2:00 PM October 12, 2011
1. A consumer electric meter reads 211245 on July 30, 2010 and on August 30, 2010 it reads 211357. The reading indicates the total energy consumed by the consumer for the month in kilowatt hours. If energy rate is P 5.75 per kilowatt hour, determine the total electric bill of the consumer for the month.
2. A ball with a mass of 500 grams is thrown with a velocity of 12 m/s. Determine the impulse imparted to the ball. If the throwing was done in 0.3 second, what is the force imparted to the ball ?
3. An 10 g bullet was fired to a 790 g block of wood suspended on a long cord. If the bullet is embedded in the block, and the center of mass of the block-bullet is raise to a height of 12 cm, determine the initial speed of the bullet.
4. Mercury has a specific gravity of 13.6. What is the mass of 800 cm3 of mercury? What is the volume in cubic meter of 2000 g of mercury ?
5. A swimmer dives into a lake, and was submerge to a height of 8 m from the surface. What is the pressure experience by the swimmer at this depth of the water ?
6. A 500 g ball is drop from a height of 40 m. What is the momentum of the ball upon reaching the level ground? What is the maximum potential energy of the ball?
7. What is the impulse of the ball in problem # 6, 1.75 seconds after it was released ? What is the potential and kinetic energy of the ball at this level above the ground?
8. The volume of an irregular nugget of iron is to be determined but no weighing scale is available. The iron is place into a cylinder with water. If the diameter of the cylinder is 12 cm, and the water level rises to a height of 15 cm, what is the volume of the nugget in cubic meter ?
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CONSERVATION OF MOMENTUM
1. Momentum is the product of mass and velocity. It is a vector quantity and its direction is towards the direction of the velocity.
2. Impulse is the product of force and time during which the force acts. The impulse is equal to the change in
momentum.
3. Elastic bodies are bodies which return to their original shapes after a temporary deformation during collision.
4. Law of Conservation of momentum. If tow or more bodies interact, the momentum after the interactions is equal to the momentum before interaction. The total momentum of any system of bodies is unchanged by any interactions between the different members of the system.
5. Resilience is the ability of a body to undergo compression or rapid deformation without the development of permanent deformation.
6. Restitution is the vigor (energy) with which a body restores to its original shape and size after deformation.
7. Coefficient of restitution ( r ) the ratio of the velocity with which the two bodies separate after collision to the velocity of approach before collision.
r = 1 for perfectly elastic collision
r = 0 for perfectly inelastic collision
r = between 0 – 1 for semi-elastic collision.
COLLISION PHENOMENA
1. In a perfectly elastic collision, kinetic energy as well as momentum is conserved. The velocity of approach is equal to the velocity of separation in magnitude but opposite in direction. ( NOTE : No perfectly elastic collision for macroscopic bodies ). Ex. Collision between atomic nuclei, atoms, molecules and electrons
2. Inelastic collision is any collision for which the final kinetic energy is less than the initial kinetic energy.
3. Perfectly inelastic collision is any collision wherein the two colliding bodies sticks together indefinitely. The colliding bodies are permanently deformed and never separate, hence both have the same final velocity and the velocity of separation is zero since the two bodies sticks together.
4. Semi-elastic collision is a type of collision that is not perfectly elastic.
NOTE :
1. In every collision momentum is conserved, kinetic energy is conserved only in perfectly elastic collision.
LIQUIDS AT REST
Fluid is a term applied to liquids and gases, for they flow readily and do not resist shearing stress.
a) Liquid occupies a well-defined volume having no shape of its own, but takes the shape of the
vessel containing it. A liquid readily changes shape in response to forces. It offers large
resistance to efforts to change its volume. Most liquids are incompressible.
b) Gas completely occupies any enclosed volume in which it is placed. Gases are readily
compressible and it is relatively easy to reduce the volume of a gas to any desired size.
Density is the ratio of the mass of a substance to its volume.
Weight density is the ratio of the weight of an object to its volume.
Specific gravity is the ratio of the density of any substance to the density of water.
Pressure is the ratio of the force to the area where it is applied. The unit of pressure is Pascal, 1 Pascal = 1 N/ m2
In meteorology, the unit of pressure is millibar and 1 bar = 1.0 x 105 N/ m2
FLUIDS AND PRESSURE
Fluid is any substance that cannot maintain its own shape or has no rigidity. Fluid is a term applied to liquids and gases for they flow readily and they do not resist shearing stress. Fluids can flow and alter its shape to conform to the outlines of its container. Both liquids and gases have many characteristics in common. Liquids are nearly incompressible, while gases are easily compressed. Liquids tend to have greater density than gases. The gaseous phase of a substance usually exists at higher temperature, hence gas molecules are able to break free from one another. Gases are able to escape from an open container, whereas liquids cannot.
PRESSURE
Pressure is the force applied per unit area, P = F/A where F is the force applied and A is the area on which it is applied. Pressure is one of the important concepts in fluids. Examples of pressure: record player needle exerts pressure on a disc record, water near the bottom of a pool exerts pressure on a swimmer’s eardrum, and atmospheric pressure changes weather conditions. Measurements of pressure are common : Tires must be inflated to correct pressure, blood pressure should stay within normal range, and too much pressure in the eye ( glaucoma ) can cause blindness.
The SI unit for pressure is Pascal ( Pa ); 1 Pa = 1 newton / m2. An atmosphere ( atm ) is also a unit of pressure, where one atm = the average pressure due to the weight of the atmosphere at sea level.
Pressure is as important as the force creating it. If someone pokes you with his finger, you will certainly feel it. If, however, a nurse pokes you with a hypodermic needle using the same amount of force, you just don’t feel it – the needle breaks the skin. The same force applied to a smaller area creates a larger pressure and has a much different effect.
Fluids as well as solids, can exert pressures. Consider the water in a square container. If the water has a mass of 10 kg, its weight of 98 N must be supported by the bottom of the container. If the bottom has an area of 0.1 m2, then the pressure due to the weight of the water on the bottom is 980 N / m2. This computation is valid only for containers with straight sides. Another example is atmospheric pressure, which is caused by the weight of air. Atmospheric pressure is 1.013 x 105 Pa at sea level which means a column of air 1m on a side extending to the top of the atmosphere weighs 1.013 x 105 N. Stationary fluids always exert forces perpendicular to surfaces whether that direction is up or down, left or right. The reason the force is always perpendicular to the surface is that fluids cannot withstand shearing or sideways forces and therefore cannot exert sideways forces.
1. A woman wearing high-heeled shoes places about 50 % of its full weight on the single heel when walking. If the woman has a mass of 50 kg, determine the pressure on the ground under one heel if the area of contact is
2.5 cm2. How would this pressure compare with the pressure exerted underneath the elephant’s foot which is circular with diameter of 30 cm assuming that it is standing on four legs and with mass of 3 776 kg.
2. Determine the pressure in N / m2 and atm exerted by a phonograph needle on a record surface if the needle
supports 2.4 g on a circular area 0.48 mm in diameter.
3. Calculate the force exerted on one side of a 3 m by 9 m wall assuming normal atmospheric conditions.
CONVERSION FACTORS FOR VARIOUS UNITS OF PRESSURE
Conversion to N / m2 | Conversion to atmosphere ( atm ) | ||||
1.0 atm | = | 1.013 x 105 N/m2 | 1.0 atm | = | 1.013 x 105 N/m2 |
1.0 dyne/cm2 | = | 0.1 N/m2 | 1.0 atm | = | 1.013 X 106dyne/cm2 |
1.0 kg/cm2 | = | 9.8 x 104 N/m2 | 1.0 atm | = | 1.03 kg /cm2 |
1.0 lb/in2 | = | 6.9 x 103 N/m2 | 1.0 atm | = | 14.7 lb/in2 |
1.0 mm Hg | = | 133 N/m2 | 1.0 atm | = | 760 mm Hg |
1.0 cm Hg | = | 1.33 x 103 N/m2 | 1.0 atm | = | 76 cm Hg |
1.0 cm water | = | 98.1 N/m2 | 1.0 atm | = | 1.03 x 103cm water |
1.0 bar | = | 1.0 x 105 N/m2 | 1.0 atm | = | 1.013 bar |
PASCAL’S PRINCIPLE
One pioneer in the Physics of fluids was the French philosopher and scientist Blaise Pascal ( 1623 – 1662 ). He discovered an important property of stationary fluids: They can be used to transmit pressure to a place other than where the pressure is created and is not diminished while in transit. This is Pascal’s principle : Any pressure applied to a confined fluid will be transmitted undiminished to all parts of the fluid.
Gauge pressure is the pressure above or below atmospheric pressure. Total pressure or absolute pressure is gauge pressure plus atmospheric pressure.
Pt = Pg + Patm, where Pt = total pressure, Pg = gauge pressure, Patm = atmospheric pressure
Pascal’s principle is applicable to hydraulic systems. It is important to note that pressure, not force is transmitted undiminished to all parts of the fluid.
P1 = P2 ==> F1 / A1 = F2 / A2
Pascal’s principle holds for gases as well as for liquids, with some minor modifications due to the change in volume of a gas when the pressure is changed.
Hyperbaric medicine treats many physical problems through the application of high-pressure air or air-oxygen mixtures. Patients are enclosed in chambers pressurized up to 6 atmospheres. By Pascal’s principle, the pressure is distributed throughout the hyperbaric chamber. Patients breathing air at 6 atmospheres takes in six times the amount of oxygen with each breath. The increased oxygen intake is useful in treating a variety of problems such as carbon monoxide poisoning, slow-healing of wounds, and burns.
1. The large piston of a hydraulic press supports a dentist’s chair and the dentist wants to lift the patient by
stepping on the pedal directly on top of the small piston. Find the force exerted by the dentist if the patient plus
chair have a mass of 125 kg and the large piston has radius of 6 cm and the small piston has radius of 1.2 cm.
PRESSURE DUE TO WEIGHT OF A COLUMN OF FLUID
P = F / A = mg /A , but m = pV => P = pVg / A => V = Ah => P = pAhg /A => P = hpg
The pressure due to the weight of a column of fluid depends only of the depth in the fluid and the density of the fluid. Pressure at the surface of a fluid is zero since h is zero. As it goes deeper the pressure increases in proportion to the increase in height.
Another manifestation of how pressure depends only on depth and density, is found in the intravenous ( IV ) administration of fluids. The pressure due to the IV fluid at the entrance of the needle is proportional to h, the height of the surface above the needle. Paths of the fluid has nothing to do with pressure but is dependent on h, P = hpg. Applied pressure can be adjusted by raising or lowering the IV bottle relative to the patient.
Problems :
1. A nurse administers medication in a saline solution to a patient by infusion into a vein in the patient’s arm. The
density of the solution is 1000 kg/m3, and the gauge pressure inside the vein is 2.7 x 103 Pa. How high above the
insertion point must the container be hung so that there is sufficient pressure to force the fluid into the patient ?
2. A scuba diver searches for treasure at a depth of 22 meters below the surface of the sea. At what pressure
must the scuba device deliver air to the diver? How many atmospheres would this pressure be ?
Note : The pressure at the diver’s depth is greater than atmospheric pressure because of the weight of the
water above the diver. If the air breathed in is not at the same pressure as the external pressure on the
diver’s chest, the excess pressure will collapse the chest. Thus, the breathing apparatus must deliver
air to the diver at the pressure of the surrounding water. ( use P = Patm + hpg )
3. A tank is filled with water to a depth of 175 cm. What is the pressure at the bottom of the tank due to the
water alone ? What is the total pressure ?
MEASUREMENT OF PRESSURE BASED ON PASCAL’S PRINCIPLE AND P = hpg.
Pascal’s principle states that any pressure applied to a confined fluid is transmitted undiminished to all parts of the fluid. Thus, fluid can be use to transmit pressure to a gauge at a convenient location. As an example, blood pressure can be measured without putting a gauge into the body. Furthermore, pressure is transmitted undiminished, so the measurement can be very accurate. In a blood pressure measurement, an inflatable cuff is placed on the upper arm, and inflated until blood flow is cut off in the brachial artery. Pressure is created by squeezing the bulb and is transmitted by air in the tubes (a confined fluid) to the cuff and to the gauge. The wall of the cuff transmits the pressure to the arm and through it to the artery . When the applied pressure exceeds the heart’s output pressure , the artery collapses. The person making the measurement slowly releases the air in the cuff, lowering its pressure, and listens for flow to resume when pressure in the cuff becomes lower than the maximum heart output.
Systolic blood pressure, ( when the heart is contracted ) the maximum blood pressure is recorded together with the lower pressure called the diastolic pressure (when the heart is relaxed between beats). Diastolic pressure is the minimum pressure the circulatory system experiences and can be detected as a change in the sound of the blood flow through the partially restricted artery. The main point here, based on Pascal’s principle, is that all these pressures are transmitted undiminished and the pressure read by the gauge is truly representative of the pressure in the heart. It is important for the cuff to be at the same level as the heart in a blood pressure measurement. Any effect due to the weight of the air in the cuff and connecting tubes is negligible because the density of air is very small. The gauge can be placed at any convenient location, such as on the wall, a table, the floor, or anywhere in between, but the cuff must be at the same level as the heart.
4. Determine the maximum force in newtons exerted by the blood on an aneurysm, or ballooning, by
the aorta, given that the maximum blood pressure is 140 mm Hg and the area of the aneurysm is 25 cm2.
F = PA = ( hpg ) A = ( 0.140 m )( 13 600 kg/ m3 ) ( 9.8 m /s2 )( 25 cm2 x 1 m2/ 10 000 cm2 ) = 46.648 N
Note that the fluid involve is mercury, hence we use 13 600 kg/ m3 for the density of mercury and convert 140 mm to meter. The area 25 cm2 is likewise converted to m2 .
>>> This is a large force for the vessel to withstand, and there is a considerable risk that the aneurysm will burst.
5. [ 6.10 / 176 ] Water towers are used to store water above the level of homes. If a user observes that
the static water pressure at home is 3 x 105 N /m2, how high above the home is the surface of the
water in the tower ?
BUOYANT FORCE AND ARCHIMEDES’ PRINCIPLE
A story has been told that Archimedes ( 287 – 212 B. C. ) conceived of the principle that bears his name after King Hiero of Syracuse asked him to determine the actual composition of the King’s crown, which was alleged to be pure gold. Archimedes was ordered to do so without damaging the crown.
According to legend, the Greek scientist’s inspiration came to him as he lay partially submerged in his bath. On getting into the tub, he observed that the more his body sank into the tub, the more water ran out over the top. He immediately jumped out of the tub and rushed through the streets naked, shouting excitedly in a loud voice “Eureka” (“I have found it”). Archimedes’ principle states that : A body whether completely or partially
submerged in a fluid, is buoyed upward by a force that is equal to the weight of the displaced fluid.
Buoyant Force = weight of displaced fluid ==> FB = wd.l.
The principle applies both to liquids and gases and to objects which are completely or partially submerged. How this principle allowed Archimedes to solve the problem of the king’s crown is shown in the following problem:
The King’s crown is said to be solid gold but maybe made of lead and covered with gold. When it is weighted
in air, the scale reads 0.475 kg. When it is submerged in water, the scale reads 0.437 kg. Is it solid gold ? If not, what percentage by mass is gold?
BOYLE’S LAW
Robert Boyle ( 1627 – 1691 ) advanced the study of gases using an air pump made by Robert Hooke. Boyle observed the relationship between the pressure and the volume of an enclosed gas at constant temperature. Boyle’s law states that at constant temperature the pressure exerted by a gas is inversely proportional to the volume in which it is enclosed.
P a1/ V => PV = k or PV = constant, where P is the gas pressure, V is the volume and the value of the constant depends on the initial conditions. A complete statement of Boyle’s law includes the
condition that both the temperature and the amount of gas must be held constant. Alternatively, Boyle’s
law can be written in the form P1V1 = P2V2 ; where the subscripts 1 and 2 refer to the different physical states of the same sample of gas with the temperature held constant.
1. A cylinder with height of 20 cm and cross sectional area of 0.40 m2 has a close-fitting piston that may be moved to change the internal volume of the cylinder. Air at 1.013 x 105 N/m2 fills the cylinder. If the piston is pushed until it is within 8 cm from the end of the cylinder, what is the new pressure of the air? Assume that the temperature of the gas remains constant and that the volume of gas in the gauge is small compared with the volume of the cylinder.
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