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by: Charles Lind

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# Physics 2001 Exam 3 Phys2001

Charles Lind
UC
GPA 3.2

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Overview to Exam 3
COURSE
ENGINEERING PHYSICS I
PROF.
Dr. Wagner
TYPE
Study Guide
PAGES
83
WORDS
CONCEPTS
Physics 2001 Study Guide
KARMA
50 ?

## Popular in Engineering and Tech

This 83 page Study Guide was uploaded by Charles Lind on Sunday April 10, 2016. The Study Guide belongs to Phys2001 at The University of Cincinnati taught by Dr. Wagner in Spring 2016. Since its upload, it has received 21 views. For similar materials see ENGINEERING PHYSICS I in Engineering and Tech at The University of Cincinnati.

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Date Created: 04/10/16
12 Static Equilibrium and Elasticity Zero Torque and Static Equilibrium Static equilibrium occurs when an object is at rest – neither rotating nor translating.  Fx 0  Fy 0    0 If the net torque is zero, it doesn’t matter which axis we consider rotation to be around; we are free to choose the one that makes our calculations easiest. F1 F 2mg  0  2 F L2  mg  mg( L) 4 F Lmg( L)  0 2 4 3 1 F2 m4; F  m1 4 A5.00 m long diving board of negligible mass is supported by two pillars. One pillar is at the left end of the diving board; the other is 1.50 m away. Find the forces exerted by the pillars when a 90 kg diver stands at the far end of the board. (1)Determine the magnitude of the normal force n exerted by the support on the board. (2) Determine where the father should sit to balance the system at rest. Center of Mass and Balance m gx m gx  0 m  m (x / x ) 1 1 2 2 2 1 1 2 •Elastic Properties of Solids •Stress – Is proportional to the force causing the deformation – It is the external force acting on the object per unit cross-sectional area •Strain – Is the result of a stress – Is a measure of the degree of deformation The elastic modulus is the constant of proportionality between the stress and the strain. stress elastic modulus  strain Young’s Modulus Measures the resistance of a solid to a change in its length F Y  tensile stress A tensile strain L Li Shear Modulus Measures the resistance of motion of the planes within a solid parallel to each other F S  shear stress A shear strain x h Bulk Modulus Measures the resistance of solids or liquids to changes in their volume volume stress F A P B  volume strain V  V Vi Vi 14 Fluid Mechanics Density The density of a material is its mass per unit volume: Pressure Pressure is force per unit area: Unit: 1 Pa Pressure Pressure is force per unit area: Unit: 1 Pa Find the pressure exerted on the skin of a balloon if you press with a force of 2.1 N using (a) your finger or (b) a needle. Assume the area of your finger tip is 1.0 x 10 m , and the 2 -7 2 area of the needle tip is 2.5 x 10 m . Atmospheric Pressure Atmospheric pressure is due to the weight of the atmosphere above us. The pascal (Pa) is 1 N/m . The standard unit of pressure is pascal. Atmospheric pressure acts uniformly in all directions. We don’t usually notice it. F = P A = 810 N at palm Gauge pressure: pressure above atmosphere pressure Static Equilibrium in Fluids The increased pressure as an object descends through a fluid is due to the increasing mass of the fluid above it. Acubical box 20.00 cm on a side is completely immersed in a fluid.At the top of the box the pressure is 105.0 kPa; at the bottom the pressure is 106.8 kPa. What is the density of the fluid? Barometer A barometer compares the pressure due to the atmosphere to the pressure due to a column of fluid, typically mercury. The mercury column has a vacuum above it, so the only pressure is due to the mercury itself. Pat h g 760 mm Hg Fluids Seek their Own Level AU-shaped tube is filled with water but a small amount of vegetable oil has been added to one side. The density of the vegetable oil is 9.20 x 10 kg/m . If the depth of the oil is 5.00 cm, what is the difference in level h between the top of the oil on one side of the U and the top of the water on the other side? Pascal’s Principle An external pressure applied to an enclosed fluid is transmitted unchanged to every point within the fluid. Hydraulic Lift F  PA  F1 A  F    F d d 2 2 A 2 A 1  1 2 1 1 1   Archimedes’Principle and Buoyancy A fluid exerts a net upward force on any object it surrounds, called the buoyant force. This force is due to the increased pressure at the bottom of the object compared to the top. Weight of the fluid that is displaced by the cube Archimedes’Principle: An object completely immersed in a fluid experiences an upward buoyant force equal in magnitude to the weight of fluid displaced by the object. F   gV b fluid Applications ofArchimedes’Principle An object floats when it displaces an amount of fluid whose weight is equal to the weight of the object. An object made of material that is denser than water can float only if it has indentations or pockets of air that make its average density less than that of water. The fraction of an object that is submerged when it is floating depends on the densities of the object and of the fluid. V sub V (s / s ) f What percentage of a floating chunk of ice projects above the level 3 of the water?Assume the density of ice is 917 kg/m . Fluid Flow and Continuity Bernoulli’s Equation When a fluid moves from a wider area of a pipe to a narrower one, its speed increases; therefore, work has been done on it. Change in Speed The kinetic energy of a fluid element is: Equating the work done to the increase in kinetic energy : Water travels through a 9.6 cm diameter fire hose with a speed of 1.3 m/s. The water pressure in the fire hose is 350 kPa.At the end of the hose, the water flows out through a nozzle whose diameter is 2.5 cm. (a) What is the speed of the water coming out of the nozzle? (b) Find the pressure in the nozzle Change in Height If a fluid flows in a pipe of constant diameter, but changes its height, there is also work done on it against the force of gravity. Equating the work done with the change in potential energy gives: General Case The general case, where both height and speed may change, is described by Bernoulli’s equation: This equation is essentially a statement of conservation of energy in a fluid. Applications of Bernoulli’s Equation Bernoulli Effect Applications of Bernoulli’s Equation Bernoulli Effect Torricelli’s Law v  2gh 2 In designing a backyard fountain, a gardener wants a stream of water to exit from the bottom of one tub and land in a second one. The top of the second tub is 0.500 m below the hole in the first tub, which has water in it to a depth of 0.150 m. How far to the right of the first tub must the second be placed to catch the stream of water. 19. Temperature Thermodynamics The Zeroth Law of Thermodynamics Measuring Temperature TCT 273.15 The Celsius and Fahrenheit Scales Thermal Expansion L  L T V V  T    Macroscopic Description of an Ideal Gas PV  nRT m N n   M N A R  8.31JolK Problem An ideal gas contains a volume of 100 cm at 20 C and 100 Pa. Find the number of moles of gas in the container. Problem Aspray can containing a propellant gas at twice atmospheric pressure (202 kPa) and having a volume of 125.00 cm is at 22 C.o It is then tossed into an open fire. When the temperature of the gas in the can reaches 195 C, what is the pressure inside the can. Assume any change in the volume of the can is negligible. 20 Heat and the First Law of Thermodynamics Temperature and Heat Heat is the energy that is transferred between a system and its environment because of a temperature difference between them TheAbsorption of Heat by Solids and Liquids Heat capacity Q  C T How many heat must be transferred to change the temperature of an object by a certain amount e.g. 1 K. Specific Heat Q  c m T Calorimetry Q cold hot m w w T fT ) W  m x x T fT )W A0.05 kg ingot of metal is heated at 200 C and then dropped into 0 a beaker containing 0.4 kg of water ini0ially at 20 C. If the final equilibrium temperature of the mixed system is 22.4 C, find the specific heat of the metal. Latent Heat Q  L m Q  L f s Q  LV w Work and Heat in Thermodynamic Processes   dW  F dr  PAdy V f W   PV Vi Work done on the gas: W  0 Compression. Work done by the gas: W  0 Expansion. < < < The work done on a gas in a quasi-static process that takes the gas from an < initial to a final state is the negative of the area under > the curve of a PV diagram, evaluated between the initial and final states. The First Law of Thermodynamics E  QW internal energy depending int adiabatic only on T free expansion Q = 0 →ΔE = int Q = W = 0 →ΔE = 0int Constant volume process: ΔV = 0 →W = 0;ΔE = Q int < < Constant pressure process (isobaric): W = -p (V - f);ΔEiand Q intnges Cyclical processes: ΔE = 0 int But (p, V, T, Eintchanged →Q = -W < Negative net work done by the system is equal to the heat absorbed by the system → engines Isothermal Expansion of an Ideal Gas  V W  nRT ln  i  f A1 mol sample of an ideal gas is kept at 0.0 C during an expansion from 3.0 L to 10.0 L. How much work is done by the gas during the expansion? How much energy transfer by heat occurs with the surroundings in this process? If the gas is returned to the original volume by means of an isobaric process, how much work is done on the gas? Asample of an ideal gas goes through the process shown in attached Figure. From A to B, the process is adiabatic; from B to C, it is isobaric with 100 kJ of energy entering the system by heat. From C to D, the process is isothermal; from D to A, it is isobaric with 150 kJ of energy leaving the system by heat. Determine the difference in internal energy Eint,B – Eint,A. Heat Transfer Mechanisms Thermal Conduction Q dT P  t  kAdx Convection Radiation P A  T 4 0  1:Emissivity   5.703 10 W/(m K ): 4 Stefan-Boltzmann constant 21 The Kinetic Theory of Gases Molecular Model of an Ideal Gas p x mv  xv  x P  2  1mv 2 3   2 Translational Kinetic Energy One atomic gas: no rotation, no vibration v  3RT M 1 3RT 3 RT 3 K  2m M  2m mN  2kT A 3 3 Ktot NkT  nRT  E int 2 2 The Molar Specific Heats of an Ideal Gas Constant olume Q  nCVT E int nCVT Constant Pressure Q  nCPT E int nCPT W nC T  nC T nRT V P CPC V R Degrees of Freedom and Molar Specific Heat TheAdiabatic Expansion of an Ideal Gas Pi i  P f f TV  1T V  1 i i f f 22 Heat Engines, Entropy and the Second Law of Thermodynamics Change in Entropy f S  S S  dQ f i  T i Entropy is a State Function An ideal gas is taken through a reversible process V T S  nRln  f nC ln  f  V i V  Ti  Second Law of Thermodynamics In a closed system S  0 Heat Engines Carnot Engine Born: 1 June 1796 in Paris Died: 24 Aug 1832 in Paris impossible engine QH Q L S    0 TH TL  1 TL1 C TH Refrigerators QL K  W

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