See discussions, stats, and author profiles for this publication at: https://www.researchgate.net/publication/304571176 BASIC PHYSICS 1 MODULE Raw Data · June 2016 DOI: 10.13140/RG.2.1.3639.9601 CITATIONS READS 0 52 1 author: John Maera Maasai Mara University 12 PUBLICATIONS 3 CITATIONS SEE PROFILE Some of the authors of this publication are also working on these related projects: ERP training View project All in-text references underlined in blue are linked to publications on ResearchGate, letting you access and read them immediately. Available from: John Maera Retrieved on: 23 November 2016 MAASAI MARA UNIVERSITY (MMU) SCHOOL OF SCIENCE DEPARTMENT OF PHYSICS PHY 110 MODULE FOR BSC, BED, BSC (COMPUTER) PHY110: BASIC PHYSICS 1 Maera John JAN-APRIL 2015 Page 1 of 64 SYMBOLS USED Take Note Further Reading Question Written Exercises A question: This symbol indicates that there is a …?... Activity Summary S A Congratulations Definitions of Key Words Words Self-Diagnosis Test Written Assignment Written Assignment ? 100 My score Objectives o Page 2 of 64 ................................................................................................................................................................................................................................... 18 Newton's laws of motion ....................................................................................................................................................................... 7 Dot Product or Scalar Product ............................................................................. 9 TYPES OF MOTION .......................................................................................................................................................................................... 8 Written Exercise 1...................................................................................................................................................................................................... 13 Summary ........................................................ 10 Classic version ................................................................................................................................................................................. 6 Specific objectives ...................................................................................................................................................................................... 14 Simple Harmonic Motion ....................................... 11 Projectile Motion ......................................... 16 Definitions:.............................................. 12 Introduction .............................................................................................................................. 15 Hooke's Law: .................................................................................................... 6 Physical and non-physical quantities ............................................ 19 Relationship to the conservation laws ........................................................................................................................................ 20 Kepler's laws ..............................................................................................Table of Contents Symbols used ................................................................................................................2 ....................................... 18 Impulse ............................................................. 6 Vectors and scalars ............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................. 12 The Trajectory of a Simple Projectile is a Parabola ................................................................................. 16 Dynamics of simple harmonic motion................................................................................................................................................................................................................ 13 Circular Motion ............................................ 6 Nature of physics ............................................................................................................................................................... 14 Equations of circular motion .............1 .............................. 7 Vector addition ...................................................................................................................... 6 Self diagnostic Test ......................................... 2 TOPIC 1: MECHANICS .. 9 Equations of uniformly accelerated linear motion .............................................................................................................................................................................................................. 11 Written Exercise 1........................................................................................................................................... 7 The VECTOR PRODUCT or CROSS PRODUCT ..................................................................................... 6 Introduction .......................................................................................................................................................... 20 Page 3 of 64 .......................................................................................................................................................... 10 Examples .............. 17 Energy of simple harmonic motion ................................................................................. ..... 35 Thermal expansion coefficient ................................... 33 Thermometers ............................................................................................................................................................................................................................................................................................................................................................................................................... 37 Heat capacity or specific heat ......................................................................................................................................................................................1 ................................. 26 BERNOULLI'S EQUATION: for Ideal Fluid Flow .......................................................................................................................................................................................................................................................... 25 Phases of matter............................................................................................................................................................................................................................. ..................................................................................................................................................................................................... 37 The first law of thermodynamics ............................................. 34 Coefficient of thermal expansion .................................................................................................................................................... 36 Conservation of energy ............. 36 Volumetric thermal expansion .................................................................................................................................................................................................................. 25 Six Common Phase Changes ......................................................................... 25 Self diagnostic Test ................................................................................................................................................................................... 33 Specific objectives ..................................................................................................................................................... 22 Coefficient of friction ........................................................................................................................................................................................................................................................................................................................................................................................................ 42 Page 4 of 64 ....................... 31 TOPIC THREE: THERMAL PHYSICS ....................................................................... 41 Postulates of kinetic theory of gases ......................................................Examples of Kepler's Third Law .................................................................................................................................................................................................................................. 33 Self diagnostic Test ......................................................................................................................... 23 Summary ............................................................................................................................................................................................................................................................ 23 TOPIC TWO: PROPRTIES OF MATTER ..................................... 25 Specific objectives ............................... 33 Rankine scale ............................................................................. 35 Linear thermal expansion ................................................................................ 38 Kinetic theory of gases ................................................................................................ 22 Reducing friction............................................................ 21 Friction ............................................ 41 Pressure .............................................................................. 35 Area thermal expansion ............................................................................................................................................................................................................ 26 Density and Elasticity ........................................................................................................... 33 Introduction ..... 34 Calibration of thermometers ................................................................. 30 Written Exercise 2.............................................. 25 Introduction .............................................................. 21 Coulomb friction ......................... ................. 56 Longitudinal and transverse waves ....................... 47 Radiation ......................................................... 48 Newton's law of cooling ................................. 56 Speed of sound ..................................................................................................................................................................... 55 Perception of sound...... 45 RMS speeds of molecules ................................................ 51 Critical insulation thickness............................................................................................................... 55 Self diagnostic Test ............................................................................................................................................................................. 62 Page 5 of 64 ....................................................................................................................................................................................................................................................... 52 TOPIC FOUR: SOUND.................................................................... using thermal circuits .................................................................................................. 61 Appendix 1: COURSE outline Phy110 ......................................... 46 Heat transfer .............. 57 Sound pressure level ..... 52 Blackbody Radiation .................................................................................................................. 55 Introduction ....................................................................................................................................................................................................................................................................... 55 Sound ....................................................................................................................................................................... 44 Number of collisions with wall ..................................................................................................................................................................................................................... 46 Conduction...................... 58 References .Temperature and kinetic energy .................................................................. 55 Physics of sound.......................................................................................................................................................................................................................................... 56 Acoustics and noise.......... 49 Insulation and radiant barriers...................................................................................................................................................................................................... 55 Specific objectives .............................................................................................................................................................................................................................................................................................................................................................................................................................................. 57 Examples of sound pressure and sound pressure levels ................................................................................................................................................................................................................................................................................ 56 Sound wave properties and characteristics .................................... 46 Convection ................................................................ 48 Clothing and building surfaces........................................................................................................................................................................................................ 57 Equipment for dealing with sound .......................................................................................................................................... 49 One dimensional application................ and radiative transfer............................................................................................................ circular motion and simple Harmonic motion. Finally we shall relate Newton’s law to Kepler’s laws. We shall then derive the equations of linear. rotational. Relate Newton’s laws to Kepler’s laws of motion INTRODUCTION In this topic we shall discuss the physical quantities. the forces they exert on one another. Sometimes in modern physics a more sophisticated approach is taken that incorporates elements of the three areas listed above. charge. A laboratory is not a physical quantity but its length. and parity. An example is smell. dealing with the fundamental constituents of the universe. PHYSICAL AND NON-PHYSICAL QUANTITIES Dfn 1: A physical quantity is one that can be measured and a number assigned to it. Add. subtract and multiply vectors 3. Distinguish between vectors and scalars giving five examples for each 4. State the conservation of energy and momentum NATURE OF PHYSICS Physics. One can only say good. What is a physical quantity? 3. it relates to the laws of symmetry and conservation. Dfn 2: A non-physical quantity is one that cannot be measured and a number given to value. SPECIFIC OBJECTIVES o At the end of this TOPIC you should be able to: 1. Derive the equations of various types of motion 4.TOPIC 1 : MECHANICS SELF DIAGNOSTIC TEST Answer all questions ? 1. is a major science. amount of substance. their measurement units and classify them into scalars and vectors. State Newton’s laws of motion 5. mass. width and height are physical quantities. An example is time. power. such as those pertaining to energy. average but not give a number objectively. force. Operation of vectors is core in dealing with mechanical systems. volume. What is physics? 100 2. Page 6 of 64 . momentum. Distinguish between vectors and scalars 2. bad. and the results produced by these forces. Discuss the types of motion 5. we imagine moving the start point of one vector to the endpoint of the other vector. This brevity is a nice aspect to vector algebra. Examples: acceleration. We should also remember that multiplication of a vector by scalar is multiplication of each component by the same scalar. j. A and B. the three of them can describe all vectors possible in the space. and electric and magnetic fields. Vy. Examples: Time. electric field Vectors are central to the study of physics. Besides displacement and velocity. Vy. other examples of vectors include acceleration. By this it is clear that in order to add two vectors. force. Two vectors. To specify V. When we visualize this in space. 8). Az = Bz. j. Classify the basic physical quantities into scalars and vectors? VECTOR ADDITION The unit vectors i. 1) = (8. and k which. 7) + (4. cAz) The length of a vector is called its magnitude and is usually denoted by |A|. when summed together. they must have the same dimension . Hence. In a two dimensional case. a three dimensional vector is an ordered set of three numbers. and Mass Dfn: Scalars quantities have only size. torque. The directionality of the vector is lost for this quantity so it is a scalar. it is sufficient to specify its three components (Vx.otherwise the operation is undefined. namely cA = (cAx. When we add vectors. velocity. DOT PRODUCT OR SCALAR PRODUCT One way of combining two vectors is through an operation called the dot product. A seventy dimensional vector is an ordered set of 70 numbers. make V. cAy. we have mathematically A + B = (4. The sum vector is the resultant. momentum. is described by specifying the amounts of i. are equal if and only if each of their components are equal: Ax = Bx. The components of Vare Vx. and Vz . temperature. Ay = By. gravitational field. and k are chosen so that through the addition of multiples of themselves with each other. It is interesting to observe that 1 vector equation (A = B) is equivalent to three scalar equations.VECTORS AND SCALARS Dfn: Vectors are quantities that have a size and a direction. Early on in this study you will encounter types of vector quantities. An arbitrary vector V. It is written as: A B B A A B COS Another form of the equation is A B (Ax i Ay j Az k) (Bx i By j Bzk) A x Bx A y By Az Bz Page 7 of 64 . and Vz). we add each of their components separately. a 2 b1)kˆ ˆi a × b = a1 ˆj a2 kˆ a3 b1 b2 b3 Take a look at the order of the subscripts in the result and you will see a cyclical appearance of each one. normalized vectors or orthonormal. for short. but we are unable to visualize "right angles" in a 70 dimensional space! With this information it is clear to see that this provides a quick route to the magnitude of a vector. It is a little more involving mathematically to remember the form of the operation so we cast it in the form of a determinant (is that any easier to remember?) Vector multiplication yielding another vector Yields a vector which has a direction determined by the right hand rule Yields a vector perpendicular to the plane containing the other two vectors The cross product DOES NOT commute ˆ + a ˆj×(b ˆi + b ˆj + b k) ˆ ˆ ˆ ˆ ˆ a × b = a1ˆi ×(b1ˆi + b2 ˆj + b3k) 2 1 2 3 + a 3k ×(b1i + b 2 j + b3k) a × b = (a 2 b3 . we can characterize this condition as having the vectors placed at right-angles to each other. Because they are fixed at 90 degrees from each other we have: i i j j k k COS0 1 and i j j k k i COS 0 2 This property of these three vectors (the dot product with themselves produces 1 and the dot product with each other produces 0) is what defines them as being a set of orthogonal. Learn to appreciate the order in this for it will appear time and again. and k. j.a3b2 )iˆ + (a3b1 . There are a couple of other properties worth noting here. In a 2 or 3 dimensional space. A A A 2 A x2 A y 2 A z 2 A A x2 A y 2 A z 2 1/ 2 THE VECTOR PRODUCT OR CROSS PRODUCT There is another way to combine vectors. Page 8 of 64 . The same concept holds in higher order spaces.a1b3 )jˆ + (a1b2 .This last form can be seen clearly when we consider the dot product of the unit vectors i. rather than a scalar as with the dot product. This is the cross product but in this case the result is another vector. namely to take the dot product of a vector with itself. 2i . Newton's second law).ˆj + 2k) Revision Exercises 1. WRITTEN EXERCISE 1.k) ˆ (ii) (5iˆ . and sometimes to the solutions to those equations. equations of motion describe the behavior of a system (e. the motion of a particle under an influence of a force) as a function of time. Page 9 of 64 . When the two original vectors are orthogonal to each other. 1.1 1. Write down five physical quantities and five non-physical quantities apart from the ones given above 2. This same happens when the two initial vectors rotate away from each other.. As the two vectors are rotated in towards each other.g.25 What displacement must be added to a 50 cm displacement in the +x-direction to give a resultant displacement of 85 cm at 25°? Ans. 1.1 Ans. Sometimes the term refers to the differential equations that the system satisfies (e.24 What displacement at 70° has a component of 450 m? What is its y-component? 1. difference.3 km. scalar product and cross product of the following vectors (i) A 3i 5 j 10k and C 3 j .2km 1. Find sum. the cross product vector has the greatest magnitude (it is at its longest). the concept to remember is that the cross product produces a vector which is perpendicular to both vectors making up the argument of the product. This means it is orthogonal to both (though the two argument vectors need not be orthogonal to each other).g.6k ˆ and (2iˆ + 3jˆ . the resultant vector shortens until it disappears when the two overlap. Using the definition of cross product and right hand rule: iˆ iˆ ˆj ˆj kˆ kˆ 0 ˆi ˆj kˆ ˆj ˆi ˆj kˆ ˆi kˆ ˆi kˆ ˆi ˆj ˆi kˆ Graphically. the resultant disappearing when the point opposite each other. 45 cm at 53° TYPES OF MOTION In physics.. Often.the position at the end of the interval (displacement) Δt--. v u at. Classic version The above equations are often written in the following form: v u at. Integration of the velocity yields a quadratic relationship for position at the end of the interval. s (u v)t and v 2 u 2 2as 2 2 where s = the distance between initial and final positions (displacement) (sometimes denoted R or x) u = the initial velocity (speed in a given direction) v = the final velocity a = the constant acceleration t = the time taken to move from the initial state to the final state Page 10 of 64 .the time interval between the initial and current states a---. Note that each of the equations contains four of the five variables. 1 s si ut a( t)2 and v 2 u 2 2a(s si ) 2 where. and several applications of the equations are required.Equations of uniformly accelerated linear motion The body is considered between two instants in time: one "initial" point and one "current".The velocity at the end of the interval s---. ui ----is the body's initial velocity. adt. or in the case of bodies moving under the influence of gravity. a = g. problems in kinematics deal with more than two instants. Thus. si-----is the body's initial position and its current state is described by: v--. 1 1 s ut at 2 ... may be integrated over an interval from 0 to Δt (Δt = t − ti). a is constant.the constant acceleration. in this situation it is sufficient to know three out of the five variables to calculate the remaining two. a differential. to obtain a linear relationship for velocity. If acceleration. How motion under gravity affect the three equations? Revision Exercises 1. Ans.32 A car is accelerating uniformly as it passes two checkpoints that are 30 m apart.0 s. the acceleration a must be in fact −g. 10 m/s 2.4 s Page 11 of 64 . The acceleration is local acceleration of gravity g.33 An auto's velocity increases uniformly from 6. and the car's speed at the first checkpoint is 5. 1.0 s has elapsed. Derive the three equations of linear motion 2. Ans. The acceleration due to gravity there is 1.60 m/s 2 downward. The time taken between checkpoints is 4. we have: s v 2 u2 u2 Substituting and cancelling minus signs gives: s 2g 2g WRITTEN EXERCISE 1. (a) 0.6 m/s2.0 m/s 2.2 2.0 m/s to 20 m/s while covering 70 m in a straight line. Given initial speed u.30 A truck starts from rest and moves with a constant acceleration of 5.2 1. Find (a) the acceleration and (b) the distance moved in the first 6. 40 m 2. Find the ball's original speed.3 m/s2. Using the fourth equation.0 s.90 m/s2. It starts from rest and attains a speed of 2. Find its speed and the distance traveled after 4. for example a ball thrown upwards into the air. (b) 16 m 2. 2. At the highest point. Choosing s to measure up from the ground. Find the acceleration and the time taken.0 s.Examples Many examples in kinematics involve projectiles.0 m/s. the direction of displacement. Ans.31 A box slides down an incline with uniform acceleration. one can calculate how high the ball will travel before it begins to fall. 5. 2. speed and acceleration is important. They could in fact be considered as uni-directional vectors.0 s. the ball will be at rest: therefore v = 0. since the force of gravity acts downwards and therefore also the acceleration on the ball due to it. 20 m/s. At this point one must remember that while these quantities appear to be scalars.7 m/s in 3. Find the car's acceleration and its speed at the second checkpoint.15 A ball that is thrown vertically upward on the Moon returns to its starting point in 4. Ans. A tossed heliumfilled balloon is not normally considered a projectile as the drag and buoyant forces on it are as significant as the weight. Horizontal Vertical Quantity Page 12 of 64 . pitched. and rocket-propelled spacecraft are sometimes also said to follow a trajectory. heaved. flung. gravity is what really drives a crashing airplane. fired. By assuming a constant value for the acceleration due to gravity. guided missiles. Every projectile problem is essentially two one-dimensional motion problems … The kinematic equations for a simple projectile are those of an object traveling with constant horizontal velocity and constant vertical acceleration. just that their effect is minimal in comparison. This is not to say that other forces do not exist. Some examples of projectiles include … a bullet the instant it exits the barrel of a gun or rifle a moving airplane in the air with its engines and wings disabled the space shuttle or any other spacecraft after main engine cut off (MECO) The force of primary importance acting on a projectile is gravity. The laws of physics are assumed universal until it can be demonstrated otherwise. The normal amounts of drag and buoyancy just aren't large enough to save the passengers on a doomed flight from an unfortunate end. horizontal velocity whose acceleration is due to gravity alone. In contrast. or thrown. A simple projectile is made mathematically simple by an idealization (basically a lie of convenience). Even though the drag and buoyant forces acting on it are much greater in absolute terms than they are on the balloon. hurled. A projectile and a satellite are both governed by the same physical principles even though they have different names. Airplanes. we make the problem easier to solve and (in many cases) do not really lose all that much in the way of accuracy.PROJECTILE MOTION Introduction A projectile is any object that is cast. a crashing airplane would be considered a projectile. they are not really projectiles. The path of a projectile is called its trajectory. The only relevant quantities that might vary from projectile to projectile then are initial velocity and initial position This is where we run into some linguistic complications. To get around this dilemma. Helium-filled balloons can't be thrown long distances and don't normally fall. it is common to use the term ballistic trajectory when dealing with projectiles. A projectile is any object with an initial non-zero. Since these devices are acted upon by the lift of wings and the thrust of engines in addition to the force of gravity. tossed. An essential characteristic of a projectile is that its future has already been preordained. The unification of physical law is a theme that surfaces from time to time in physics. horizontal velocity o whose acceleration is due to gravity alone. equal ranges for launch angles that exceed and fall short of 45° by equal amounts (ex. he horizontal distance traveled by a projectile is called its range. Revision Exercises 1. o a satellite if it follows a closed path that never brings it in contact with a celestial body (like the earth). 40° & 50°. (Identical projectiles launched at complementary angles have the same range. The kinematic equations for a simple projectile are those of an object traveling with … o constant horizontal velocity and o constant vertical acceleration. 30° & 60°.) o will have a maximum range when θ = 45°. 0° & 90°) Summary S A projectile is any object … o with an initial non-zero.43 A marble. projectile launched on level ground with an initial speed u at an angle θ above the horizontal … o will have the same range as a projectile launched with an initial speed u at (90° − θ). o a general projectile no matter where its trajectory may take it. rolls off the edge of a table that is 80 cm high. A projectile is said to be … o a simple projectile if the acceleration due to gravity may be assumed constant in both magnitude and direction throughout its trajectory. rolling with speed 20 cm/s. The path of a projectile is called its trajectory.ay = −g ax = 0 acceleration vx = ux vy = uy − gt velocity-time x = x0 + uxt y = y0 + uyt − ½ gt2 displacement-time vy2 = uy2 − 2g(y − u) velocity-displacement The Trajectory of a Simple Projectile is a Parabola max range at 45°.3 2. (a) How long does it Page 13 of 64 . 40 s. (a) 6. (b) 0.2 s.16 km. (a) How long will it take before striking the ground? (b) How far from the foot of the building will it strike? (c) At what angle with the horizontal will it strike? Ans.take to drop to the floor? (b) How far. (a) How long will it take to hit the ground? (b) How far from the starting point will it strike? (c) At what angle with the horizontal will it strike? Ans. (a) 0. horizontally. (b) 8. Equations of circular motion The analogues of the above equations can be written for rotation: 1 2 1 2 1 2 ω = ωo + αt. the centripetal acceleration of the motion can be derived. (a) 4.15 km. θ = ωo t + αt 2 . (b) 0. θ = (ω + ωo )t. Since in radian measure. Page 14 of 64 .45 A body is projected downward at an angle of 30° with the horizontal from the top of a building 170 m high. (c) 60° CIRCULAR MOTION For circular motion at a constant speed v. (c) 50° 2.44 A body projected upward from the level ground at an angle of 50° with the horizontal has an initial speed of 40 m/s.3 s.αt 2 where: α is the angular acceleration ω is the angular velocity θ is the angular displacement ω0 is the initial angular velocity.1 cm 2. from the table edge does the marble strike the floor? Ans. ω2 = ωo2 + 2αθ and θ = ωt . Its initial speed is 40 m/s. and (b) the tangential acceleration of a point on its rim. Each oscillation is identical. Find (a) the constant angular acceleration in rad/s 2 .1 m/s 9. If the equilibrium position is taken to be zero. The motion is periodic: the body oscillates about an equilibrium position in a sinusoidal pattern. Ans. In the case of the spring–mass system. The orbit is periodic. (b) 1 8 m 9. (a) 6.Revision questions 1. and thus the period. which is a force that tends to restore the mass to the equilibrium position. and Φ is the phase. which is a mass attached to a spring.20 It is desired that the outer edge of a grinding wheel 9. It will.00 h as a result of the Earth's rotation? What is the speed of a point on the equator? Take the radius of the Earth to be 6370km. (a) 67 rad/s.22 A wheel 25.28 rad/s 2 . that is. 463 m/s 9. the system is in mechanical equilibrium). Idea: Any object that is initially displaced slightly from a stable equilibrium point will oscillate about its equilibrium position. (b) 157 cm/s 2 SIMPLE HARMONIC MOTION Simple harmonic motion (SHM) is the motion of a simple harmonic oscillator.21 Through how many radians does a point on the Earth's surface move in 6. which is given by Hooke's Law. 1. and amplitude of the motion are constant. The frequency of the motion is determined by the intrinsic properties of the system (often the mass of the body and a force constant). A body in simple harmonic motion experiences a single force which is given by Hooke's law. 50. (a) Determine the angular speed of the wheel.4 9.19 A flywheel turns at 480 rpm. 15. there is no net force on the mass (that is.0 cm in radius turning at 120 rpm increases its frequency to 660 rpm in 9. The kinetic and potential energies of the system are also determined by these properties and conditions. Compute the angular speed at any point on the wheel and the tangential speed 30. Ans. the displacement x of the body at any time t is given by x(t) A cos( 2ft ) Where A is the amplitude.0 m/s. f is the frequency. If the spring is outstretched. Ans.57 rad. the spring will exert a restoring force. a motion that is neither driven nor damped. frequency. if the mass is displaced from equilibrium. experience a restoring force that depends linearly on the displacement x from equilibrium: Page 15 of 64 . However. this force is the elastic force. Introduction Simple harmonic motion showed both in real space and phase space.3 rad/s. while the amplitude and phase are determined by the initial conditions (displacement and velocity) of the system.0 cm in radius move at a rate of 6.0 s when it is turning at this rate? Ans.00 s. (Here the velocity and position axes have been reversed from the standard convention in order to align the two diagrams) A typical example of a system that undergoes simple harmonic motion is an idealized spring–mass system. the force is directly proportional to the displacement x and points in the opposite direction. (b) What length of thread could be wound on the rim of the wheel in 3. in general.0 cm from the center. A simple harmonic oscillator moves back and forth between the two positions of maximum displacement. The units of k are: Newtons k (2) metres Definitions: Amplitude (A): The maximum distance that an object moves from its equilibrium position. Frequency ( f ): The number of cycles (or oscillations) the object completes per unit time. at x = A and x = .e. Page 16 of 64 . Note: The negative sign in Hooke's law ensures that the force is always opposite to the direction of the displacement and therefore back towards the equilibrium position (i. Period ( T ): The time that it takes for an oscillator to execute one complete cycle of its motion. Simple Harmonic Oscillator: Any object that oscillates about a stable equilibrium position and experiences a restoring force approximately described by Hooke's law. f 1 T (3) The unit of frequency is usually taken to be 1 Hz = 1 cycle per second. Examples of simple harmonic oscillators include: a mass attached to a spring. then it gets back to x = A after one full period at t = T .Hooke's Law: Fs -kx (1) where the equilibrium position is chosen to have x -coordinate x = 0 and k is a constant that depends on the system under consideration. a car stuck in a ditch being ``rocked out'' and a pendulum.A. a molecule inside a solid. a restoring force). If it starts at t = 0 at x = A . Each of these constants represents an important physical property of the motion: A is the amplitude. and since T = 1/f. Page 17 of 64 . The motion of any simple harmonic oscillator is completely characterized by two quantities: the amplitude. and Φ is the phase. x is its displacement from the mean position. The solutions to this differential equation are sinusoidal. Dynamics of simple harmonic motion For oscillation in a single dimension. Now since ma = −mω2x = −kx. and the period (or frequency). . ω = 2πf is the angular frequency. Position. and the equilibrium position is chosen to be the origin. combining Newton's second law (F = m d2x/dt2) and Hooke's law d2 x (F = −kx. ω. and Φ are constants. Then since ω = 2πf. one solution is x(t) A cos( 2ft ) where A. the velocity and acceleration as a function of time can be found: Acceleration can also be expressed as a function of displacement: . These equations demonstrate that period and frequency are independent of the amplitude and the initial phase of the motion. . velocity and acceleration of an harmonic oscillator Using the techniques of differential calculus. and k is a constant. The constant k in Hooke's law is traditionally called the spring constant for the system. as above) gives the second-order differential equation m kx where m is the mass dt 2 of the body. even when the restoring force is not provided by a simple spring. 45 Hz [11. k =300 3000300 m = 2. that is. [11.0 cm. it takes 1 s for half a cycle. (c) the maximum speed of the mass. Find: (a) the frequency. (b) 24.2] A spring makes 12 vibrations in 40 s.0 cm. Find the period and frequency of the vibration. find (a) the amplitude of the resulting oscillation and (b) its center or equilibrium point. and the period is 1.0 kg N/m [11.80 m/s ? (b) What is the length there of a pendulum for which T = 1. [Ans: 0.00 s? Ans.5] A 50-g mass vibrates in SHM at the end of a spring. (e) the speed when the displacement is 6. below the 2. Neglecting the inertia and friction of the pulley and the mass of the spring and string. Ans. Objects weigh 0. (b) the spring constant.3 cm. The amplitude of the motion is 12 cm. (d) the maximum acceleration of the mass.16] In Fig. (a) What is the length of a simple "seconds pendulum" at a place where g = 9.40 as much on Mars as on the Earth.5 [11.27] Find the frequency of vibration on Mars for a simple pendulum that is 50 cm long.28] A "seconds pendulum" beats seconds. 0.Energy of simple harmonic motion The kinetic energy K of the system at time t is and the potential energy is The total mechanical energy of the system therefore has the constant value Revision Exercises 1.70 s. and (f) the acceleration when x = 6.30 Hz] [11. (a) 99.8 cm NEWTON'S LAWS OF MOTION Page 18 of 64 .0-kg mass is released when the spring is unstretched. this law is often stated as." Note: These laws describe the relationship between the forces acting on a body and the motion of that body. They were first compiled by Sir Isaac Newton in his work Philosophiæ Naturalis Principia Mathematica. "To every action there is an equal and opposite reaction. They are: First law There exists a set of inertial reference frames relative to which all particles with no net force acting on them will move without change in their velocity. Page 19 of 64 . Newton showed that these laws of motion. his is understood. it follows that I p mv This relation between impulse and momentum is closer to Newton's wording of the second law. B simultaneously exerts a force on A with the same magnitude in the opposite direction. "Force equals mass times acceleration (F = ma): the net force on an object is equal to the mass of the object multiplied by its acceleration. Newton used them to explain and investigate the motion of many physical objects and systems." Newton's first law is often referred to as the law of inertia. and happens along the straight line on which that impulse is impressed. in modern terms. According to modern ideas of how Newton was using his terminology. and it is given by I t Fdt Since force is the time derivative of momentum. IMPULSE An impulse I occurs when a force F acts over an interval of time Δt. as an equivalent of: The change of momentum of a body is proportional to the impulse impressed on the body. explained Kepler's laws of planetary motion. in the third volume of the text.Newton's laws of motion are three physical laws that form the basis for classical mechanics." Third law Whenever a particle A exerts a force on another particle B. This law is often simplified as "A body persists its state of rest or of uniform motion unless acted upon by an external unbalanced force. combined with his law of universal gravitation. the net force on a particle is equal to the time rate of change of its linear momentum: F = d(mv)/dt. 1687. The strong form of the law further postulates that these two forces act along the same line. Second law Observed from an inertial reference frame. first published on July 5. When mass is constant. For example. This law is often simplified into the sentence. Impulse is a concept frequently used in the analysis of collisions and impacts. and to both classical and non-classical physics. Find the horizontal recoil speed of the gun.46 m/s KEPLER'S LAWS LAW 1: The orbit of a planet/comet about the Sun is an ellipse with the Sun's center of mass at one focus This is the equation for an ellipse: x2 y 2 1 2 2 a b LAW 2: A line joining a planet/comet and the Sun sweeps out equal areas in equal intervals of time Page 20 of 64 . Ans. The coal initially has zero horizontal velocity.21 An empty 15000-kg coal car is coasting on a level track at 5. the long delay occurring because of the difficulty in understanding the role of microscopic and invisible forms of energy such as heat and infra-red light." Conservation of energy was discovered nearly two centuries after Newton's lifetime.75 m/s. the laws of conservation of momentum. Ans. energy. Ans.0° above the horizontal. respectively.27 A 1200-kg gun mounted on wheels shoots an 8. since they apply to both light and matter. suddenly 5000 kg of coal is dumped into it from directly above it.22 Sand drops at a rate of 2000 kg/min from the bottom of a hopper onto a belt conveyer moving horizontally at 250 m/min.00 m/s. 3. Determine the force needed to drive the conveyer.Relationship to the conservation laws In modern physics.6 8. neglecting friction.26 Two bodies of masses 8 kg and 4 kg move along the x-axis in opposite directions with velocities of 11 m/s — POSITIVE ^-DIRECTION and 7 m/s — NEGATIVES-DIRECTION . 5 m/s — POSITIVE S-DIRECTION 8. "Momentum. 3. Ans. This can be stated simply. energy and angular momentum cannot be created or destroyed. 8. Revision Exercises 1. and angular momentum are of more general validity than Newton's laws. Find the final speed of the car. Find their velocity just after collision. 139 N 8.00-kg projectile with a muzzle velocity of 600 m/s at an angle of 30. They collide and stick together. and R = semi-major axis Example .20 142 141 Saturn 29.24 0.06 0.88 1.n LAW 3: The squares of the periods of the planets are proportional to the cubes of their semimajor axes: Ta2 / Tb2 = Ra3 / Rb3 Square of any planet's orbital period (sidereal) is proportional to cube of its mean distance (semi-major axis) from Sun Mathematical statement: T = kR3/2 .72 0. Friction between solid objects is often referred to as dry friction or sliding friction and between a solid and a gas or liquid as fluid friction. It is not a fundamental force.g.00 1.54 870 868 Friction Friction is the force resisting the relative motion of two surfaces in contact or a surface in contact with a fluid (e. but instead must be found empirically. or heat. and so cannot be calculated from first principles.06 Venus 0. where T = sideral period. and the mathematical relation becomes T2 =k R3 Examples of Kepler's Third Law Planet P (yr) a (AU) T2 R3 Mercury 0.39 0.9 5. as it is derived from electromagnetic forces between atoms and electrons.62 0.If a is measured in astronomical units (AU = semi-major axis of Earth's orbit) and sidereal period in years (Earth's sidereal period). When contacting surfaces move relative to each other.00 1. then the constant k in mathematical expression for Kepler's third law is equal to 1.00 1. air on an aircraft or water in a pipe). Both of these types of friction are called kinetic friction.52 3. Page 21 of 64 .51 Jupiter 11. the friction between the two objects converts kinetic energy into thermal energy.5 9.00 Mars 1.39 0.37 Earth 1.53 3. or. Coulomb friction Coulomb friction. where μk is the coefficient of kinetic friction. however. Friction should not be confused with traction. ice on steel has a low coefficient of friction (the two materials slide past each other easily). the Coulomb approximation provides a threshold value for this force.and nano-scale objects where surface area forces dominate inertial forces. The coefficient of friction depends on the materials used. where μs is the coefficient of static friction. Coefficient of friction The coefficient of friction (also known as the frictional coefficient) is a dimensionless scalar value which describes the ratio of the force of friction between two bodies and the force pressing them together. For surfaces in relative motion μ = μk. This is usually larger than its kinetic counterpart. and the frictional force on each surface is exerted in the direction opposite to its motion relative to the other surface. Fn is the normal force exerted between the surfaces For surfaces at rest relative to each other μ = μs. It is described by the equation: Ff = μFn where Ff is either the force exerted by friction. it balances the net force tending to cause such motion. above which motion would commence. in the static case. Surface area does not affect friction significantly. a tire on concrete may have a coefficient of friction of 1. The Coulomb friction is equal to Ff. is a model to describe friction forces. rather than providing an estimate of the actual frictional force. in the case of equality. while rubber on pavement has a high coefficient of friction (the materials do not slide past each other easily). The Coulomb friction may take any value from zero up to Ff. for example.7. the maximum possible magnitude of this force. sliding friction is caused not by surface roughness but by chemical bonding between the surfaces. Internal friction is the motion-resisting force between the surfaces of the particles making up the substance. which is an empirical property of the contacting materials. Thus. named after Charles-Augustin de Coulomb. and the direction of the frictional force against a surface is opposite to the motion that surface would experience in the absence of friction. In this case. Page 22 of 64 . do affect sliding friction for micro. Surface roughness and contact area. the frictional force is exactly what it must be in order to prevent motion between the surfaces.Contrary to many popular explanations. μ is the coefficient of friction. but in traction it is essential. Coefficients of friction range from near zero to greater than one – under good conditions. A common way to reduce friction is by using a lubricant. a recently-discovered effect. acoustic lubrication actually uses sound as a lubricant. air cushion or roller bearing can change sliding friction into a much smaller type of rolling friction.Reducing friction Devices such as tires. - There are threeNewton’s laws of motion Viz: Every body persists in its state of being at rest or of moving uniformly straight forward. turbulent fluids or powdery solids such as graphite and talc. which is placed between the two surfaces. Lubricants to overcome friction need not always be thin. approaching zero levels. HDPE and PTFE are commonly used for low friction bearings. To every action there is always an equal and opposite reaction: or the forces of two bodies on each other are always equal and are directed in opposite directions. There are three Kepler’s laws of motion Viz: LAW 1: The orbit of a planet/comet about the Sun is an ellipse with the Sun's center of mass at one focus LAW 2: A line joining a planet/comet and the Sun sweeps out equal areas in equal intervals of time LAW 3: The squares of the periods of the planets are proportional to the cubes of their semimajor axes: Page 23 of 64 . especially to industrial or commercial objectives. or grease. and happens along the straight line on which that impulse is impressed. Many thermoplastic materials such as nylon. ball bearings. A very small amount of frictional energy would still be dissipated. Superlubricity. Lubricant technology is when lubricants are mixed with the application of science. such as oil. The change of momentum of a body is proportional to the impulse impressed on the body. except insofar as it is compelled to change its state by force impressed. has been observed in graphite: it is the substantial decrease of friction between two sliding objects. The science of friction and lubrication is called tribology. They are especially useful because the coefficient of friction falls with increasing imposed load. water. Summary S We have basically tackled three major issues in this TOPIC. often dramatically lessening the coefficient of friction. A vector quantity can not be added to scalar quantity. Box 861. the horizontal distance the book traveled before hitting the ground d. Discuss the methods of reducing friction. Written Assignments PHY110/1 Written Assignment 1.15 m above the ground. Narok. Do the following assignment and post it to: The Head Department of PHYSICS MMU P.Note: 1. the time the book was in the air c. 1. 3.8 m/s (60. the horizontal and vertical components of the book's velocity the instant I released it b.0 mph) I accidentally dropped the Encyclopedia of Physics out the window.Kenya While driving on the interstate one day at 27. Newton’s and Kepler’s laws are applicable to macroscopic bodies. 2. the horizontal and vertical components of the book's velocity the instant it hit the ground Page 24 of 64 .O. Determine the following … a. What is the difference between ductile and melliable materials? PHASES OF MATTER. so the solid keeps its shape. The atoms or molecules are tightly packed. The arrangement of particles in a solid are in a regular. There are four phases of matter. SPECIFIC OBJECTIVES o At the end of this TOPIC you should be able to: 1. 1. What is Bernaulli’s equation? 4. This enables us to select materials for different purposes and to deeply understand properties of materials such as elasticity. Plasma The state of matter depends on the motion of the molecules that make it up. Gas 4.TOPIC TWO : PROPRTIES OF MATTER SELF DIAGNOSTIC TEST Answer all questions ? 100 1. Identify the phases of matter 2. Page 25 of 64 . Solids: Solids are objects that have definite shapes and volumes. Explain laminar and turbulent motion of a fluid INTRODUCTION In this topic we shall discuss the properties of materials. Define Hooke’s laws 2. Liquid 3. Solid 2. surface tension and turbulency. Define elasticity. 3. Explain the term elasticity 3. brittleness and ductility. We wil the consider fluid flow through pipes and relationships that the flow obeys. Explain factors that afffect surface tension. rigidity. repeating pattern called a crystal. viscosity and streamline flow of fluid. Six Common Phase Changes 1. THE SPECIFIC GRAVITY (sp gr) of a substance is the ratio of the density of the substance to the density of some standard substance. while for gases. Plasma: Plasma is a gas-like mixture of positively and negatively charged particles. The particles of a gas have much more energy than either solids or liquids and can move around freely. Gases: Agas does not have a definite shape or volume.) 4. and in fluorescent lighting. Evaporation – substance changes from a liquid to a gas. Freezing – temperature at which a substance changes from a liquid to a solid.substance changes from a gas or vapor to a liquid. such as the sun. The standard is usually water (at 4°C) for liquids and solids. The density of water is close to 1000 kg/m3. (Heat of Vaporization. making up 99% of all matter. Liquids have no definite shape and take on the shape of the container that they are in. it is usually air. Page 26 of 64 . Condensation. Plasma is the most commonly found element in the universe. 3. 5. 2.energy a substance must absorb in order to change from a liquid to a gas. Melting. but are able to move around more freely than in a solid. although g/cm3 is also used: 1000 kg/m3 = 1 g/cm3. It is found in stars. Deposition – substance changes directly into a solid without first changing to a liquid (exothermic) DENSITY AND ELASTICITY THE MASS DENSITY (ρ) of a material is its mass per unit volume: mass of body volume of body The SI unit for mass density is kg/m3.Liquids: The particles in a liquid are close together. Sublimation – substance changes from a solid to a gas or vapor without changing to a liquid first (endothermic) 6. Plasma occurs when temperatures are high enough to cause particles to collide violently and be ripped apart into charged particles.temperature at which a substance changes from solid to liquid. A large modulus means that a large stress is required to produce a given strain . ELASTICITY is the property by which a body returns to its original size and shape when the forces that deformed it are removed. and not on its dimensions or configuration. It is measured as the ratio of the change in some dimension of a body to the original dimension in which the change occurred. is defined as modulus of elasticity= stress strain The modulus has the same units as stress. THE STRESS (σ) experienced within a solid is the magnitude of the force acting (F). the normal strain under an axial load is the change in length (AL) over the original length L 0: = L Lo Strain has no units because it is a ratio of like quantities. Unlike the constant k in Hooke's Law. Page 27 of 64 . Young's modulus is an important basic measure of the mechanical behavior of materials. the body will not return exactly to its original state after the stress is removed. YOUNG'S MODULUS (Y) or the modulus of elasticity. if a cane supports a load the stress at any point within the cane is the load divided by the cross-sectional area at that point. the value of Y depends only on the material of the wire or rod. Thus. STRAIN (‹ε) is the fractional deformation resulting from a stress. the narrowest regions experience the greatest stress. F Y= A FLo L AL Lo Its SI unit is Pa. C onsequently. it has the same value for all systems of units. divided by the area (A) over which it acts: stress= force area of surface on which force acts Its SI unit is the pascal (Pa).sp gr= standard Since sp gr is a dimensionless ratio. where 1 Pa = 1 N/m2. strain= change in dimension original dimension Thus. Accordingly. The exact definition of strain for various situations is given later. When a stress in excess of this limit is applied.the object is rigid. THE ELASTIC LIMIT of a body is the smallest stress that will produce a permanent distortion in the body. Page 28 of 64 . These shearing forces distort the block as indicated. as shown in Fig. 12-1. We then define volume stress=ΔP and volume strain=- Then Bulk modulus = V Vo V P stress P and then B = o V strain V Vo The minus sign is used so as to cancel the negative numerical value of ΔF and thereby make B a positive number. Suppose that a uniformly distributed compressive force acts on the surface of an object and is directed perpendicular to the surface at all points. where ΔV will be negative. shearing strain= distancesheared distance between surfaces shearing modulus= F stress A FLo or S L strain AL Lo Definition Of Mechanical Properties Tensile Strength Compressive Strength This is the ability of a material to withstand tensile loads without rupture when the material is in tension This is the ability of a material to withstand Compressive (squeezing) loads without being crushed when the material is in compression .THE BULK MODULUS (B) describes the volume elasticity of a material. we define pressure on A=P= F A The SI unit for pressure is Pa. THE SHEAR MODULUS (S) describes the shape elasticity of a material. but its volume remains unchanged. Suppose. The reciprocal of the bulk modulus is called the compressibility K of the substance. We define s F A. The pressure increase causes a volume change ΔV. Suppose that the pressure on an object of original volume F 0 is increased by an amount A/ 3 . The bulk modulus has the units of pressure. Then if F is the force acting on and perpendicular to an area A. s L Then shear modulus Lo . that equal and opposite tangential forces F act on a rectangular block. 33 Determine the fractional change in volume as the pressure of the atmosphere (1 x 10 5 Pa) around a metal block is reduced to zero by placing the block in vacuum. All materials which are formed by drawing are required to be ductile This is the property of a material to deform permanently under the application of a compressive load. This is the ability of a material to withstand shatter.Shear Strength Toughness Elasticity Plasticity Ductility Malleability Fatigue Strength Hardness This is the ability of a material to withstand offset or traverse loads without rupture occurring .1 12.32 A platform is suspended by four wires at its corners. Ans. when subjected to a pressure of 20 MPa. How far will the platform drop (due to elongation of the wires) if a 50-kg load is placed at the center of the platform? Ans. A ductile material combines the properties of plasticiy and tensile strength. A material which easily shatters is brittle. -10mm 3 . 73 μ m 12. Page 29 of 64 Ans. The bulk modulus for copper is 125 GPa. 40 mm on each edge.0 mm. Toughness indicates the ability of a material to absorb energy This is the ability of a material to deform under load and return to its original size and shape when the load is removed. Young's modulus for the material of the wires is 180 GPa. 0. The property is required for springs This is the property of a material to deform permanently under the application of a load. The wires are 3.g Diamonds are very hard.60 cm in diameter. This is the property of a material to withstand continuously varying and alternating loads This is the property of a material to withstand indentation and surface abrasion by another hard object.0 m long and have a diameter of 2.34 Compute the volume change of a solid copper cube. How much will the rod stretch? Y = 190 GPa for steel. It is an indication of the wear resistance of a material. 8 x 10-7 12. This is the exact opposite to elasticity. Ans. This is ability of a material to stretch under the application of tensile load and retain the deformed shape on the removal of the load. Plastacine is plastic.31 A load of 50 kg is applied to the lower end of a steel rod 80 cm long and 0. The bulk modulus for the metal is 125 GPa. Revision Exercises 2.e. A material which is forged to its final shape is required to be malleable.65 mm 12. m)/(m2 . i. There is no heat energy transferred across the boundaries of the pipe to the fluid as either a heat gain or loss. Note that the unit of pressure even can be expressed at a unit of energy density. it has stored energy. Page 30 of 64 . There is no energy loss due to friction between the fluid and the wall of the pipe. The fluid flow is laminar and steady state. This is the same as saying that the energy of a unit mass of the fluid does not change as it flow through the pipe system. There are no pumps in the section of pipe under consideration. The magnitude of the pressure P is equal to the Potential Energy per unit volume due to the Hydrostatic Pressure in the fluid. Pa = N/m2=(N.BERNOULLI'S EQUATION: FOR IDEAL FLUID FLOW Assumptions: The fluid is incompressible and nonviscous.e. 1 1 p1 1v12 1gh1 p2 2 v22 2 gh2 2 2 Bernoulli's Equation is basically a statement of the conservation of energy per unit volume along the pipe. E Energy Density or Energy per unit Volume (SI: J/m3): V For an ideal fluid flow the energy density is the same at all locations along the pipe. 1 Cons tan t P v 2 gh 2 * A compressed fluid or gas has the ability to do work if it is allowed to expand. m) = J/m3. 6 m/s. Explain the equation of continuity. has a mass of 1.0 m /s. What are the theoretical (a) velocity of the water at the point of discharge and (b) power delivered by the pump.31 A pump lifts water at the rate of 9. Ans.0 liters/s from a lake through a 5.195 13.6 g/cm3)? Ans. At one place the pressure is 130 kPa and the speed is 0. 21 m/s 14. Which is it? Ans. Ans. Page 31 of 64 . which may be either bronze (sp gr 8. is just the pressure of a fluid due to its weight.0 hp 14.11 g in water.0 m below the surface of water in a large tank. Ans: 1000cm3 Revision Questions (fluid in motion) 2. (b) 2.29 Calculate the theoretical velocity of efflux of water from an aperture that is 8. What are the factors which affect surface tension of a fluid? 2.32 Water flows steadily through a horizontal pipe of varying cross-section. ρgh.d.* The kinetic energy density can be though of a pressure exerted by the fluid due to its motion.42 A spring.2 13. 39 hp 14.26 g when measured in air and 1. 90 k Pa. Ans..0 cm i.0 m 3 of water per minute into a water main at a pressure of 220 kPa? Ans. the cube rises 2.43 What fraction of the volume of a piece of quartz (p = 2. Determine the pressure at another place in the same pipe where the speed is 9. Revision Questions (fluid at rest) 2.3 14.4).8) or brass (sp gr 8. When the mass is removed. * We have already seen that gravitational potential energy density.47 A cube of wood floating in water supports a 200-g mass resting on the center of its top face. if an added pressure of 140 kPa is applied to the surface of the water.00 cm. WRITTEN EXERCISE 2. Pipe and discharges it into the air at a point 16 m above the level of the water in the lake. (a) 4.60 m/s.65 g/cm3) will be submerged when it is floating in a container of mercury (p = 13.30 What horsepower is required to force 8.1 1. 0. brass 13. Determine the volume of the cube. Explain why tree branches bend towards the road on a busy tarmac road used by motor vehicles 3. Liquid . modulus of elasticity= . Note: Written Assignments PHY110/1 Do the following revision exercices and post it to: Written Assignment The Head Department of PHYSICS MMU P.The specific gravity (sp gr) of a substance is the ratio of the density of the substance to the density of some standard substance. is defined as stress strain Bernoulli's Equation is basically a statement of the conservation of energy per unit volume along the pipe.Summary S We have basically tackled three major issues in this TOPIC.Strain has no units because it is a ratio of like quantities.O. Box 861. the exact definition of strain for various situations is given later. the standard is usually water (at 4°c) for liquids and solids. Narok. it is usually air . Gas and Plasma The mass density (ρ) of a material is its mass per unit volume: mass of body volume of body Young's modulus (Y) or the modulus of elasticity. There are four phases of matter: Solid .Kenya Page 32 of 64 . while for gases. Discuss construction of temperature scales 3. while a volume expansion coefficient is more useful for a liquid or a gas. What is the difference between radiation and conduction methods of heat transfers? INTRODUCTION Thermal expansion. and the crystal will change shape as the temperature changes. SPECIFIC OBJECTIVES o At the end of this TOPIC you should be able to: 1. there may be different expansion coefficients for different crystallographic directions. In a solid or liquid. The symbol is °R). zero on the Rankine scale is absolute zero. who proposed it in 1859.Explain the mechanisms of heat transfer Rankine scale Rankine is a thermodynamic (absolute) temperature scale named after the Scottish engineer and physicist William John Macquorn Rankine.67 Page 33 of 64 . the expansion will be uniform in all dimensions of the crystal.TOPIC THREE : THERMAL PHYSICS SELF DIAGNOSTIC TEST Answer all questions ? 100 1. If it is not isometric.Distinguish between Cp and Cv 4. the general increase in the volume of a material as its temperature is increased. 2. rather than the one degree Celsius used by the Kelvin scale. Show how you can convert temperature from celcius to Rankin scale 3. there is a dynamic balance between the cohesive forces holding the atoms or molecules together and the conditions created by temperature. higher temperatures imply greater distance between atoms. a linear expansion coefficient is usually employed in describing the expansion of a solid.67°R is precisely equal to 0°F. Describe expansion in matter 2. Different materials have different bonding forces and therefore different expansion coefficients. A temperature of 459. Rankine temperature conversion formulas To find Fahrenheit From Formula Rankine °F = °R − 459. It is usually expressed as a fractional change in length or volume per unit temperature change. State the assumptions of the kinetic theory of gases 4. arrange the metals you know in order of their expansivities. If a crystalline solid is isometric (has the same structural configuration throughout). But the Rankine degree is defined as equal to one degree Fahrenheit. As with the Kelvin scale (symbol: K). the bulb on a mercury thermometer) in which some physical change occurs with temperature.g.65 K (−272.) Page 34 of 64 . The most recent official temperature scale is the International Temperature Scale of 1990. For primary thermometers the measured property of matter is known so well that temperature can be calculated without any unknown quantities. Thermometers increasingly use electronic means to provide a digital display or input to a computer.8 Celsius Rankine °C = (°R ÷ 1.15 Rankine Celsius °R = (°C + 273. triple points and superconducting transitions. 1.5 °C. It extends from 0. 1 °F = 1 °R and 1 kelvin = 1. (Note that the boiling point of water varies with pressure.8 Rankine kelvin °R = K × 1.8) – 273. Thermometers can be divided into two separate groups according to the level of knowledge about the physical basis of the underlying thermodynamic laws and quantities.g. A thermometer has two important elements: the temperature sensor (e. the scale on a mercury thermometer). They have to be calibrated against a primary thermometer at least at one temperature or at a number of fixed temperatures.15) × 1. for example.67 kelvin Rankine K = °R ÷ 1.8 °R THERMOMETERS A thermometer is a device that measures temperature or temperature gradient using a variety of different principles. CALIBRATION OF THERMOMETERS Thermometers can be calibrated either by comparing them with other certified thermometers or by checking them against known fixed points on the temperature scale. For secondary thermometers knowledge of the measured property is not sufficient to allow direct calculation of temperature.085 °C. they are often much more sensitive than primary ones. There is an absolute thermodynamic temperature scale. plus some means of converting this physical change into a value (e. Secondary thermometers are most widely used because of their convenience.8 For temperature intervals rather than specific temperatures.985 °F).5 °F) to approximately 1.358 K (1. so this must be controlled.Rankine Fahrenheit °R = °F + 459. Such fixed points. The best known of these fixed points are the melting and boiling points of pure water. based on fixed points and interpolating thermometers. occur reproducibly at the same temperature. −458. Internationally agreed temperature scales are designed to approximate this closely. Also. Immerse the sensing portion in a stirred mixture of pure ice and water and mark the point indicated when it had come to thermal equilibrium. The linear thermal expansion can only be defined for solids. solids typically expand in response to heating and contract on cooling. Thermal expansion Some substances expand when cooled. As a result. When the stored energy increases. 2. 760. 3. The volumetric thermal expansion coefficient can be defined for both liquids and solids.The traditional method of putting a scale on a liquid-in glass or liquid-in-metal thermometer was in three stages: 1. and is common in engineering applications. the new length. It relates the change in temperature to the change in a material's linear dimensions. such as freezing water.Immerse the sensing portion in a steam bath at 1 standard atmosphere (101. so does the length of the molecular bonds. It is the fractional change in length per degree of temperature change. so they have negative thermal expansion coefficients.0 mmHg) and again mark the point indicated. COEFFICIENT OF THERMAL EXPANSION When the temperature of a substance changes. dL = L0 x ( alpha x dT ) where is the original length. Different coefficients of thermal expansion can be defined for a substance depending on whether the expansion is measured by: Specific heat linear thermal expansion area thermal expansion volumetric thermal expansion Compressibility These characteristics are closely related. . and Linear thermal expansion Page 35 of 64 the temperature.Divide the distance between these marks into equal portions according to the temperature scale being used. the energy that is stored in the intermolecular bonds between atoms changes. this dimensional response to temperature change is Material Properties expressed by its coefficient of thermal expansion. Thermal expansion coefficient The thermal expansion coefficient is a thermodynamic property of a substance.3 kPa. Thus. derivatives are taken at constant pressure . Note that the partial derivative of volume with respect to length as shown in the above equation is exact. however. Proof: This ratio arises because volume is composed of three mutually orthogonal directions.The linear thermal expansion is the one-dimensional length change with temperature. The volumetric thermal expansion coefficient can be written where is the temperature. Area thermal expansion The change in area with temperature can be written: For exactly isotropic materials. the area thermal expansion coefficient is very closely approximated as twice the linear coefficient. measures the fractional change in density as temperature increases at constant pressure. Volumetric thermal expansion The change in volume with temperature can be written. the volumetric thermal expansion coefficient is very closely approximated as three times the linear coefficient. is the density. one-third of the volumetric expansion is in a single axis (a very close approximation for small differential changes). in Page 36 of 64 . For exactly isotropic materials. is the volume. in an isotropic material. Therefore. while the internal energy U is a particular form of energy associated with the system. and is almost universally ignored. the term "work energy" for δW means "that amount of energy lost as the result of work". The only thing that can happen with energy in a closed system is that it can change form. but also shows a secondary term that scales as . The δ's before the heat and work terms are used to indicate that they describe an increment of energy which is to be interpreted somewhat differently than the dU increment of internal energy. when combined with a large change in temperature the differential change in length can become large enough that this factor needs to be considered. As the change in temperature increases. in an isolated system. Likewise. Although the coefficient of linear thermal expansion can be quite small. The last term. THE FIRST LAW OF THERMODYNAMICS For a thermodynamic system with a fixed number of particles. and as the value for the linear coefficient of thermal expansion increases. they must lose mass and thus eventually disappear over perpetual time. where δQ is the amount of energy added to the system by a heating process. which shows that a large change in temperature can overshadow a small value for the linear coefficient of thermal expansion. conservation of “energy” refers to the conservation of the total mass-energy. Today. If such machines produce more energy than is put into them. neither mass nor pure energy are conserved separately. δW is the amount of energy lost by the system due to work done by the system on its surroundings and dU is the change in the internal energy of the system. mass and "pure energy" can be converted to one another. A consequence of this law is that energy cannot be created nor destroyed. Albert Einstein's theory of relativity shows that energy can be converted to mass (rest mass) and mass converted to energy. for instance kinetic energy can become thermal energy. but the total amount of energy (which includes the energy of the mass of the system) remains constant. . the first law of thermodynamics may be stated as: . CONSERVATION OF ENERGY The law of conservation of energy states that the total amount of energy in a closed system remains constant. For non-negligible changes in volume: Note that this equation contains the main term. which includes energy of the rest mass. or equivalently. . Another consequence of this law is that perpetual motion machines can only work perpetually if they deliver no energy to their surroundings. and are therefore impossible. the error in this formula also increases.. The most significant result of this distinction is the fact that one can clearly state the amount of Page 37 of 64 .practice it is important to note that the differential change in volume is only valid for small changes in volume (i. as it was understood in pre-relativistic physics. Work and heat are processes which add or subtract energy. the expression is not linear).e. is vanishingly small. Thus the term "heat energy" for δQ means "that amount of energy added as the result of heating" rather than referring to a particular form of energy. Therefore. or specific heat. Temperature and entropy are also system variables. The heat energy may be written . We can overcome this problem by only allowing the body in question to absorb a very small amount of heat. each of which are system variables. For a simple compressible system. In the limit as the amount of absorbed heat becomes infinitesimal. we have tacitly assumed that the specific heat of a body is independent of its temperature. it is usual to define two specific heats. where P is the pressure and dV is a small change in the volume of the system.. only converted from one form to another. where T is the temperature and dS is a small change in the entropy of the system. this is not true. and its specific heat remains approximately constant. denoted cv 1 dQ v dT V and. this means that energy cannot be created or destroyed. Firstly. HEAT CAPACITY OR SPECIFIC HEAT Suppose that a body absorbs an amount of heat ΔQ and its temperature consequently rises byΔT. the work performed by the system may be written . the molar specific heat at constant pressure. In simple terms. but one cannot tell how much energy has flowed into or out of the system as a result of its being heated or cooled. secondly. nor as the result of work being performed on or by the system. of the body is C Q T If the body consists of moles of some substance then the molar specific heat (i. In general. so that its temperature only rises slightly. denoted 1 dQ cp v dT P Page 38 of 64 .internal energy possessed by a thermodynamic system. the molar specific heat at constant volume. the specific heat of one mole of this substance ) is defined c 1 Q v T In writing the above expressions. we obtain c 1 Q v T In classical thermodynamics. The usual definition of the heat capacity.e. and the first law of thermodynamics reduces to It follows that Now. for an ideal gas the internal energy is volume independent. Since is a function only of . and so the gas does work on its environment.Consider the molar specific heat at constant volume of an ideal gas. Thus. if the pressure is kept constant then the volume changes. the above expression implies that the specific heat at constant volume is also volume independent. Since . In general. the temperature changes by . and the pressure remains constant. The equation of state of an ideal gas tells us that if the volume changes by . According to the first law of thermodynamics. then The previous two equations can be combined to give Now. no work is done by the gas. we can write The previous two expressions can be combined to give for an ideal gas. Let us now consider the molar specific heat at constant pressure of an ideal gas. by definition Page 39 of 64 . 302 1.214 Revision Questions 3. less heat is available to increase the temperature of the gas. What is a for the material of the rod? Ans.1 1. Thus.666 Argon Ar 12.2 1.666 1.666 1.666 Nitrogen 20. In fact.5 1.) From Reif.13 15.so we obtain for an ideal gas.6 1.5 1.407 Oxygen 21. its temperature.7 x 10 -3 0C. whereas at constant pressure some of the absorbed heat is used to do work on the environment as the volume increases. γ is very easy to measure because the speed of sound in an ideal gas is written where ρ is the density.396 1. a = 1.7cm 15.3 1. as indicated by the above formula. Table 2: Specific heats of common gases in joules/mole/deg. 1. The ratio of the two specific heats cp cv is conventionally denoted γ. Ans. This means that. Note that at constant volume all of the heat absorbed by the gas goes into increasing its internal energy.220 1. hence. Table below lists some experimental measurements of cv and γ for common gases.1 x 10-6 °C Page 40 of 64 . we expect the specific heat at constant pressure to exceed that at constant volume.298 Ethane 39. The extent of the agreement between γ calculated andexperimental γ is quite remarkable. and.14 Compute the increase in length of 50 m of copper wire when its temperature changes from 12 °C to 32 °C. This is a very famous result. in the latter case. We have for an ideal gas. 5. (at 15 C and 1 atm. For copper. Gas Symbol (experiment) (experiment) (theory) Helium He 12.091 cm in length after a temperature rise of 60 °C.0 m long is found to have expanded 0.405 1.1 15.15 A rod 3.397 Carbon Dioxide 28. These room-temperature atoms have a certain. a= l.2 g/cm3 KINETIC THEORY OF GASES The temperature of an ideal monatomic gas is a measure related to the average kinetic energy of its atoms as they move.l0x -5 °C Ans 221 oC 15. The number of molecules is large such that statistical treatment can be applied. Page 41 of 64 . a bare wheel has a diameter of 30.30 g/cm 3 at 20. or volume. and the coefficient of linear expansion is 14. These molecules are in constant.19 Calculate the increase in volume of 100 cm3 of mercury when its temperature changes from 10 °C to 35 °C. They exert no forces on one another except during collisions. The collisions of gas particles with the walls of the container holding them are perfectly elastic. such as pressure. by considering their molecular composition and motion.45 cm . Kinetic theory (or kinetic theory of gases) attempts to explain macroscopic properties of gases. Ans. To what temperature must the rim be heated so as to slip over the wheel? For this type of steel.15 At 15.000 cm. Ans. The rapidly moving particles constantly collide with the walls of the container. In this animation. Essentially.0 °C. but due to collisions between molecules moving at different velocities.0°C.930 cm. and the inside diameter of a steel rim is 29. 19. The interactions among molecules are negligible. as was Isaac Newton's conjecture.000 18 °C-1 15. POSTULATES OF KINETIC THEORY OF GASES The theory for ideal gases makes the following assumptions: The gas consists of very small particles. average speed (slowed down here two trillion fold).23 The density of gold is 19.3 x Compute the density of gold at 90. the size of helium atoms relative to their spacing is shown to scale under 1950 atmospheres of pressure. temperature. random motion. all with non-zero mass. 0. Kinetic theory is also known as the kinetic-molecular theory (of gases) or the collision theory. the theory posits that pressure is due not to static repulsion between molecules.15.0 °C. The volume coefficient of expansion of mercury is 0. The average kinetic energy of the gas particles depends only on the temperature of the system. The equations of motion of the molecules are time-reversible. The molecules are perfectly spherical in shape. The particle impacts the wall once every 2l/vx time units (where l is the length of the container). This means that the inter-particle distance is much larger than the thermal de Broglie wavelength and the molecules are treated as classical objects. Relativistic effects are negligible. The time during collision of molecule with the container's wall is negligible as comparable to the time between successive collisions. Although the particle impacts a side wall once every 1l/vx time units. Quantum-mechanical effects are negligible. Pressure Pressure is explained by kinetic theory as arising from the force exerted by gas molecules impacting on the walls of the container. then the momentum lost by the particle and gained by the wall is: where vx is the x-component of the initial velocity of the particle. This is equivalent to stating that the average distance separating the gas particles is large compared to their size. When a gas molecule collides with the wall of the container perpendicular to the x coordinate axis and bounces off in the opposite direction with the same speed (an elastic collision). enclosed in a cuboidal container of volume V. and elastic in nature. The force due to this particle is: The total force acting on the wall is: where the summation is over all the gas molecules in the container. each of mass m. Consider a gas of N molecules. Page 42 of 64 . only the momentum change on one wall is considered so that the particle produces a momentum change on a particular wall once every 2l/vx time units. The total volume of the individual gas molecules added up is negligible compared to the volume of the container. in this case an average over all particles and where N is the number of particles in the box. adding the contributions from each direction we have: where the factor of two arises from now considering both walls in a given direction. we have the following expression for the pressure where V is the volume.The magnitude of the velocity for each particle will follow: Now considering the total force acting on all six walls. Thus the force can be written as: Pressure. which is force per unit area. Then we have Page 43 of 64 . where the bar denotes an average. as cross-sectional area multiplied by length is equal to volume. Assuming there are a large number of particles moving sufficiently randomly. This quantity is also denoted by where vrms is the root-mean-square velocity of the collection of particles. the force on each of the walls will be approximately the same and now considering the force on only one wall we have: The quantity can be written as . Thus. of the gas can then be written as: where A is the area of the wall of which the force exerted on is considered. As Nm is the total mass of the gas. Note that the product of pressure and volume is simply two thirds of the total kinetic energy. and from the above result we have then the temperature takes the form (2) which leads to the expression of the kinetic energy of a molecule The kinetic energy of the system is N time that of a molecule K 1 Nmv 2rms 2 The temperature becomes (3) Page 44 of 64 . 1 to the average (translational) kinetic energy per molecule mv 2rms which is a microscopic 2 property. a macroscopic property. Temperature and kinetic energy From the ideal gas law (1) where is the Boltzmann constant. and the absolute temperature. Then the V pressure is This result is interesting and significant. the density is mass divided by volume Nm . because it relates pressure. 2 meV Number of collisions with wall One can calculate the number of atomic or molecular collisions with a wall of a container per unit area per unit time. From Eq.(1) and Eq.(3)1. the product of pressure and volume per mole is proportional to the average (translational) molecular kinetic energy.7 yJ = 129 μeV At standard temperature (273.(1) and Eq.15 K). Assuming an ideal gas. Thus the kinetic energy per kelvin (monatomic ideal gas) is: per mole: 12. . the (5) In the kinetic energy per dof. we get: per mole: 3406 J per molecule: 5. the constant of proportionality of temperature is 1/2 times Boltzmann constant. Eq. a derivation results in an equation for total number of collisions per unit time per area: Page 45 of 64 . This result is related to the equipartition theorem. but the lighter gases act as if they have only 5.Eq.65 zJ = 35.(4) are called the "classical results". which could also be derived from statistical mechanics. we have (4) Thus. diatomic gases should have 7 degrees of freedom.47 J per molecule: 20. As noted in the article on heat capacity. Since there are degrees of freedom (dofs) in a monoatomic-gas system with kinetic energy per dof is particles.(3)1 is one important result of the kinetic theory: The average molecular kinetic energy is proportional to the absolute temperature. 10 mg of virus per mL? Ans. 1170 K « 900 °C HEAT TRANSFER Heat transfer is the transition of thermal energy from a hotter object to a cooler object ("object" in this sense designating a complex collection of particles which is capable of storing energy in many different ways).18 At what temperature will the molecules of an ideal gas have twice the rms speed they have at 20 °C? Ans.16 17.0 x 107 kg/kmol.15 A certain strain of tobacco mosaic virus has M = 4. The transfer of energy could be primarily by elastic impact as in fluids or by free electron diffusion as predominant in metals or phonon vibration as predominant in insulators. heat transfer between them can never be stopped. In liquids (except liquid metals) and gases. (b) 8 x 1019 17. or heat exchange.13 Find the mass of a neon atom. Heat transfer always occurs from a higher-temperature object to a cooler temperature one as described by the second law of thermodynamics or the Clausius statement. 1. Where there is a temperature difference between objects in proximity. Conduction is greater in solids. Conduction Conduction is the transfer of heat by direct contact of particles of matter.4 x 10 -5 kg/m3 17. 4. In other words. The most probable speed is 81.36 x 10 kg 17. When an object or fluid is at a different temperature than its surroundings or another object. and the mean speeds 92.5 x 1012 17.) Ans. How many molecules of the virus are present in 1.0 kg/kmol. Page 46 of 64 . heat is transferred by conduction when adjacent atoms vibrate against one another. and R is the gas constant. (a) What is the mass in kilograms of such a molecule? (b) How many such molecules would make up 2 g of polymer? Ans.2 17. If the temperature of the gas is 20 °C.1% (distribution of speeds).2 kg/kmol. T in kelvins. The molar mass is given as kg/mol. Revision questions 3. where atoms are in constant contact. Ans. also known as heat transfer. or as electrons move from atom to atom.200 mmHg. giving a lower chance of molecules colliding and passing on thermal energy.17 The pressure of helium gas in a tube is 0.6% of the rms speed.0 mL of a solution that contains 0. it can only be slowed. what is the density of the gas? (Use MHe = 4. transfer of thermal energy.RMS speeds of molecules From the kinetic energy formula it can be shown that with v in m/s. the molecules are usually further apart.14 A typical polymer molecule in polyethylene might have a molecular mass of 15 x 103. (a) 7. 3. The atomic mass of neon is 20.5 x 10^23 kg. occurs in such a way that the body and the surroundings reach thermal equilibrium. A heat pipe is a passive device that is constructed in such a way that it acts as though it has extremely high thermal conductivity. and molecular bonding. to ensure more nearly instantaneous effects of a change in comfort level setting. Conductivity of gases increases with temperature. temperature. platinum. The presence of bulk motion of fluid enhances the heat transfer between the solid surface and the fluid. ventilating and air-conditioning. in a direction normal to a surface of area (A). There are two types of Convective Heat Transfer: Natural Convection: is when the fluid motion is caused by buoyancy forces that result from the density variations due to variations of temperature in the fluid. conductivity increases only slightly with increasing pressure and density. Also it can be applied to HVAC (heating. This mode of analysis has been applied to forensic sciences to analyse the time of death of humans. This is due to the way that metals are chemically bonded: metallic bonds (as opposed to covalent or ionic bonds) have free-moving electrons which are able to transfer thermal energy rapidly through the metal.) are usually the best conductors of thermal energy. Thermal conductivity is a material property that is primarily dependent on the medium's phase. When this happens. Internal and external flow can also classify convection. whereas mass diffusion is mostly limited to fluids. fluids (and especially gases) are less conductive. k. Convection Convection is the transfer of heat energy between a solid surface and the nearby liquid or gas in motion. Internal flow occurs when the fluid is enclosed by a solid boundary such as a flow through a pipe. An external flow occurs when Page 47 of 64 . Metals (eg. only in as much as it can occur in solids. engineers employ the thermal conductivity. due to a temperature difference (ΔT). In thermal conductivity k is defined as "the quantity of heat. Conductivity increases with increasing pressure from vacuum up to a critical point that the density of the gas is such that molecules of the gas may be expected to collide with each other before they transfer heat from one surface to another. copper. Thus the hotter volume transfers heat towards the cooler volume of that fluid. This is due to the large distance between atoms in a gas: fewer collisions between atoms means less conduction. gold. the fluid is displaced vertically or horizontally while the cooler fluid gets denser and the fluid sinks. density. in the situation where there are no fluid currents.Heat conduction is directly analogous to diffusion of particles into a fluid. After this point in density. etc. As density decreases so does conduction. transmitted in time (t) through a thickness (L). This type of heat diffusion differs from mass diffusion in behaviour. Therefore. For example in the absence of a external source when the mass of the fluid is in contact with the hot surface its molecules separate and scatter causing the mass of fluid to become less dense. Q. To quantify the ease with which a particular medium conducts. As fluid motion goes more quickly the convective heat transfer increases. also known as the conductivity constant or conduction coefficient. iron. or building climate control). Forced Convection: is when the fluid is forced to flow over the surface by external source such as fans and pumps. It creates an artificially induced convection current. Therefore. Clothing and building surfaces. The h is the constant heat transfer coefficient which depends upon physical properties of the fluid such as temperature and the physical situation in which convection occurs. For laminar flows the heat transfer coefficient is rather low compared to the turbulent flows. Formulae and correlations are available in many references to calculate heat transfer coefficients for typical configurations and fluids. Both these convections. q = hA(Ts − Tb) A is the surface area of heat transfer. the two kinds of emission are simply different "colours" of electromagnetic radiation. either natural or forced. For example. fresh snow. However Tb varies with each situation and is the temperature of the fluid “far” away from the surface. For any body the reflectivity depends on the wavelength distribution of incoming electromagnetic radiation and therefore the temperature of the source of the radiation. The emissivity depends on the wave length distribution and therefore the temperature of the body itself. is 0. and radiative transfer Lighter colours and also whites and metallic substances absorb less illuminating light. this is due to turbulent flows having a thinner stagnant fluid film layer on heat transfer surface. All objects with a temperature above absolute zero radiate energy at a rate equal to their emissivity multiplied by the rate at which energy would radiate from them if they were a black body. can be internal or external as they are independent of each other. The energy from the Sun travels through the vacuum of space before warming the earth. The formula for Rate of Convective Heat Transfer. Both reflectivity and emissivity of all bodies is wavelength dependent. but rather in the far infrared. The temperature determines the wavelength distribution of the electromagnetic radiation as limited in intensity by Planck’s law of black-body radiation. at a temperature of about -5C. Radiation Radiation is the transfer of heat energy through empty space. since the dominant emitted wavelengths are nowhere near the visible spectrum.the fluid extends indefinitely without encountering a solid surface. but otherwise colour makes little difference as regards heat transfer between an object at everyday temperatures and its surroundings. The difference between visible light and the radiation from objects at conventional temperatures is a factor of about 20 in frequency and wavelength. Visible light is simply another form of electromagnetic radiation with a shorter wavelength (and therefore a higher frequency) than infrared radiation. which is highly reflective to visible light. Ts is the surface temperature and while Tb is the temperature of the fluid at bulk temperature.90) appears white due to reflecting sunlight with a peak energy wavelength of about 0. peak energy wavelength of about 12 micrometres. Gases absorb and emit energy in characteristic wavelength patterns that are different for each gas. however. radiation works even in and through a perfect vacuum. and thus heat up less. No medium is necessary for radiation to occur.99. Emissivities at those wavelengths have little to do with Page 48 of 64 . Its emissivity.5 micrometres. the heat transfer coefficient must be derived or found experimentally for every system analyzed. (reflectivity about 0. visual emissivities (visible colours); in the far infrared, most objects have high emissivities. Thus, except in sunlight, the colour of clothing makes little difference as regards warmth; likewise, paint colour of houses makes little difference to warmth except when the painted part is sunlit. The main exception to this is shiny metal surfaces, which have low emissivities both in the visible wavelengths and in the far infrared. Such surfaces can be used to reduce heat transfer in both directions; an example of this is the multi-layer insulation used to insulate spacecraft. Low-emissivity windows in houses are a more complicated technology, since they must have low emissivity at thermal wavelengths while remaining transparent to visible light. Newton's law of cooling A related principle, Newton's law of cooling, states that the rate of heat loss of a body is proportional to the difference in temperatures between the body and its surroundings, or environment. The law is Q = Thermal energy in joules h = Heat transfer coefficient A = Surface area of the heat being transferred T0 = Temperature of the object's surface Tenv = Temperature of the environment This form of heat loss principle is sometimes not very precise; an accurate formulation may require analysis of heat flow, based on the (transient) heat transfer equation in a nonhomogeneous, or else poorly conductive, medium. An analog for continuous gradients is Fourier's Law. In such cases, the entire body is treated as lumped capacitance heat reservoir, with total heat content which is proportional to simple total heat capacity C , and T, the temperature of the body, or Q = C T. From the definition of heat capacity C comes the relation C = dQ/dT. Differentiating this equation with regard to time gives the identity (valid so long as temperatures in the object are uniform at any given time): dQ/dt = C (dT/dt). This expression may be used to replace dQ/dt in the first equation which begins this section, above. Then, if T(t) is the temperature of such a body at time t , and Tenv is the temperature of the environment around the body: where r = hA/C is a positive constant characteristic of the system, which must be in units of 1/time, and is therefore sometimes expressed in terms of a characteristic time constant t0 given by: r = 1/t0 = ΔT/[dT/dt] . Thus, in thermal systems, t0 = C/hA. (The total heat capacity C of a system may be further represented by its mass-specific heat capacity cp multiplied by its mass m, so that the time constant t0 is also given by mcp/hA). One dimensional application, using thermal circuits A very useful concept used in heat transfer applications is the representation of thermal transfer by what is known as thermal circuits. A thermal circuit is the representation of the Page 49 of 64 resistance to heat flow as though it were an electric resistor. The heat transferred is analogous to the current and the thermal resistance is analogous to the electric resistor. The value of the thermal resistance for the different modes of heat transfer are calculated as the denominators of the developed equations. The thermal resistances of the different modes of heat transfer are used in analyzing combined modes of heat transfer. The equations describing the three heat transfer modes and their thermal resistances, as discussed previously are summarized in the table below: In cases where there is heat transfer through different media (for example through a composite), the equivalent resistance is the sum of the resistances of the components that make up the composite. Likely, in cases where there are different heat transfer modes, the total resistance is the sum of the resistances of the different modes. Using the thermal circuit concept, the amount of heat transferred through any medium is the quotient of the temperature change and the total thermal resistance of the medium. As an example, consider a composite wall of cross- sectional area A. The composite is made of an L1 long cement plaster with a thermal coefficient k1 and L2 long paper faced fiber glass, with thermal coefficient k2. The left surface of the wall is at Ti and exposed to air with a convective coefficient of hi. The Right surface of the wall is at To and exposed to air with convective coefficient ho. Page 50 of 64 Using the thermal resistance concept heat flow through the composite is as follows: Insulation and radiant barriers Thermal insulators are materials specifically designed to reduce the flow of heat by limiting conduction, convection, or both. Radiant barriers are materials which reflect radiation and therefore reduce the flow of heat from radiation sources. Good insulators are not necessarily good radiant barriers, and vice versa. Metal, for instance, is an excellent reflector and poor insulator. The effectiveness of an insulator is indicated by its R- (resistance) value. The R-value of a material is the inverse of the conduction coefficient (k) multiplied by the thickness (d) of the insulator. The units of resistance value are in SI units: (K·m²/W) ; Rigid fiberglass, a common insulation material, has an R-value of 4 per inch, while poured concrete, a poor insulator, has an R-value of 0.08 per inch. The effectiveness of a radiant barrier is indicated by its reflectivity, which is the fraction of radiation reflected. A material with a high reflectivity (at a given wavelength) has a low emissivity (at that same wavelength), and vice versa (at any specific wavelength, Page 51 of 64 Page 52 of 64 . the larger the corresponding thermal resistance (R) value. not dependent upon the type of radiation which is incident upon it. one would add insulating materials i. The units of thermal conductivity(k) are W·m-1·K-1 (watts per meter per kelvin). In simple cylindrical pipes: BLACKBODY RADIATION "Blackbody radiation" or "cavity radiation" refers to an object or system which absorbs all radiation incident upon it and re-radiates energy which is characteristic of this radiating system only. therefore increasing width of insulation (x meters) decreases the k term and as discussed increases resistance. outer radius increases and therefore surface area increases.e with low thermal conductivity (k). satellites use multi-layer insulation which consists of many layers of aluminized (shiny) mylar to greatly reduce radiation heat transfer and control satellite temperature. In space vacuum. The radiated energy can be considered to be produced by standing wave or resonant modes of the cavity which is radiating. An ideal radiant barrier would have a reflectivity of 1 and would therefore reflect 100% of incoming radiation. Critical insulation thickness To reduce the rate of heat transfer. The smaller the k value. This follows logic as increased resistance would be created with increased conduction path (x).emissivity). However. adding this layer of insulation also has the potential of increasing the surface area and hence thermal convection area (A). The point where the added resistance of increasing insulation width becomes overshadowed by the effects of surface are is called the critical insulation thickness. Vacuum bottles (Dewars) are 'silvered' to approach this. An obvious example is a cylindrical pipe: As insulation gets thicker.reflectivity = 1 . Ans. k T is 0.3 19.The amount of radiation emitted in a given frequency range should be proportional to the number of modes in that range. But the predicted continual increase in radiated energy with frequency (dubbed the "ultraviolet catastrophe") did not happen.4 x 10 3 kcal/h 19. Why are heat and work not thermodynamic quantity? Revision Questions 3.46 mm Summary S We have basically tackled three major issues in this TOPIC. Note: Page 53 of 64 .15 A sphere of 3.80 W/K-m). Ans. It is in equilibrium with its surroundings and absorbs 30 kW of power radiated to it from the surroundings. and both have the same area. Nature knew better. . How thick is the glass if the glass-brass interface is at 65 °C? Ans.16 A 2.10 A single-thickness glass window on a house actually has layers of stagnant air on its two surfaces. Explain why water has the highest heat capacity? 1.6 x 103 K 19. Thermometers have been built which utilise a range of physical effects to measure temperature. Most thermometers are originally calibrated to a constant-volume gas thermometer.20 cm thick.33 kg/h-cm 2 19.84 W/K-m.0 mm window each hour on a day when the outside temperature was precisely 0°C and the inside temperature was 18°C? For glass. 0.0 cm thick brass plate (kT = 105 W/K-m) is sealed to a glass sheet (kT = 0. 0. 2. or equivalently. if the temperature difference between the plate faces is 100 °C? For steel. Convert 456oR to oC 3. The exposed face of the brass plate is at 80 °C.11 How many grams of water at 100 °C can be evaporated per hour per cm 2 by the heat transmitted through a steel plate 0. 1. The best of classical physics suggested that all modes had an equal chance of being produced. how much heat would flow out of an 80 cm x 40 cm x 3. But if it did not. kT is 42 W/K-m. What is the temperature of the sphere? Ans.0 cm radius acts like a blackbody. while the exposed face of the glass is at 20 °C. What is black body radiation? 2. and that the number of modes went up proportional to the square of the frequency. . the liquid expands more thean the glass.O.In a Liquid in glass thermometer. Narok. Box 861.Kenya Page 54 of 64 . Written Assignments PHY110/1 Written Assignment Do the following assignment and post it to: The Head Department of PHYSICS MMU P. or gas. or earthquake. Distinguish between sound and light waves. transmit. Distinguish between sound and light waves 2. such as fire. Many species. hearing is normally limited to frequencies between about 12 Hz and 20. Furthermore. these have evolved to produce song and speech. dogs can perceive vibrations higher than 20 kHz. surf. although these limits are not definite. INTRODUCTION SPECIFIC OBJECTIVES o At the end of this TOPIC you should be able to: 1. For example. Earth's atmosphere. wind. SOUND Sound is a travelling wave which is an oscillation of pressure transmitted through a solid. In some species. composed of frequencies within the range of hearing and of a level sufficiently strong to be heard. liquid. have also developed special organs to produce sound. marine and terrestrial mammals. navigation. and broadcast sound. State the properties of sound 2. What is ultrasonic? 3. and communication.TOPIC FOUR: SOUND SELF DIAGNOSTIC TEST Answer all questions ? 100 1. The upper limit generally decreases with age. such as frogs. birds. PERCEPTION OF SOUND For humans. or the sensation stimulated in organs of hearing by such vibrations. produces (and is characterized by) its unique sounds. humans have developed culture and technology (such as music. Other species have a different range of hearing. predation. Page 55 of 64 . sound is used by many species for detecting danger.000 Hz (20 kHz). rain. As a signal perceived by one of the major senses. Identify three characteristics of sound waves. 3. and virtually any physical phenomenon. water. telephone and radio) that allows them to generate. Explain the characteristics of sound waves. record. or wavelength and direction are combined as a wave vector). speed. have an additional property of polarization. period. the speed of sound is approximately 343 m/s (1. however. wavelength. Through solids. causing local regions of compression and rarefaction. Speed of sound The speed of sound depends on the medium through which the waves are passing. the bottom waves have higher frequencies than those above. Sound characteristics can depend on the type of sound waves (longitudinal versus transverse) as well as on the physical properties of the transmission medium. the speed of sound in gases depends on temperature. Matter in the medium is periodically displaced by a sound wave. and liquids as longitudinal waves. The horizontal axis represents time. The energy carried by the sound wave converts back and forth between the potential energy of the extra compression (in case of longitudinal waves) or lateral displacement strain (in case of transverse waves) of the matter and the kinetic energy of the oscillations of the medium. Page 56 of 64 . Sound cannot travel through vacuum. which are frequency. Transverse waves. liquids. In 20 °C (68 °F) air at the sea level. Longitudinal sound waves are waves of alternating pressure deviations from the equilibrium pressure. and is often quoted as a fundamental property of the material. Those physical properties and the speed of sound change with ambient conditions. while transverse waves (in solids) are waves of alternating shear stress at right angle to the direction of propagation. Sound wave properties and characteristics Sound waves are characterized by the generic properties of waves. also called compression waves. solids.PHYSICS OF SOUND The mechanical vibrations that can be interpreted as sound are able to travel through all forms of matter: gases. Sound is transmitted through gases. In general. amplitude. intensity. LONGITUDINAL AND TRANSVERSE WAVES Sinusoidal waves of various frequencies. For example. and thus oscillates. the speed of sound is proportional to the square root of the ratio of the elastic modulus (stiffness) of the medium to its density.230 km/h. The matter that supports the sound is called the medium. also known as shear waves. and direction (sometimes speed and direction are combined as a velocity vector. it can be transmitted as both longitudinal and transverse waves. and plasmas. plasma. In fresh water. noise is an undesirable component that obscures a wanted signal. In steel.1-1994. Sound pressure level Sound pressure is defined as the difference between the average local pressure of the medium outside of the sound wave in which it is traveling through (at a given point and a given time) and the pressure found within the sound wave itself within that same medium. A square of this difference (i. and reflection of sound waves is called acoustics. Examples of sound pressure and sound pressure levels Source of sound Theoretical limit for undistorted sound at RMS sound sound pressure pressure level Pa dB re 20 µPa 101. 3. The speed of sound is also slightly sensitive (a second-order anharmonic effect) to the sound amplitude. which means that there are nonlinear propagation effects. Commonly used reference sound pressures. and a square root of such average is taken to obtain a root mean square (RMS) value.325 191 Page 57 of 64 .482 m/s (5. such as the production of harmonics and mixed tones not present in the original sound . also at 20 °C.6 and 101326. 13. and can cause hearing damage. Such a tiny (relative to atmospheric) variation in air pressure at an audio frequency will be perceived as quite a deafening sound.4 Pa. As the human ear can detect sounds with a very wide range of amplitudes. The sound pressure level (SPL) or Lp is defined as where p is the root-mean-square sound pressure and pref is a reference sound pressure. The International Electrotechnical Commission (IEC) has defined several weighting schemes. defined in the standard ANSI S1. are 20 µPa in air and 1 µPa in water.460 km/h. sound pressures are often frequency weighted so that the measured level will match perceived levels more closely.335 km/h.e.6T) m/s". Noise is a term often used to refer to an unwanted sound.960 m/s (21. sound pressure is often measured as a level on a logarithmic decibel scale. Since the human ear does not have a flat spectral response. absorption. C-weighting is used to measure peak levels. 1 Pa RMS sound pressure (94 dBSPL) in atmospheric air implies that the actual pressure in the sound wave oscillates between (1 atm Pa) and (1 atm Pa). the speed of sound is about 5.315 mph). a square of the deviation from the equilibrium pressure) is usually averaged over time and/or space. Without a specified reference sound pressure. Acoustics and noise The scientific study of the propagation. the speed of sound is approximately 1. according to the table below.767 mph) using the formula "v = (331 + 0. that is between 101323. a value expressed in decibels cannot represent a sound pressure level.330 mph). In science and engineering. For example. A-weighting attempts to match the response of the human ear to noise and A-weighted sound pressure levels are labeled dBA. Many of these use electroacoustic transducers such as microphones and loudspeakers. calm human breathing 0. because the molecules of the material in which the waves are traveling cannot pass the vibration along rapidly enough.02 very calm room 0. hearing aids.2–0. 10 m distant 0. Frequency range of hearing for humans and selected animals animal frequency (hertz) low high Page 58 of 64 . indeed. Many animals have the ability to hear sounds in the human ultrasonic frequency range.1 atmosphere environmental pressure 1883 Krakatoa eruption Stun grenades rocket launch equipment acoustic tests threshold of pain 100 hearing damage during short-term effect 20 jet engine. 1 m distant 0. sonar systems and sound reproduction and broadcasting equipment. Some ranges of hearing for mammals and insects are compared with those of humans in the Table. 60 40–60 20–30 10 0 Equipment for dealing with sound Equipment for generating or using sound includes musical instruments. 100 approx. even in a liquid or a solid. 100 m distant 6–200 jackhammer. above a frequency of about 1.002–0.2 TV set – typical home level. A presumed sensitivity of roaches and rodents to frequencies in the 40 kilohertz region has led to the manufacture of “pest controllers” that emit loud sounds in that frequency range to drive the pests away. 165 134 approx.0006 quiet rustling leaves. it is impossible for longitudinal waves to propagate at all. 1 m distant / discotheque 2 hearing damage from long-term exposure 0.02–0. 120 110–140 approx. but they do not appear to work as advertised. sometimes called praetersound or microsound. 10 m distant 0. Vibrations of frequencies greater than the upper limit of the audible range for humans—that is.6 traffic noise on major road.25 × 1013 hertz. 1 m distant 0.00002 approx 180 at 100 miles 170-180 approx. 85 80–90 60–80 approx. At such high frequencies it is very difficult for a sound wave to propagate efficiently.6 moving automobile.02 normal talking.00006 auditory threshold at 2 kHz – undamaged human ears 0. Hypersound. greater than about 20 kilohertz. is sound waves of frequencies greater than 1013 hertz. The term sonic is applied to ultrasound waves of very high amplitudes.0002–0. The advantage of Migatron's ultrasonic sensors is that both wide and narrow areas can be covered.000 12.000 150. the ultrasonic method has unique advantages over conventional sensors.000 150. Non contact distance measuring: Because sound can be timed from when it leaves the transducer to when it returns.002 of an inch at a distance of 4 inches.000 whales and dolphins 70 seals and sea lions 200 20. All it takes is the proper ultrasonic transducer selection. Broad area detection: While some photo electric sensors can detect over long distances they lack the ability to detect over a wide area without using a large number of sensors. Page 59 of 64 .000 55. Ultrasonic sensors detect over long ranges up to forty feet..000 50. an ultrasonic sensor can be designed to solve many application problems that are cost prohibitive or simply cannot be solved by other sensors. The target material can be clear. Long range detection: In industrial sensing.000 40.000 46. solid.000 32..000 grasshoppers and locusts 100 rodents 1.000 40. What are the advantages of ultrasonics? A: When used for sensing functions.000 What do you think the term noise means? How do ultrasonic sensors service the marketplace? A: Ultrasonic sensors service the market by providing a cost effective sensing method with unique properties not possessed by other sensing technologies. Measures and detects distances to moving objects. surface and color. while limit switches and inductive sensors do not.05% of range which equates to +or. Solid-state units have virtually unlimited. more and more applications require detection over distance. maintenance-free lifespan.000 100. distance measuring is easy and accurate to .. wood and any color because all can be detected. By using a wide variety of ultrasonic transducers and several different frequency ranges.humans 20 cats 100 dogs 40 horses 31 elephants 16 cattle 16 bats 1. Widest range of target materials: Only ultrasonic sensors are impervious to target material composition. porous. Impervious to target materials. liquid. soft. 26 km/s Helium is a monatomic gas that has a density of 0. Ans. intensity. State threeof sound waves Why is sound heard over corners more than light waves? A modern mehod of controlling moscuitoes in the house is production of vibrations from a coil. 0.35 A person riding a power mower may be subjected to a sound of intensity 2. Detects small objects over long operating distances.0 cm of mercury and a temperature of precisely 0 °C. 0. Explain three applications of sound waves. the speed of sound in air is 331 m/s.34 A sound has an intensity of 5. 3. liquid. Compute the average horizontal velocity of the shell. To two significant figures.00 x 10 -2 W/m2.30 km/s At S. what is the sound intensity at 107 dB? Ans. Explain howthis is possible? 4. 1.29 23. Ans.0690 and if 7 = 1.30 What is the speed of sound in air when the air temperature is 31 °C? Ans. dirt or high-moisture environments.. The scientific study of the propagation. 23. The air temperature is 20 °C.35 km/s A shell fired at a target 800 m distant was heard to strike it 5.T. Noise is a term often used to refer to an unwanted sound. Ans.24 23. absorption. ambient noise and EMI radiation. speed. infrared radiation.23 23. 970 m/s. wavelength.0 x 10 57 dB 23. 7 W/m2. and reflection of sound waves is called acoustics. if the specific gravity of hydrogen relative to air is 0. In science and engineering. 1.P. period. Resistant to external disturbances such as vibration. Sound is a travelling wave which is an oscillation of pressure transmitted through a solid.1 23. What is its intensity level? Ans.36 A rock band might easily produce a sound level of 107 dB in a room. 0. and direction . 2. Determine the speed of sound in hydrogen at S. Revision Questions 4. Sound waves are characterized by the generic properties of waves. 103 dB 23. What is the intensity level to which the person is subjected? Ans. or Page 60 of 64 .0 s after leaving the gun. noise is an undesirable component that obscures a wanted signal.P. Find the speed of compression waves (sound) in helium at this temperature and pressure. amplitude.0500 W/m2 Summary S We have basically tackled three major issues in this TOPIC. which are frequency.40 for both gases. Ultrasonic sensors are not affected by dust.T.179 kg/m3 at a pressure of 76. Wiley. Physics for Scientists and Engineers. 2. Box 861. In O'Mara. David.). London: Hodder & Stoughton. Narok. or the sensation stimulated in organs of hearing by such vibrations. (5th Ed.gas. USA.. Addison-Wesley ltd. Written Assignments PHY110/1 Written Assignment Do the following assignment and post it to: The Head Department of PHYSICS MMU P. 4. KIE (2008). Nelkon. Lee P. "Chapter 6". Val. Raymond A.. Porkess. New Jersey: Noyes Publications. Additional Mathematics for OCR. Secondary school physics books 1-4. Hanrahan. 431. R (2003). pp. YoungD.W.USA 7.Kenya References 1. NY. (1998) Theory and Problems in college Physics 9th edition. Jearl (2004-06-16).). Halliday. Fredrick J. Hunt. p. Gene Mosca (2008). Page 61 of 64 .O. Walker. Jewett (2006) Physics for scientists and Engineers 6th edition.A Freedman (1998) International students Edtn University physics with modern physics 9th edtn. p. 219. Paul A. composed of frequencies within the range of hearing and of a level sufficiently strong to be heard. Serway and J. (1978) Mechanics and Properties of Matter.. Note: Sound is a longitudinal wave while light is transverse wave.H & R. 6.. 8. Jomo Kenyatta Foundation. M. Volume 1 (6th ed. Herring. ISBN 1-4292-0132-0. ISBN 0-340-86960-7. Resnick. W. William C. Thomson. Murray Bullis (1990).Schaum’s series. Handbook of semiconductor silicon technology.USA. Eugene H. Robert B. Fundamentals of Physics (7 Sub ed. ISBN 0-8155- 3. Robert. New York. Park Ridge.) London: 9. 666–670. Nairobi. ISBN 0471232319.B. Tipler. 5. NY: Worth Publishers. Elasticity . five and six: Properties of Matter .Derivation of equations of motion of each kind and applications . liquids and gases Temperature scales Page 62 of 64 . Simple Harmonic motion(SMH) .Vectors: classification. iv) State and derive Newton’s and Kepler’s laws of motion.Surface tension . viii) Discuss the velocity of sound through media and ix) Discuss ultrasonics and its aaplications.Sit in CAT one . Cell phone: +254-722-325306 Consultation hrs: _Tues 2-4pm and Frid 2-5pm____Office No: PHY LAB Purpose: To introduce the student to basic concepts in physics Objectives: At the end of the course the learner should be able to: i) Resolve vectors ii) Describe the various types of motion iii) Explain the properties of matter.Fluid flow Week seven: CAT one . vector product .ac.Newton’s laws of motion .APPENDIX 1: COURSE OUTLINE PHY110 MAASAI MARA UNIVERSITY School of Science Department of PHYSICS Course: PHY 110: BASIC PHYSICS 1 Credit: 4 Lecture: Maera John (Mr) E-mail:
[email protected] of CAT one Week Eight.Physical and non-physical quantities . scalar product.ke. vii) Explain the kinetic theory of gases.Types of motion: Linear. COURSE OUTLINE Week one.Friction: causes and its prevention Week four. vi) Explain temperature scales and thermometers. two and Three: Mechanics . v) Discuss expansionof matter and heat transfer mechanisms. addition and subtraction. nine and ten: Thermal physics - Expansion of solids.Viscosity .Conservation of Energy and momentum . Circular. Projectile. Schaum’s series. convection and Radiation) Week Eleven and twelve: Sound - Characteristics of sound Genaral equation of a wave Velocity of sound in medium Waves on a string Velocity and elasticity of medium .Kinetic theory of gases .) London: Secondary school physics books 1-4. Serway and J. Jewett (2006) Fredrick J.Mechanism of heat transfer (conduction.First law of thermodynamics .Individual learner revision . (1998) Nelkon.USA Theory and Problems in college Physics 9th edition. NY.H & R.USA. USA. Addison-Wesley ltd. Thomson. Nairobi.B.Ulatrsonics and its applications Week Thirteen: - Sit in CAT two Revision of CAT two Week fourteen and fifteen: End of semester Examinations . M..W. Jomo Kenyatta Foundation. (1978) KIE (2008). Eugene H.A Freedman (1998) Raymond A.Sit-in examinations Course assessment: ASSIGNMENTS: 5% PRACTICALS: 10% CATS: 15% EXAM: 70%. International students Edtn University physics with modern physics 9th edtn. Mechanics and Properties of Matter. Physics for scientists and Engineers 6th edition. End Page 63 of 64 .Thermometers . Assessment schedule First test: Second Test: ________7th week ________ 13th week References YoungD.Specific heat capacities of gases . 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