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

Television Violence :: essays research papers

Violence and Television In today’s society, television violence is shaping the way our children behave, making them prone to violence and abuse as they get older. Living in a world where the majority of our entertainment is television, it is very likely that we would become more immune to the physical and damaging acts of damaging force committed. Violence is all over our media but mostly on the TV. Parents should monitor what their children are watching closely, so that their behavior does not become more destructive at; even at the age of 5. History shows that some of the first violent acts were noticed in the 1950’s. They say that even back then a lot of television was filled with destructive acts. From talking to my grandparents television was not as much of a necessity as we believe it to be today. Being the society that has to be entertained around the clock, we just do not think about what is going into our children’s heads. It seems to affect children more, because their behavior patterns are still developing, and are very vulnerable, and we don’t want them to develop into what they see on TV. Children who watch television a lot of the time become less aware to the pain and suffering of other people around them. If they are always in front of the tube, and just sit there, and soak in all of that bad stuff they are seeing, they become not as aware as they should be. When terrible acts of violence happen in the children’s life that do watch a lot of TV, they are not as aroused by the acts going on in front of them. It is not as disturbing as it would be for a child who does not watch a lot of violent television. A study was shown that a child who had watched a violent television show such as The Power Rangers, rather than a nonviolent show like My Little Pony, were slower getting involved when they saw a younger child getting beat up or playing destructively. Instead of children taking action, or getting involved if they are old enough, they are more fearful of the things going on around them. If a child watches a lot of violence, when a violent or destructive situation came up they might be afraid to take action. They could be fearful that they would get hurt. Television Violence :: essays research papers Violence and Television In today’s society, television violence is shaping the way our children behave, making them prone to violence and abuse as they get older. Living in a world where the majority of our entertainment is television, it is very likely that we would become more immune to the physical and damaging acts of damaging force committed. Violence is all over our media but mostly on the TV. Parents should monitor what their children are watching closely, so that their behavior does not become more destructive at; even at the age of 5. History shows that some of the first violent acts were noticed in the 1950’s. They say that even back then a lot of television was filled with destructive acts. From talking to my grandparents television was not as much of a necessity as we believe it to be today. Being the society that has to be entertained around the clock, we just do not think about what is going into our children’s heads. It seems to affect children more, because their behavior patterns are still developing, and are very vulnerable, and we don’t want them to develop into what they see on TV. Children who watch television a lot of the time become less aware to the pain and suffering of other people around them. If they are always in front of the tube, and just sit there, and soak in all of that bad stuff they are seeing, they become not as aware as they should be. When terrible acts of violence happen in the children’s life that do watch a lot of TV, they are not as aroused by the acts going on in front of them. It is not as disturbing as it would be for a child who does not watch a lot of violent television. A study was shown that a child who had watched a violent television show such as The Power Rangers, rather than a nonviolent show like My Little Pony, were slower getting involved when they saw a younger child getting beat up or playing destructively. Instead of children taking action, or getting involved if they are old enough, they are more fearful of the things going on around them. If a child watches a lot of violence, when a violent or destructive situation came up they might be afraid to take action. They could be fearful that they would get hurt.

Basic Economic Question

Through market research and analysis it has been found out that there are several economic choices that every firm must face. This is in line with the core objective of the firm which basically entails producing standardized product to meet the customers demand and making profits for the firm.Research from surveys and focus groups have indicated that our investment bank needs to develop a product which best suits the clients establishing good consumer behaviour and meeting the impacts of government policies.No longer are banks interested in their own interest of making profits but what is of fundamental concern is whether the customer is satisfied. Our investment bank is not an exceptional one. We need to address the issues of developing a product that will make the bank outstand despite its rivals in the market. Development of product Economic choices imply that the forces of demand and supply need to be put into consideration when choosing a particular product for the firm.The cent ral question therefore that I considered in developing this product was the ease at which the customers will learn on the use and the reliability of the product when it’s introduced into the market. As an investment bank, a structured deposit is one of the products which is fundamental in the bank. This is because they allow customers to attain higher yields and also be able to take market risks upto a certain degree will be set by the bank.The bank through this product will be able to meet the needs and expectations of the consumers and the management team of the bank too. The product is introduced into the market just like any other products which have been introduced before. The customer is taken into an orientation so that he gets to learn that this product involves saving just like any other saving accounts. They will be informed that this service provides more services beyond the traditional savings such as mutual funds.Reasons for choosing this product One of the reaso ns why a structured deposit would be important in our investment bank is that it provides cash to the customer ant time he needs it. The customer does not need to wait upto a certain maturity date or pay a penalty for withdrawal of the money in his account. Another important issue which is fundamental is that through this product the bank can get more profits by soliciting so much money at a specific time so that it can invest at a go and get maximum returns.The customer will also benefit in that incase he requires some loan, then he will be granted since the pool from the other investors in this type of fund will have contributed. Another major reason that is important is that a structured deposit is a very good method of investment of ones money. An individual who has an account with a structured fund will find it easier in the future to get his money in a lumpsum amount. This therefore is a good investment solution to an individual. ConclusionFor our bank to achieve the demands o f the product there is need for awareness to be created to customers so that they can be in apposition to get the products and benefit from it. Reference: California Institute of Technology (2008, February 16). Insights on Economic Choices and challenges facing banks, an article pp. 35-56. Products of investment banks Retrieved on 8th April 2009 from http://www. citibank. ro/romania/corporate. Sample memo for business executives retrieved on 8th April 2009 from http//www. purdue. edu/owl/resource

The Hidden Secrets Of Our Earth - 2024 Words

The world is a majestical place, filled with the unheard of, the unknown, the lost, and the very small percentage that is seen as it wants to be. There are many things the human mind cannot comprehend at this point in time like they used to. Humans are too focused on the large picture to see the smaller effects, too impatient to really try and look for the hidden secrets of our Earth. If they knew what I did, they would understand why all of this happens. In a world filled with unparalleled powerful beings, writhing in gods in whom have pantheons that cover the world; there is no surprise any more to why phenomenon happens. I have spoken too much already though, I should tell you my story before I ruin my own fate and get crushed by one†¦show more content†¦I was a perfect embodiment of a child; maybe that is why I am here. It was soon after my fourteenth birthday when everything had changed. People around me were becoming more quite, less friendly; I had thought I had done something wrong. It wasn’t unusual for people to go unnoticed at my school, though. It was a very tight knit community and when I entered, I was definitely viewed as an outcast and unwelcomed. Almost what had been happening those few days after my birthday. I had to go under an entirely new alias whenever I was at school, always finding ways to alter my appearance so that they wouldn’t know they were bothering a celebrity child. It was one day, two days ago, right before the end of my last day of middle school. That was the day that most everything had changed. My parents were gone on their usual spit, half-way across the world shooting a movie that they said I could not be a part of, so I was stuck at our penthouse in Atlanta, Georgia with my personal butler and bodyguard, Marcel Arnold. Marcel wasn’t bad at all, no, he was a strikingly handsome, brunet young man that was no more than the age of twenty (He would never tell me his age), and he was extremely qui et and easy to talk to. Though this scenario was pretty common, it had changed when I received a Skype  ©call from my parents. It was the usual spiel of â€Å"Be careful, honey!†, â€Å"We love you so much!†, â€Å"See you soon, babe.†, but it felt as if there was an undertone, and apparently there was.