies. From the previous analysis we found that trace metals concentration in sediments at shipbreaking area are so much higher than recommendation by GESAMP (Joint Group of Experts on the Scientific Aspects of Marine Pollution).But the researchers found that Sandwip which is significantly very lower than that of sediments at affected area. The values of Lead (Pb), Cadmium(Cd) and Mercury(Hg) found six and half; eight and half and ninety four times higher than that of certified values respectively. These could be attributed to the combined effects of oil and oil spillage, petroleum hydrocarbons from ships, tankers, mechanized boats etc. During the investigation all the researcher found water qualities such as Hydrogen Ion H Concentration (p ), Dissolve Oxygen (D. O), Biochemical Oxygen Demand (B. O. D), Chemical Oxygen Demand (C. O.D), Total Suspended Solid (TSS), Total Dissolved Solid (TDS), Oil and Ammonia (NH3) were concentrated as a higher value in affected area than control area according to the standard value of water quality for the coastal water of Bangladesh (EQS, 1991). All the parameters observed such a higher that they exceeded the value of EQS. But the water parameters in Sandwip channel were optimum and near to the value of EQS standard. Water qualities in affected area exceeded the EQS standard which reveals that the water body of the adjacent area of ship breaking yards is not suitable for the existence of flora and fauna.The higher concentration was due to the discharge of various refuse oils and oily substances, dyes, chemicals, iron pieces, various types of metal rusts, solids, dyes, erosion of soil dust etc. from the ship breaking yards. Pollutants are also discharged from the Sitakunda industrial area into the run-off open t o the Bay. Impact of Marine Biodiversity: Biodiversity, which is sort for biological diversity, is the term used to describe the whole variety of life on earth. In popular usage, the word biodiversity is often used to describe all the species living in a particular area.Biodiversity can be summarized as â€Å"Life on earth. † It is defined as â€Å" The varieties of life on earth at all its levels, from genes to ecosystem, and the ecological and evolutionary processes that sustain it. † The total biodiversity of an area can be broken down into two hierarchical components: the number of functional types of organisms (animals and plants) or ecosystems (forest, prairie, tundra and marine intertidal) and the number of functionally equivalent organisms within each functional type.There are three types of aspects to biodiversity: species diversity, genetic diversity and ecosystem diversity. All three interact and change over time and from place to place. Phytoplankton is the primary food producers of the aquatic habitat and plays an important role in the food chain. Phytoplankton is the best index of the biological productivity. Analysis of phytoplankton showed that during monsoon, Aanabaena , Clostratrum(10. 98%) and Coscinodiscus(21. 97%), Euglena (9. 89%) and Zygnema (30. 76%) and during post monsoon Coscinodiscus (97. 5%) and Euglena (2. 5%) dominated in the affected site.In the control site of Sandwip, these were dominated as 30. 41%, 19. 46%; 17. 03%; 9. 73%; 23. 35% in the monsoon and 94. 73% and 5. 26% respectively in the post monsoon. Throughout the study period the abundance of phytoplankton at affected site was 91 cells/ l in monsoon season and 80 cells/ l in post monsoon season and in control site it was 411 cells/l in monsoon season and 190 cells / l in post monsoon season. Drifting small floating animals, in the water body are collectively known as zooplankton on which the whole aquatic life depends directly or indirectly.As zooplankton i s very sensitive to optimum condition, so the coastal pollution due to ship breaking activities may have profound affects on its survival and occurrence. Analysis of zooplankton showed that Calanoida, Cyclopedia, Sagitta, Lucifer etc in the monsoon and Calanoida; Acetes shrimp; Lucifer and Zoea in the post monsoon were dominated in the affected site as revealed during zooplankton analysis whereas in the control site the dominant zooplankton were found as Calanoida, Cylpclpedia, Sagitta and Zoea during the post monsoon and Calanoida; Acetes shrimp, Lucifer, Cladocera and Zoea in the post monsoon respectively.The bottom living organisms –the benthos play an important role in the food chain especially in the inter tidal zone and it is also well recognized that the richest fisheries of the world are closely related to the benthic communities. Among the macro benthos, Amphipods, Polychaetes, Nemertina and Fish egg in the monsoon and Nemertina, Cladocera, Cyclopoida and Calanoida w ere found to be dominated at the affected site.But at the control site, Amphipod, Polychaete, Nemertina, Fish egg during the monsoon and Cladocera, Nemertina , Calanoida and Polychaete in the post monsoon were dominated So, the abundance of macro benthos in affected site was 118. 46 ind. / m3 in monsoon season and 4186. 74 ind. / m3 in post monsoon season, while in in the control site 368. 28 ind. / m3 in monsoon season and 14204. 41 ind. / m3 in post monsoon season.The fishery resources of the area seems to be affected by the ship breaking activities as revealed by increased fishing efforts, reduced species diversity, increased amount of trash fish. Horizontal expansion of the ship breaking yards has posed threat not only to the diversified coastal resources but also on the adjacent coastal inhabitants specially the fisher folks. The fishing hamlets of this study were found backward in the field of primary education and health that are the basic needs for them.Communication and dri nking water supply were observed satisfactory but the sanitation status was found to be very poor. Participatory Rural Appraisal (PRA) study showed that about 90% of them were local and full time fishermen and 10% were migratory of different districts including Bhola, Barisal, Mymensingh and others. The fisher folks are dissatisfied with different NGO activities working in this area. No government aid was found to be available for the welfare of fishermen communities.The catch has declined in the tune of at least 50 to 60% of what was two decades ago. This incident has got serious implication in the context of survival of such a disgraced community. It was found that about 70% of the fishermen had either nets or boats or both of them. They use both mechanized and non-mechanized boats and some traditional fishing crafts (Dinghi) for fishing. Among the fishing nets Set Beg Net (Behundi Jal) and Gill Net (Ilish Jal) were found to be widely used.Though the gears are available for fishin g, they can catch a very little amount in every effort. They uniquely reported that the fish catch had been reduced more than half of the previous time. Analysis of catch composition indicates that some commercially important fishes like Indian salmon (Polynemus indicus) commonly known as Lakhua, grouper (Epinephelus lanceolatus) known as bole coral; Long jew fish (Otolithoides brunneus) locally known as lombu fish; spanish mackerel (Cybium guttatum) known as maitta and butter fish (Psenes indicus) etc are in endangered position.Some other commercially important fishes like River shad (Tenualusa ilisha) Jwelled shad (Ilisha filigera) locally known as choikka; mud skipper (Gobies); mango fish (Polynemus paradysius) known as ‘Hriska Machh’; silver pomfret (Stromateus chinensis); bombay duck (Herpodon nehereus); mullet ( Mugil cephalus); Sea bass (Lates calcarifer); Anchovy(Coilia dussumeri; Coilia ramkorati; Setipinna taty) etc are reduced in catch. Many coastal fishermen are leaving their hereditary profession and moving around everyday as ‘environmental refugees’ from a state of unemployed and poverty to underemployment and grim poverty.Due to the deterioration of the water body, fishes are moving away from this area into the deep sea. But the poor fishermen with small fishing boats can hardly fish at deep seas the creditors and swindlers are taking this chances and rush to lend money and thus make them run into debt. The fishermen also reported that while they catch fish at sea they face piracy. They also face the muscle man, middle man and swindlers when they return with fewer amounts of fish.. These criminals snatch away the fishes forcefully. The fishermen are exploited by the dealers in dadon (earnest money).It is made obligatory that the middleman determines the price of fish in the season of fish. They are to sell fish to the lenders at a nominal or throwaway price. The middlemen indulge in maintaining miscreants to exercise th eir authority over them. Before the season of fish they borrow 4/5 thousand Taka from the dadonders (Buyers cum earnest money lenders) to repair the boat which lead them run into debt. Generally the months of MarchApril-May (Falgoon-Chaitra-Baishakh) are the â€Å"season of scarcity† as reported by the fishermen during PRA survey.At this time they require at least Tk. 4000/5000 to repair their boats, nets and for other incidental expenses. Most of the fishermen opined that Ship Breaking Activity (SBA) creates problems to them and the rest did not respond. Besides, 90% of the villagers of the vicinity were anxious about their existence in future. Their nets for the catching fish become stuck with oil and impurities. They suffer from respiratory difficulties, sonic booms, explosions, lightning, fumes, toxic chemicals and skin diseases.It is clear from the sociological study that coastal fish species diversity has been reduced due to the ship breaking activities. The fishermenâ €™s hereditary profession is now at the vulnerable position due to the Ship Breaking Activity (SBA); less access to credit etc. But the most interesting thing is that they are not conscious about their rights and deprivation. The increasing SBA is depleting the fishery resources which simultaneously decreasing catch per unit effort (CPUE). So this trend is provoking the fisher folks to change their livelihoods for what they never feel comfort and safe.Land grabbing by the yard owners also occurs sometime. Expansion of the yard shrinks the area of the fishing villages. They have to leave their space with a very nominal price. They have rights over only 200 feet of the seashore where they dry nets and anchor boats. Erosion of the village by sea wave action threatened the fishing village’s decade after decade. It started in sixties and nearly half a kilometer of the village has disappeared into the sea. Now at least 20,000 people engaged in catching fish are at the stake of their existence.Bangladesh is a party to the Convention on Biological Diversity (CBD), 1992 and the Biosafety Protocol. Bangladesh is also a party to the Convention on international Trade in Endangered Species of Wild fauna and Flora (CITES). Bangladesh has a Marine Fisheries Ordinance 1983 (ordinance No XXXV) and under this ordinance government may declare any area of Bangladesh fisheries waters and any adjacent or surrounding land to be a marine reserve (Part VII). So, if government will declare Sandwip channel and its adjacent area as a marine reserve area for fisheries biodiversity it will be.Besides, there are many laws, rules and polices in Bangladesh to conserve marine biodiversity such as The territorial water and maritime zones act (1974) and rules (1977), The forest act (1917), Environment Conservation Act (1995), Protection and Conservation of Fish rules (1985), national fish policy (1996), the water policy (1999), the environment policy (1992). If we conserve our diversi fy marine biodiversity it will be urgent need to establish environment friendly Shipbreaking activity in Bangladesh. Related article: Padma Bridge

Gender Roles in Things Fall Apart Essay

In Chinua Achebe’s novel Things Fall Apart, the Ibo society has a strict system of behavioral customs that are assigned by gender. These customs restrict the freedom of Ibo woman and help to reinforce generation after generation the notion that Ibo men are superior to women. In Achebe’s essay An Image of Africa: Racism in Conrad’s Heart of Darkness, he claims that Joseph Conrad’s Heart of Darkness, despite it’s insights, ought to be eradicated from literature as an appropriate piece of work on the argument that it is racist. Achebe focuses on gender roles and avoiding stereotypes to dismiss the racist attitude towards Africans in his novel by bringing the reader down to the level of his unbiased narration of a historical fiction novel. In Things Fall Apart, Achebe is cautious to avoid typical stereotypes describing white men. When the first white man is introduced, he is described as â€Å"†¦ not an Albino. He was quite different† his presence is foreign and not understood, but not dangerous. Achebe creates no account of the man being atrocious. In fact, throughout the entire story Achebe consistently attributes those same qualities to the white men later on in the novel; with the exception of The District Commissioner, who is described as â€Å"strict and unreasonable†. Achebe even puts white men in a good light; Mr. Brown constantly reminds the people of Umuofia that they should send their children to his church to learn to read and write so that other white men will not come to institute a new government and destroy their culture. By not negatively chronicling that white men are evil through narration and characterization, Achebe successfully lets loose the bounded stereotype of evil white men that are against African culture. Achebe affirms that in Ibo society, the condition of weakness is strongly associated with women. Therefore, a man being declared â€Å"woman-like† is an extreme insult. Unoka, Okonwko’s father, embodies the counter-values that stand in opposition to the rigid social ideal of the tribe. Okonkwo is ashamed when he learns that â€Å"agbala† was not only another name for woman, it could also mean a man who had taken no title when this insult is applied to his father. By stereotyping based on titles that are indigenous to Ibo culture rather than stereotypes that are demeaning to a race, Achebe is able to maintain an unbiased narrative in Things Fall Apart. Achebe does not completely demean woman in his story; it only seems as though that this is the case because he creates distractions to shy away from racism. In Ibo society, women marry to add to their father’s wealth, become subservient to their husbands, continue on their husband’s legacy through child birth and emanate the essence that women know the secret of life, since they are the source of it. This is evident when Uchendu ask â€Å"Can you tell me, Okonkwo, why it is that one of the commonest names we give to our children is Nneka, or ‘Mother is Supreme’? We all know that a man is the head of the family and his wives do his bidding. A child belongs to its father and his family and not to its mother and her family. A man belongs to his fatherland and not to his motherland. And yet we say Nneka–‘Mother is Supreme. ‘ Why is that? ‘ ‘A man belongs to his fatherland when things are good and life is sweet. But when there is sorrow and bitterness he finds refuge in his motherland. Your mother is there to protect you. She is buried there. And that is why we say that mother is supreme. ‘† Although the characters in the story never recognize the full capacities of women, they are described as emotionally strong. Achebe has an amazing gift for spotting bias that hides in plain sight; adapting against those anecdotes and historical asides, he creates something resembling a coherent story that necessitates dropping much that is important. Things Fall Apart possesses very deep meanings beyond that of avoiding stereotypes. Achebe purposely degrades women to not come across to the reader as racist, however his main priority is conveying lessons that are forgotten through literature. People who pick up a copy of Things Fall Apart will embark on journey that will let them see the subjugation of the African natives from their own perspective.