Physics Quiz 1 PDF

Replacement Class 28 October 2024 Chapter 4 8.30 pm Online  Quiz 1 Date: 5/11/202 (Tuesday) Time: The Quiz will be open from 9.00 am - 5.00 pm Chapters: 1 & 2 Duration of Quiz: 15 minutes General Multiple Choice Questions Ecampus  Trial Quiz Date...

Replacement Class 28 October 2024 Chapter 4 8.30 pm Online  Quiz 1 Date: 5/11/202 (Tuesday) Time: The Quiz will be open from 9.00 am - 5.00 pm Chapters: 1 & 2 Duration of Quiz: 15 minutes General Multiple Choice Questions Ecampus  Trial Quiz Date: 28/10/2024 (Monday) Time: The Quiz will be open from 9.00 am - 3.00 pm General Multiple Choice Questions Ecampus CHAPTER 3 Kinematic (Motion) in Two Dimensions  Kinematics in One Direction Vertical motion Horizontal motion Kinematics in Two Dimensions PROJECTILE MOTION  Definition: Projectile Motion Projectile motion refers to the 2-D motion of an object that is given an initial velocity and projected into the air at an angle. The only force acting upon the object is gravity. Projectile Motion?  Introduction Revision – Vector and Its Component In 2-D;  Factors Affecting Projectile Motion What two factors would affect projectile motion? u An object projected at an arbitrary angle θ° relative to the horizontal with initial velocity, u; Sy v Sx the path of motion : parabolic arc analyzed by considering horizontal & vertical components separately.  An object projected at an arbitrary angle, θ° relative to the horizontal with initial velocity, u - Resolved into 2 separate components : (i) Horizontal velocity component, ux (x-component) If air resistance is negligible, ux is a constant – no force acting on the object in the horizontal direction ux = u cos  ax = 0 vx = ux sx = u x t (ii) Vertical velocity component, uy (y-component) uy = u sin  ay = - g vy = uy – gt sy = uy t – ½ g t2 vy2 = uy2 – 2 gsy where : u = initial velocity u x = horizontal / x-component of the initial velocity u y = vertical / y-component of the initial velocity θ° = angle of projection vx = horizontal final velocity vy = vertical final velocity sx = horizontal displacement at various time sy = vertical displacement at various time  Consider a general case : An object projected at an arbitrary angle, θ° relative to the horizontal with initial velocity, u  Consider a general case : An object projected at an arbitrary angle, θ° relative to the horizontal with initial velocity, u  Consider a general case : An object projected at an arbitrary angle, θ° relative to the horizontal with initial velocity, u  Consider a general case : An object projected at an arbitrary angle, θ° relative to the horizontal with initial velocity, u  Consider a general case : An object projected at an arbitrary angle, θ° relative to the horizontal with initial velocity, u  Maximum Height , H -- the maximum vertical displacement / height reached. -- at the top of the arc, vy = 0 when vy = 0, sy = H, using : vy2 = u y2 – 2g sy 0 = ( u sin θ ) 2 – 2g H 2g H = ( u sin θ ) 2 u 2 sin 2  H= 2g Time to reach the maximum height, tH If tH = Time taken by the object to reach the max height, H using : tH = u sin vy = u y – g t 0 = ( u sin θ ) – g tH g g tH = u sin θ  Range , R -- the maximum horizontal distance traveled. -- vertical displacement, sy = 0 horizontal displacement, sx = R at t = tR Using : Note : sx = uxt The maximum range is & it occurs R = u cos θ ( tR ) when 2 = 90 2u sin   = 45  R = u cos  ( ) g u 2 (2 sin  cos ) R= g u 2 sin 2 R= g Example A cannonball is fired with an initial velocity of 30.0 m s-1 at an angle of 35° to the horizontal. Determine (a) the horizontal and vertical component of the initial velocity Ux = 24.6 m s-1 Uy = 17.2 m s-1 (b) the maximum height reached by the ball ? H = 15.1 m (c) its range ? 86.21 m Solution; Given : u = 30.0 ms-1 ; θ = 35° ; ay = g = 9.81 ms-2 H u= 30m/s θ=35° R (a) From u = 30 ms-1 ; compute ux & uy : ux = u cos 35 = 30(0.819) = 24.6 m s-1 uy = u sin 35 = 30(0.574) = 17.2 m s-1 (b) At maximum height, H : vy = 0 From kinematics equation : vy2 = uy2 – 2gSy (0) = ( 17.2 )2 – 2 ( 9.81 )(H) 295.84 H= = 15.1 m 19.62 Solution; Given : u = 30.0 ms-1 ; θ = 35° ; ay = g = 9.81 ms-2 H u= 30m/s θ=35° R (a) From u = 30 ms-1 ; compute ux & uy : ux = u cos 35 = 30(0.819) = 24.6 m s-1 uy = u sin 35 = 30(0.574) = 17.2 m s-1 (b) At maximum height, H : vy = 0 From kinematics equation : vy2 = uy2 – 2gSy (0) = ( 17.2 )2 – 2 ( 9.81 )(H) 295.84 H= = 15.1 m 19.62 @ use the equation we have derived : u 2 sin 2  (30) 2 (sin 35) 2 H= H= 2g 2(9.81) H = 15.1 m (c) Range = ? u 2 sin 2 R= g = (30)2 (sin 2 (35)) / 9.81 = (30)2 (sin 2 (35)) / 9.81 = 86.21 m Example 2 u Solution; ux uy Example A ball is projected from a height of 25.0 m above the ground and is thrown with an initial horizontal velocity of 8.25 m s-1. (a) How long is the ball in flight before striking the ground ? (b) How far from the building does the ball strike the ground ? Solution; (a) At maximum height, H : Uy = 0 Sy = uyt – ½ gt2 - 25 = 0 – ½ (9.81)t2 t = 2.26 s (b) The ball travels in the x-direction for the same amount of time it travels in y-direction. Sx = ux(t ) = (8.25)(2.26) = 18.6 m  CHAPTER 1A INTRODUCTION & REVISION IN MATHEMATICS MEASUREMENTS Units Basic Unit and Derived Unit Unit Conversion 1.1 MEASUREMENTS Units Definition – standard size of measurement of physical quantities Examples - 1 second, 1 kilogram, 1 meter and etc. 1.1 MEASUREMENTS Importance of units In 11 December 1998, the $125 million Mars Climate Orbiter was launched to Mars to study the Martian climate Unfortunately, it burned up in the Martian atmosphere instead of entering a safe orbit from which it could perform observations Main reason? Mars Climate Orbiter  1.1 MEASUREMENTS Importance of units Main reason? Faulty units!! An engineering team had provided critical data on spacecraft performance in English units, but the navigation team assumed these data were in metric units. As a consequence, the navigation team had the spacecraft fly too close to the planet, and it burned up in the atmosphere 1.1 MEASUREMENTS Physical Quantity Definition – quantity which can be measured. Categorised into 2 components; (i) Basic Quantity (ii) Derived Quantity 1.1 MEASUREMENTS Physical Quantity Definition – quantity which can be measured. Physical Quantity Derived Basic Quantity Quantity 1.1 MEASUREMENTS The unit of basic quantity – called Base Unit The unit of derived quantity – called Derived Unit The common system of units used today are S.I unit (System International/metric system) 1.1 MEASUREMENTS Basic Quantity Definition – which cannot be derived from any physical quantities The table below listing examples of base quantities: 1.1 MEASUREMENTS Derived Quantity Definition – a physical quantity which can be derived in term of base quantity The table below listing examples of derived quantities: Basic Quantity Derived Quantity MEASUREMENTS used for presenting larger and smaller value. PREFIX VALUE SYMBOL Tera 1012 T Giga 109 G Mega 106 M Kilo 103 k Deci 10 -1 d Centi 10 -2 c Mili 10 -3 m Micro 10 -6 μ Nano 10 -9 N Pico 10 -12 P MEASUREMENTS MEASUREMENTS The Evolution Of Storage IBM 350 – 5 MB – price? What’s next? Unit Conversion 1 km = m ? 1m=s? 1 cm = m ? Unit Conversion 240 s = m ? 1.1 MEASUREMENTS Example; 58000 m → km 0.000009 s → µs 1.1 MEASUREMENTS Exercises 1.1 MEASUREMENTS Exercises  CHAPTER 1B INTRODUCTION & REVISION IN MATHEMATICS 1.3.TRIGONOMETRY 1.4.INTRODUCTION TO VECTOR Vectors Representation Standard Basis Vector, Vector Component 1.3 TRIGONOMETRY SOH CAH TOA 1.3 TRIGONOMETRY 1.3 TRIGONOMETRY Example; The roof of a house has the form of a right triangle with sides of length as showed in the figure below. 5.0 m x 4.0 m Find; (a) x (b) the interior angle of the triangle (Express your answers in degrees)  5.0 m x 4.0 m (a) Pythagoras Theorem (b) Interior angle 52 = 42 + x2 cos θ = 4/5 52 = 42 + x2 θ = 36.86° 52 – 42 = x2 x2 = 9 x=3 1.3 INTRODUCTION TO VECTORS Scalar & Vector Scalar Vector Scalar Vector Magnitude Magnitude Direction 1.3 INTRODUCTION TO VECTORS Scalar Definition - quantities which have magnitude without direction No directions involve! 1.3 INTRODUCTION TO VECTORS Scalar Movement of F1 car on track with speed of 300 km/h No directions involve! 1.3 INTRODUCTION TO VECTORS Vector Definition - quantities with magnitude & direction A plane flies 357 km/h at 14º east of north What are the differences between Distance & Displacement DISTANCE DISPLACEMENT What are the differences between Distance & Displacement DISTANCE DISPLACEMENT -scalar -vector -magnitude -magnitude and direction -meter -meter WIND DIRECTION Movement along North-South (y-axis) N Upward (+ve value) Downward (-ve value) S Movement along East-West (x-axis) Toward East (+ve value) W E Toward West (-ve value)  Example En Ali walks 10 m towards East and then 5 m towards West. Calculate: (i)Distance (ii)Displacement  10 m towards East 5 m towards West (i) Distance = 10 m + 5 m = 15 m (SCALAR) (ii) Displacement = 10 m - 5 m (VECTOR) = 5 m (East) We can also do the online simulation to study the displacement https://phet.colorado.edu/sims/html/vec tor-addition/latest/vector- addition_en.html  Example Anna walks 90 m towards east and then 50 m due north. Calculate her distance and displacement from the starting point.  displacement, d 50 m 50 m 90 m 90 m (i) Distance = 90 m + 50 m (ii) Displacement = 140 m = 902 + 502 = 102.96 m (magnitude) Θ = tan-1 (50/90) = 29° (East) – direction CHAPTER 2 Kinematics in One Direction  Introduction Kinematics is defined as the studies of motion of an objects without considering the forces that produce the motion. Kinematics is quantities state to i)position (displacement) ii)velocity iii)acceleration One dimension means motion along a line are considering horizontal motion is for x-axis or vertical motion is y-axis  examples of motion i)water drop(vertical) ii)car on highway(horizontal) Distance & Displacement Linear motion (1-D) DISTANCE is the length of the actual path taken by an object. Consider travel from point A to point B in diagram below: Distance, d is a scalar quantity (no direction) contains magnitude only. SI Unit; meter i.e The length of the path from A to B is 15 m  DISPLACEMENT is the straight line separation of two points in a specified direction. a vector quantity contain magnitude and direction (angle) SI Unit of Displacement: meter (m) Example A teacher walks 4 meters East, 2 meters South, 4 meters West, and finally 2 meters North. Calculate the: i) total distance ii) displacement Example Total distance = 12 m Displacement = 0  Example An aero plane flies 600 km north and then 400 km to the east. What is the magnitude of distance and displacement of the aero plane? Solution  SPEED is the rate of change of distance. Scalar quantity Equation Average Speed - total distance ,Σd traveled divided by the total time Σt elapsed in traveling that distance. Total distance traveled d Average speed , v = = total time elapsed for that distance  t Velocity & Average velocity To define the velocity of an object, we will use two concepts that we have already DEFINE , displacement and time DISPLACEMENT VELOCITY TIME  Velocity is the displacement per unit time. Vector quantity displacement s velocity , v = = travel time t Average velocity is the rate of change of displacement The S.I. unit for average velocity is m s-1. Equation; Example A man taking a leisurely walk takes 15 minutes to walk 100 m in an eastward direction. He then walks 50 m back (west) in 5 minutes. What is his average speed and velocity in m/s? (Note: for velocity, take eastward direction as positive) Example A man taking a leisurely walk takes 15 minutes to walk 100 m in an eastward direction. He then walks 50 m back (west) in 5 minutes. What is his average speed and velocity in m/s? (Note: for velocity, take eastward direction as positive) Average Speed Average velocity d 100 m + 50 m v speed = = t (15 min + 5 min ) 60 s  x = 100 m - 50 m = 50 m (east)  1 min  t = 20 min = 1200 s v speed = 0.12 m/s x 50 m v= = = 0.042 m/s (east) t 1200 s Example An insect crawls along the edge of a rectangular swimming pool of length 27m & width 21m. If it crawls from corner A to corner B in 30 min. (a) What is its average speed ? (b) What is the magnitude of its average velocity ? Average speed = 0.027 m/s Average velocity = 0.019 m/s Solution; Given : length L = 27m ; width, W = 21m Total distance travel, d = 27 + 21 = 48 m Displacement, s = 27 2 + 212 = 34.21 m (a) d Average speed , v = t 48 = 30(60) = 0.0267 m s −1  s (b) Average velocity, v av = t 34.21 = 30(60) = 0.019 m s −1 Average acceleration and instantaneous acceleration Acceleration is the ability to change in velocity or ratio between the change in velocity and time interval Average Acceleration is the rate of change of velocity Equation; Unit: meter per second squared (m/s2) Vector quantity  for linear motion, plus ( + ) & minus ( - ) signs will be used to indicate the directions of velocity & acceleration Deceleration - object is slowing down. ( where by the direction of acceleration is opposite to the direction of the motion ) Example: If a car moves from the rest to 5 m/s in 5 seconds, determine the average acceleration of the car. Solution; 5m / s − 0 m / s acceleration = 5s = 1 m s-2 Motion equations with linear acceleration constant “The Famous 4 Equations” for constant (uniform) acceleration 1. v = u + at u = initial velocity v = final velocity 2. s = ½(u + v) t a = acceleration s = displacement t = time taken 3. s = u t + ½ at2 4. v2 = u2 + 2as Assume a car has uniform acceleration & consider the motion between X and Y : u = initial velocity ( velocity on passing X ) v = final velocity ( velocity on passing Y ) a = acceleration s = displacement ( in moving from X to Y ) t = time taken ( to move from X to Y )  The velocity vs time graph for the car can be plotted as; v u From the graph, the gradient of the graph = acceleration v −u gradient a = t Can be arranged into final velocity, v; v = u + at … (1)  The area under the graph – the distance traveled or displacement , s s = area of trapezium s=½(u+v)t … (2) Substitute (1) into (2) : s = ½ ( u + u + at ) t s = ut + ½ at 2 … (3)  From (1) : v = u + at , get an expression for t : v−u t= a Substitute t into (2) : 1 v −u s = (v + u )( ) 2 a (v + u )(v − u ) s= 2a 2as = v 2 − u 2 v2 = u2 + 2as … (4) Example A car accelerates from rest at a constant rate of 2.0 m s-2 for 5.0s. (a) What is the final velocity of the car at the end of that time ? (b) How far does the car travel in this time ? Solution : Given : a = 2.0 m s-2 t = 5s u = 0 m s-1 (a) From Kinematics Equation : v=u+at =0+2(5) = 10 m s-1 (b) 1 2 S = ut + at 2 1 = 0(5) + (2)(5) 2 2 = 25 m Free Fall Free Fall a linear vertical motion where an object moving freely under the sole influence of gravity. only force acting on it – pull of GRAVITY Free falling objects do not encounter air resistance All free falling objects ( on Earth ) accelerate downwards with g ≈ 9.81 m s-2 g : vector quantity – directed downwards – g has a minus sign. Free Fall Motion Equation v=u–gt v2 = u2 – 2 g s s = u t – ½ g t2 s=½[u+v]t where s : vertical displacement u : initial velocity v : final velocity t : time * displacement, s, & velocity, v, may be +ve @ -ve depending on the direction of motion. Example A student drops a ball from the top of a tall building, it takes 2.8 s for the ball to reach the ground. (a) What was the ball’s final velocity just before hitting the ground ? (b) What is the height of the building ? 27.468m/s, 38.45m Solution : Given : u = 0 m s-1 ; t = 2.8 s ; free fall – a = 9.81 m s-2 (a) Using Free Fall equation : v = u − gt v = 0 − (9.81)(2.8) v = −27.47 m s −1 * ( Minus sign – indicates that v is downward ) 1 2 (b) s = ut − gt 2 1 s = 0 − (9.81)(2.8) 2 2 s = −38.46 m - ve : displacement is downward Example A boy throws a stone straight upward with an initial speed of 15 ms-1. What maximum height will the stone reach before falling back down ? Solution At the very top of the trajectory, object’s velocity is zero for an instant. → when smak , v = 0 ms-1 given : u = 15 m s-1 ; g = 9.81 m s-2 v 2 = u 2 − 2 gs (0) 2 = (15) 2 − 2(9.81) s 225 s= 19.62 s = 11.47 m CHAPTER 4 FORCE & MOTION  Introduction Why does everything in the universe move? Big, huge, massive forces! And little ones too. What is a FORCE? An interaction between TWO objects. Inertia & Newton’s First Law Garfield testing Newton’s First Law of Motion Newton’s First Law of Motion An object at rest stays at rest and an object in motion continues in motion with constant velocity if there is no net external force between the object and the environment. In equation :  Inertia Inertia is a term used to measure the ability of an object to resist a change in its state of motion. An object with a lot of inertia takes a lot of force to start or stop; an object with a small amount of inertia requires a small amount of force to start or stop.  Newton’s 2nd Law & Mass States : The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.  F = ma  S.I Unit = N = kg m s-2 The direction of the acceleration = direction of the applied net force.  Gravitational Force Gravitational force is a vector The magnitude of the gravitational force acting on an object of mass m near the Earth’s surface is called the weight w of the object w = mg The weight of an object points toward the center of the Earth. That direction does not change if the slant of the incline changes. m: Mass g = 9.8 m/s2  Normal Force: N Force from a solid surface which keeps object from falling through Direction: always perpendicular to the surface N = mg  Tension Force: T A stretched rope exerts forces on whatever holds its ends Direction: always along the cord (rope, cable, string ……) and away from the object Magnitude of tension force: the tension in the rope m2 m1 F Smooth surface m2 m1 F Smooth surface Explanation The free-body diagram and Newton's second law To keep track of how all forces are affecting a single object, it is a good idea to draw a free body diagram. A free body diagram is just a simple sketch of the object showing all the forces that are acting on it. – Draw a quick sketch of the object. – Draw an arrow showing every force acting on the object. – To calculate the net force, add any vectors acting on the same axis (x and y), making sure to pay attention to the directions. Free Body Diagram Free Body Diagram Mass & Weight Mass Weight is defined as a measure of a force exerted on an object by a body’s inertia gravitational field Scalar quantity vector quantity S.I. unit - kilogram (kg). S.I. unit - Newton (N) or kgms-2 Net Force A B C 3N 3N 3N 3N 5N Net force=3N. Net force=0N. Net force A is moving B is stationary = 5 – 3=2N to right. C is moving to right. D D is initially moving to right. A force of 5N acts on D, it 5N changes its direction. Force Components - Applications of Newton’s Second Law of Motion Example Two forces are acting on an object, P of mass 5kg on smooth surface. Find the acceleration of this object. If there is a frictional force of 2N, what is the new acceleration? 20N 10N Solution; Net force = 20-10=10N to the right. F=ma , 10=5xa, a = 2ms-2. If there is a frictional to oppose its motion , Net force=20-10-2=8N F=ma, 8=5xa, New acceleration =1.6ms-2. Example Example Newton’s 3rd Law For every action (force) there is a reaction (opposing force) of equal magnitude but opposite in direction. Explanation : Whenever one object exerts a force FAB on a second object. The second object exerts an equal and opposite force FBA on the first. FAB = -FBA Newton’s Third Law of Motion Newton’s Law and Its Application  Lecturing Notes by Dr. Leony Week 06 Chapter 5 Work and Energy Student to do! Timer  EFT 1103 Basic Physics Chapter 1. Introduction And Revision In Mathematics Chapter 2. Kinematics In One Direction Chapter 3. Motion In Two Dimensions Chapter 4. Force And Motion Chapter 5. Work And Energy Chapter 6. Momentum And Collision Chapter 7. Rotation And Gravity Chapter 8. Solid, Liquid And Gas Chapter 9. Temperature Energy is a capacity for doing work. You must have energy to accomplish work - it is like the "currency" for performing work. To do 100 joules of work, you must expend 100 joules of energy. Work is an activity involving a force and movement in the directon of the force. Work = Force x Displacement = Nm = Joule A force of 20 newtons pushing an object 5 meters in the direction of the force does 100 joules of work. Power is the rate of doing work or the rate of using energy, which are numerically the same. Power = Work / Time = Watts (N/s) If you do 100 joules of work in one second (using 100 joules of energy), the power is 100 watts. Task 5.1 Define Energy, Work and Power in Bahasa Malaysia and image. Energy is a capacity for doing work. You must have energy to accomplish work - it is like the "currency" for performing work. To do 100 joules of work, you must expend 100 joules of energy. Energy is a capacity for doing work. You must have energy to accomplish work - it is like the "currency" for performing work. To do 100 joules of work, you must expend 100 joules of energy. If you have ENERGY converter… Kinetic energy converted to electrical energy VALUES are ALWAYS the same = Conservation of energy law of conservation of energy states that the total energy of an isolated system remains constant — it is said to be conserved over time. VALUES are ALWAYS the same = Conservation of energy law of conservation of energy states that the total energy of an isolated system remains constant—it is said to be conserved over time. RM33.75 = JPY 1024.59 = THB 250.93 = USD 7.27 = PHP 411.08 SUBJECTED to available EXCHANGER!! Joule = A unit used to measure energy or work 1 Joule = 1Nm = energy by YOU lift 100g (1 apple) vertically through one metre of air if you can be converted to other kind energy E = mc2 = 48kgx(299,792,458 m/s)2= 4314024857936724500 J YOU ARE POWERFUL!! YOU can move an apple 54,600,000,000 meter (Earth to Mars) 4,314,024,857,936,724,500 meter (energy in YOU move apple) Task 5.2 If you invented an exchanger, your can convert a 0.1kg marble into electrical energy via E=mc2, what are two things you would do with these huge electrical energy? Kinetic energy is a form of energy associated with the motion of a particle, single body, or system of objects moving together. Kinetic energy (KE) of a mass m moving at a speed v. KE = 0.5 x mv2 Suppose a 30 kg package on the roller belt conveyor system is moving at 0.5 m/s. What is its kinetic energy? KE = 0.5 x mv2 KE = 0.5 x (30kg) (0.5 m/s)2 KE = 3.75 J  Work = Force x distance Wnett = _ _ _ J Wgravity = _ _ _ J Wnormal = _ _ _ J Wapplied = _ _ _ J Wfriction = _ _ _ J Suppose that you push on the 30 kg package with a constant force of 120 N through a distance of 0.8 m, and that the opposing friction force averages 5 N. (a) Calculate the net work done on the package. Wnett= 92 J (b) Solve the same problem as in part (a), this time by finding the work done by each force that contributes to the net force. Wgravity =0 J, Wnormal =0 J, Wapplied = 96 J, Wfriction = -4 J Suppose that you push on the 30 kg package with a constant force of 120 N through a distance of 0.8 m, and that the opposing friction force averages 5 N. (a) Calculate the net work done on the package. Wnett= 92 J (b) Solve the same problem as in part (a), this time by finding the work done by each force that contributes to the net force. Wgravity =0 J, Wnormal =0 J, Wapplied = 96 J, Wfriction = -4 J Potential energy is a form of energy that is stored within an object, not in motion but capable of becoming active. Potential energy (PE) example: gravitational potential energy (PEg) = mgh Task 5.3 A 60 kg person jumps onto the floor from a height of 3 m. He lands with his knee joints compressing by 0.5 cm, calculate the force on the knee joints. (W = Fd = energy = mgh, Fd = mgh, F=3.53x105N) Task 5.3 A 60 kg person jumps onto the floor from a height of 3 m. He lands with his knee joints compressing by 0.5 cm, calculate the force on the knee joints. (W = Fd = energy = mgh, Fd = mgh, F=3.53x105N) Potential energy is a form of energy that is stored within an object, not in motion but capable of becoming active. Potential energy (PE) example: potential energy of spring (PES) = 0.5kx2 k is the spring’s force constant and x is the displacement from its undeformed position Task 5.4 A spring is compressed 4 cm and has a force constant of 250 N/m. A 0.1kg car released from spring. Find the car speed at A and B. Use formula of KE = 0.5mv2, PEg = mgh, PES = 0.5kx2 (A = 2m/s, B = 0.687m/s) Task 5.4 A spring is compressed 4 cm and has a force constant of 250 N/m. A 0.1kg car released from spring. Find the car speed at A and B. Use formula of KE = 0.5mv2, PEg = mgh, PES = 0.5kx2 (A = 2m/s, B = 0.687m/s) Energy is a capacity for doing work. You must have energy to accomplish work - it is like the "currency" for performing work. To do 100 joules of work, you must expend 100 joules of energy. Work is an activity involving a force and movement in the directon of the force. Work = Force x Displacement = Nm = Joule A force of 20 newtons pushing an object 5 meters in the direction of the force does 100 joules of work. Power is the rate of doing work or the rate of using energy, which are numerically the same. Power = Work / Time = Watts (N/s) If you do 100 joules of work in one second (using 100 joules of energy), the power is 100 watts. Task 5.5 Find the power output for a 60 kg woman who runs up a 3 m high flight of stairs in 3.5 s, starting from rest but having a final speed of 2.00 m/s. (538W) Task 5.5 Find the power output for a 60 kg woman who runs up a 3 m high flight of stairs in 3.5 s, starting from rest but having a final speed of 2.00 m/s. (538W) Task 5.6 Step 1 Step 2 Choose one solar power station (actual) You are Head Engineer. Imagine you are □ Country : _ _ _ _ _ _ _ _ building a new solar power station in Malaysia □ State : _ _ _ _ _ _ _ _ □ 1 special thing/feature I want to add into to □ Year built : _ _ _ _ _ _ _ _ increase electric power generated (in 30 words) □ Capacity per year : _ _ _ _ _ _ _ ___________________________ □ 1 special thing/feature about this solar power station (in 30 words) : _ _ _ _ _ _ _ _ _ _ □ 1 special thing/feature I want to add into to ____________________ make the station becoming tourist location (in 30 words) ___________________________ W06 SBP / SEB / SEH LECTURE 24/11/2022 Thu 2.00pm - Prepare 2 piece of A4, https://prnt.sc/l8E2G3aiPVYn 2.05pm - Start Chapter 5 Work and Energy 2.35pm - Scan attendance @ Dewan Serbaguna 2.40pm - Experiment 3 Projectile Motion (Lab Manual) Lab Conduct 15% - W07, https://tinyurl.com/2022fizikgroup 2.45pm - Quiz 3 on 1/12/2022 Thu, preference, https://tinyurl.com/fizik2022 2.50pm - Start completing Task 5.1-5.6 (PDF) 3.30pm - Turn to behind, great to new friend, snap photo his/her Task 5.6 3.35pm - Submit online, https://tinyurl.com/fizik2022 3.40pm - 3 names to tell us his/her Task 5.6, https://tinyurl.com/fiziksiapa 3.45pm - Class end https://www.youtube.com/watch?v=BNGBDIH-Rfg  EFT 1103 Basic Physics Dr. Ahmad Zul Izzi Bin Fauzi Chapter 1. Introduction And Revision In Mathematics Chapter 2. Kinematics In One Direction Chapter 3. Motion In Two Dimensions Chapter 4. Force And Motion Chapter 5. Work And Energy Chapter 6. Momentum And Collision Chapter 7. Rotation And Gravity Chapter 8. Solid, Liquid And Gas Chapter 9. Temperature Momentum: linear momentum, is the mass in motion, P = mv https://www.youtube.com/watch?v=EzNfvw PCFs0 One way to think of momentum is that momentum measures how hard it will be to stop the object. m = 0.000005 kg v = 0.6 m/s Pmosquito = mv = 0.000003 kgm/s m = 755 kg v = 25 m/s PViva = mv = 18,875 kgm/s The change in momentum is called Impulse. Impulse, Δp, is the change in momentum caused by a new force: this force will increase or decrease the momentum depending on the direction of the force; towards or away from the object that was moving before. Newton’s Second Law The net external force equals the change in momentum of a system divided by the time over which it changes, Fnet = ma = m(Δv / Δt) = (mΔv)/Δt = (Δp)/Δt = (p2 – p1) / (t2 – t1) https://www.youtube.com/watch?v=xzA6IB WUEDE T6.1 2007, Venus Williams hit the fastest recorded serve in a premier women’s match, 58 m/s. A tennis ball contacted with the racquet for 0.005 s. Find the force exerted on the 0.057 kg tennis ball by her racquet. (Δp = 3.3 kg m/s, F = 660N) Fnet = ma = m(Δv / Δt) = (mΔv)/Δt = (Δp)/Δt = (p2 – p1) / (t2 – t1)  The principle of conservation of momentum states that when you have an isolated system with no external forces, the initial total momentum of objects before a collision equals the final total momentum of the objects after the collision. In other words: ∑Pbefore collide = ∑Pafter collide https://www.youtube.com/watch?v=F8DnN P1 = P2 qBhUfQ&t=59s m1v1 = m2v2 T6.2 An elastic collision happen between moving m1 and static m2 blocks. Draw and calculate the velocity of m2 after the collision, given m1 = 0.5 kg, m2 = 3.5 kg, v1 = 4 m/s, v1’ = -3 m/s, and v2 = 0. (answer v2’ = 1.00 m/s) ∑Pbefore collide = ∑Pafter collide P1 = P2 m1v1 + m2v2 = m1v1' + m2v2'  T6.3 Ball A moving VA = 3.0m/s strikes an equal-mass static ball B. The two balls are observed to move same speed off at 45° to the x axis, ball A above the x axis and ball B below. Find VA’45º and VB’45º. (VA’45º = VB’45º = 2.1m/s) ∑Pbefore collide = ∑Pafter collide P1 = P2 Horizontal : mAvAx + mBvBx = mAvA’x + mBvB’x Horizontal : mAvAx + mBvAx = mA(v A’45ºcos 45º) + mB(v B’45ºcos 45º) T6.4 A 70kg stone is falling from 0.50 m, and the collision on the floor lasts 0.082 s. Find the magnitudes of the Impulse (Δp) and Average force (F) acting on the stone from the floor during the collision? (220Ns, 2700N) T6.5 Choose a friend near you. Decide who is doing T6.5A OR T6.5Z question. Copy and solve. After 10 minutes, listen to your friend how he/she solve, then write his/her name. Then you explain. Lastly, both selfie holding 2nd page. T6.5Z T6.5A A 5.20 g bullet moving at 672 m/s A 2100 kg truck traveling north at strikes a 700 g wooden block at rest 41 km/h turns east and accelerates on a frictionless surface. The to 51 km/h. bullet emerges, traveling in the (a) What is the change in the same direction with its speed truck’s reduced to 428 m/s. kinetic energy (a) What is the resulting speed of (b) What are the magnitude and the block? (c) direction of the change in its (b) What is the speed of the bullet– momentum? block center of mass? (a) We choose +x along the initial direction of motion and apply momentum conservation: (b) It is a consequence of momentum conservation that the velocity of the center of mass is unchanged by the collision. We choose to evaluate it before the collision: Thank you  EFT 1103 Basic Physics Chapter 1. Introduction And Revision In Mathematics Chapter 2. Kinematics In One Direction Chapter 3. Motion In Two Dimensions Chapter 4. Force And Motion Chapter 5. Work And Energy Chapter 6. Momentum And Collision Chapter 7. Rotation And Gravity Chapter 8. Solid, Liquid And Gas Chapter 9. Temperature  Reason Life (you & me) Exist Oxygen Water Sun Gravity Rotation https://www.youtube.com/watch?v=H9YMg x5T9Sk https://www.youtube.com/watch?v=6BKiTZ CSyTU Rotation Angle Δθ Theta Displacement of the object at arc Δθ = rotation in angle Δθ = Δs / r (Δs = r Δθ) = 2πr / r (full circle) = 2π (unit in radian) = 2(22/7) (unit in radian) = 6.29 (unit in radian = 360o(unit in degree) 2π radian = 360o rotation angle 1 full circle = 1 revolution Angular Velocity ω Omega Rate of change of an angle v v = linear velocity v = Δs / Δt = (rΔθ) / Δt ω/a = rω (meter per second) ω = angular velocity ω = Δθ / Δt = (Δs / r) / Δt = radian per second a = angular acceleration a = Δω / Δt = radian per second2  Centripetal Acceleration ac Acceleration experienced while in uniform circular motion. It always points toward the center of rotation ac = centripetal acceleration = acceleration of an object travel across a circular path = v2 / r = (rω)2 / r = rω2 = meter per second2  Centripetal Force Fc Force that makes an object follow a curved path, toward the center of the circular motion Fc = centripetal force = mac = m(v2 / r ) = m(rω2) = Newton or kgms-2 T7.1 A car is moving from rest to a final angular velocity of 250 rpm in 5s. (a) Calculate the angular acceleration in rad/s2 (b) Brake is applied, causing an angular acceleration of – 87.3 rad/s2, calculate time taken the wheel to stop (5.24 rad/s2, 0.3s) T7.2 A motorcycle accelerates from 0 to 30.0 m/s in 4.2 s. Calculate angular acceleration of its 0.32 m radius wheels. (22.3 rad/s2)  initial final v v0 = 0m/s vf = 30m/s ω ω0 = v0/r = 0 / 0.32 = 0 rad/s ωf = vf/r = 30 / 0.32 = 93.75 rad/s a = Δω / t = (ωf – ω0) / t = (93.75 – 0) / 4.2 = 22.3 rad /s2 T7.3 Calculate centripetal acceleration ac of a car following a curve of radius 0.5km at 25m/s. Then compare ratio of ac to gravity acceleration. (1.25m/s2, 0.128, weak) Kinematics is concerned with the description of motion without regard to force or mass. Translational kinematic quantities, such as displacement, velocity, and acceleration have direct analogs in rotational motion. T7.4 A fish caught but swims away from the fishing line from fishing reel. The whole system is initially at rest and the fishing line unwinds from the reel at a radius of 4.50 cm from its axis of rotation. The reel is given an angular acceleration of 110 rad/s2 for 2.00 s. (a) Find final angular velocity of the reel? (220rad/s) (b) The speed is fishing line leaving the reel after 2s elapses (9.9m/s) (c) The revolutions does the reel make (35rev) (d) The meters of fishing line come off the reel in this time (9.9m) Newton's law of universal gravitation Any two bodies in the universe attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. T7.5 Moon mass is 7.35 x 1022 kg, 0.012 of Earth's mass. Closest distance between earth-moon is 363,104 km. Farthest distance between earth-moon is 405,696 km. (a) Calculate closest force. (b) Calculate farthest force. (c) Calculate the different force between closest and farthest distance Chapter 9 Solids and Fluids States of Matter n Solid n Liquid n Gas n Plasma Solids n Has definite volume n Has definite shape n Molecules are held in specific locations n by electrical forces n vibrate about equilibrium positions n Can be modeled as springs connecting molecules More About Solids n External forces can be applied to the solid and compress the material n In the model, the springs would be compressed n When the force is removed, the solid returns to its original shape and size n This property is called elasticity Crystalline Solid n Atoms have an ordered structure n This example is salt n Gray spheres represent Na+ ions n Green spheres represent Cl- ions Amorphous Solid n Atoms are arranged almost randomly n Examples include glass  Liquid n Has a definite volume n No definite shape n Exists at a higher temperature than solids n The molecules “wander” through the liquid in a random fashion n The intermolecular forces are not strong enough to keep the molecules in a fixed position  Gas n Has no definite volume n Has no definite shape n Molecules are in constant random motion n The molecules exert only weak forces on each other n Average distance between molecules is large compared to the size of the molecules Plasma n Matter heated to a very high temperature n Many of the electrons are freed from the nucleus n Result is a collection of free, electrically charged ions n Plasmas exist inside stars Deformation of Solids n All objects are deformable n It is possible to change the shape or size (or both) of an object through the application of external forces n when the forces are removed, the object tends to its original shape n This is a deformation that exhibits elastic behavior  Elastic Properties n Stress is the force per unit area causing the deformation n Strain is a measure of the amount of deformation n The elastic modulus is the constant of proportionality between stress and strain n For sufficiently small stresses, the stress is directly proportional to the strain n The constant of proportionality depends on the material being deformed and the nature of the deformation Elastic Modulus n The elastic modulus can be thought of as the stiffness of the material n A material with a large elastic modulus is very stiff and difficult to deform n Analogous to the spring constant n stress=Elastic modulus×strain  Young’s Modulus: Elasticity in Length n Tensile stress is the ratio of the external force to the cross- sectional area n Tensile is because the bar is under tension n The elastic modulus is called Young’s modulus Young’s Modulus, cont. n SI units of stress are Pascals, Pa n 1 Pa = 1 N/m2 n The tensile strain is the ratio of the change in length to the original length n Strain is dimensionless F DL =Y A Lo  Young’s Modulus, final n Young’s modulus applies to a stress of either tension or compression n It is possible to exceed the elastic limit of the material n No longer directly proportional n Ordinarily does not return to its original length Breaking n If stress continues, it surpasses its ultimate strength n The ultimate strength is the greatest stress the object can withstand without breaking n The breaking point n For a brittle material, the breaking point is just beyond its ultimate strength n For a ductile material, after passing the ultimate strength the material thins and stretches at a lower stress level before breaking Shear Modulus: Elasticity of Shape n Forces may be parallel to one of the object’s faces n The stress is called a shear stress n The shear strain is the ratio of the horizontal displacement and the height of the object n The shear modulus is S Shear Modulus, final F n shear stress = A Dx shear strain = h F Dx =S A h n S is the shear modulus n A material having a large shear modulus is difficult to bend Bulk Modulus: Volume Elasticity n Bulk modulus characterizes the response of an object to uniform squeezing n Suppose the forces are perpendicular to, and act on, all the surfaces n Example: when an object is immersed in a fluid n The object undergoes a change in volume without a change in shape  Bulk Modulus, cont. n Volume stress, ΔP, is the ratio of the force to the surface area n This is also the Pressure n The volume strain is equal to the ratio of the change in volume to the original volume Bulk Modulus, final DV DP = -B V n A material with a large bulk modulus is difficult to compress n The negative sign is included since an increase in pressure will produce a decrease in volume n B is always positive n The compressibility is the reciprocal of the bulk modulus Notes on Moduli n Solids have Young’s, Bulk, and Shear moduli n Liquids have only bulk moduli, they will not undergo a shearing or tensile stress n The liquid would flow instead Ultimate Strength of Materials n The ultimate strength of a material is the maximum force per unit area the material can withstand before it breaks or factures n Some materials are stronger in compression than in tension  Post and Beam Arches n A horizontal beam is supported by two columns n Used in Greek temples n Columns are closely spaced n Limited length of available stones n Low ultimate tensile strength of sagging stone beams Semicircular Arch n Developed by the Romans n Allows a wide roof span on narrow supporting columns n Stability depends upon the compression of the wedge-shaped stones Gothic Arch n First used in Europe in the 12th century n Extremely high n The flying buttresses are needed to prevent the spreading of the arch supported by the tall, narrow columns Density n The density of a substance of uniform composition is defined as its mass per unit volume: m r º V n Units are kg/m3 (SI) or g/cm3 (cgs) n 1 g/cm3 = 1000 kg/m3 Density, cont. n The densities of most liquids and solids vary slightly with changes in temperature and pressure n Densities of gases vary greatly with changes in temperature and pressure Density Slide 13-12 Specific Gravity n The specific gravity of a substance is the ratio of its density to the density of water at 4° C n The density of water at 4° C is 1000 kg/m3 n Specific gravity is a unitless ratio Pressure n The force exerted by a fluid on a submerged object at any point if perpendicular to the surface of the object F N Pº in Pa = 2 A m Reading Quiz 1. The SI unit of pressure is A. N B. kg/m2 C. Pa D. kg/m3 Slide 13-6 Answer 1. The SI unit of pressure is A. N B. kg/m2 C. Pa D. kg/m3 Slide 13-7 Measuring Pressure n The spring is calibrated by a known force n The force the fluid exerts on the piston is then measured Variation of Pressure with Depth n If a fluid is at rest in a container, all portions of the fluid must be in static equilibrium n All points at the same depth must be at the same pressure n Otherwise, the fluid would not be in equilibrium n The fluid would flow from the higher pressure region to the lower pressure region Pressure and Depth n Examine the darker region, assumed to be a fluid n It has a cross- sectional area A n Extends to a depth h below the surface n Three external forces act on the region Pressure and Depth equation n n Po is normal atmospheric pressure n 1.013 x 105 Pa = 14.7 lb/in2 n The pressure does not depend upon the shape of the container Pascal’s Principle n A change in pressure applied to an enclosed fluid is transmitted undimished to every point of the fluid and to the walls of the container. n First recognized by Blaise Pascal, a French scientist (1623 – 1662)  Pascal’s Principle, cont n The hydraulic press is an important application of Pascal’s Principle F1 F2 P= = A1 A 2 n Also used in hydraulic brakes, forklifts, car lifts, etc. Absolute vs. Gauge Pressure n The pressure P is called the absolute pressure n Remember, P = Po + rgh n P – Po = rgh is the gauge pressure  Pressure Measurements: Manometer n One end of the U- shaped tube is open to the atmosphere n The other end is connected to the pressure to be measured n Pressure at B is Po+ρgh Blood Pressure n Blood pressure is measured with a special type of manometer called a sphygmomano- meter n Pressure is measured in mm of mercury Pressure Measurements: Barometer n Invented by Torricelli (1608 – 1647) n A long closed tube is filled with mercury and inverted in a dish of mercury n Measures atmospheric pressure as ρgh The Barometer Slide 13-19 Pressure Values in Various Units n One atmosphere of pressure is defined as the pressure equivalent to a column of mercury exactly 0.76 m tall at 0o C where g = 9.806 65 m/s2 n One atmosphere (1 atm) = n 76.0 cm of mercury n 1.013 x 105 Pa n 14.7 lb/in2 Pressure Units Slide 13-20 Pressure The pressure of the F p= water behind each hole A pushes the water out. The SI unit of pressure is 1 pascal = 1 Pa = 1 N/m2. Slide 13-13 Pressure in a Liquid Increases with Depth Slide 13-14 Archimedes n 287 – 212 BC n Greek mathematician, physicist, and engineer n Buoyant force n Inventor Archimedes' Principle n Any object completely or partially submerged in a fluid is buoyed up by a force whose magnitude is equal to the weight of the fluid displaced by the object. Buoyant Force n The upward force is called the buoyant force n The physical cause of the buoyant force is the pressure difference between the top and the bottom of the object Buoyant Force, cont. n The magnitude of the buoyant force always equals the weight of the displaced fluid B = r fluidVfluid g = wfluid n The buoyant force is the same for a totally submerged object of any size, shape, or density Buoyant Force, final n The buoyant force is exerted by the fluid n Whether an object sinks or floats depends on the relationship between the buoyant force and the weight Buoyancy Slide 13-21 Archimedes’ Principle: Totally Submerged Object n The upward buoyant force is B=ρfluidgVobj n The downward gravitational force is w=mg=ρobjgVobj n The net force is B-w=(ρfluid- ρobj)gVobj Totally Submerged Object n The object is less dense than the fluid n The object experiences a net upward force Totally Submerged Object, 2 n The object is more dense than the fluid n The net force is downward n The object accelerates downward Archimedes’ Principle: Floating Object n The object is in static equilibrium n The upward buoyant force is balanced by the downward force of gravity n Volume of the fluid displaced corresponds to the volume of the object beneath the fluid level Archimedes’ Principle: Floating Object, cont n The forces balance n robj Vfluid = r fluid Vobj Reading Quiz 3. The buoyant force on an object submerged in a liquid depends on A. the object’s mass. B. the object’s volume. C. the density of the liquid. D. both B and C. Slide 13-10 Answer 3. The buoyant force on an object submerged in a liquid depends on A. the object’s mass. B. the object’s volume. C. the density of the liquid. D. both B and C. Slide 13-11 FB = rfVf g Slide 13-22 Floating When the object sinks to the point that the weight of the displaced fluid equals the weight of the object, then the forces balance and the object floats in equilibrium. No net force. The volume of fluid displaced by a floating object of density ro and volume Vo is po Vf = Vo pf The density of ice is 90% that of water. When ice floats, the displaced water is 90% of the volume of ice. Thus 90% of the ice is below water and 10% is above. Slide 13-23 How a Boat Floats Slide 13-24 Checking Understanding Two blocks of identical size are submerged in water. One is made of lead (heavy), the other of aluminum (light). Upon which is the buoyant force greater? A. On the lead block. B. On the aluminum block. C. They both experience the same buoyant force. Slide 13-25 Answer Two blocks of identical size are submerged in water. One is made of lead (heavy), the other of aluminum (light). Upon which is the buoyant force greater? A. On the lead block. B. On the aluminum block. C. They both experience the same buoyant force. Slide 13-26 Checking Understanding Two blocks are of identical size. One is made of lead, and sits on the bottom of a pond; the other is of wood and floats on top. Upon which is the buoyant force greater? A. On the lead block. B. On the wood block. C. They both experience the same buoyant force. Slide 13-27 Answer Two blocks are of identical size. One is made of lead, and sits on the bottom of a pond; the other is of wood and floats on top. Upon which is the buoyant force greater? A. On the lead block. B. On the wood block. C. They both experience the same buoyant force. Slide 13-28 Checking Understanding A barge filled with ore floats in a canal lock. If the ore is tossed overboard into the lock, the water level in the lock will A. rise. B. fall. C. remain the same. Slide 13-29 Answer A barge filled with ore floats in a canal lock. If the ore is tossed overboard into the lock, the water level in the lock will A. rise. B. fall. C. remain the same. Slide 13-30  Fluids in Motion: Streamline Flow n Streamline flow n Every particle that passes a particular point moves exactly along the smooth path followed by particles that passed the point earlier n Also called laminar flow n Streamline is the path n Different streamlines cannot cross each other n The streamline at any point coincides with the direction of fluid velocity at that point  Streamline Flow, Example Streamline flow shown around an auto in a wind tunnel Fluids in Motion: Turbulent Flow n The flow becomes irregular n exceeds a certain velocity n any condition that causes abrupt changes in velocity n Eddy currents are a characteristic of turbulent flow Turbulent Flow, Example n The rotating blade (dark area) forms a vortex in heated air n The wick of the burner is at the bottom n Turbulent air flow occurs on both sides of the blade Fluid Flow: Viscosity n Viscosity is the degree of internal friction in the fluid n The internal friction is associated with the resistance between two adjacent layers of the fluid moving relative to each other Characteristics of an Ideal Fluid n The fluid is nonviscous n There is no internal friction between adjacent layers n The fluid is incompressible n Its density is constant n The fluid motion is steady n Its velocity, density, and pressure do not change in time n The fluid moves without turbulence n No eddy currents are present n The elements have zero angular velocity about its center Atmospheric Pressure patmos = 1 atm = 103,000 Pa Slide 13-16 Constrained Flow: Continuity Slide 13-33  Equation of Continuity n A1v1 = A2v2 n The product of the cross-sectional area of a pipe and the fluid speed is a constant n Speed is high where the pipe is narrow and speed is low where the pipe has a large diameter n Av is called the flow rate Equation of Continuity, cont n The equation is a consequence of conservation of mass and a steady flow n A v = constant n This is equivalent to the fact that the volume of fluid that enters one end of the tube in a given time interval equals the volume of fluid leaving the tube in the same interval n Assumes the fluid is incompressible and there are no leaks Daniel Bernoulli n 1700 – 1782 n Swiss physicist and mathematician n Wrote Hydrodynamica n Also did work that was the beginning of the kinetic theory of gases Bernoulli’s Equation n Relates pressure to fluid speed and elevation n Bernoulli’s equation is a consequence of Conservation of Energy applied to an ideal fluid n Assumes the fluid is incompressible and nonviscous, and flows in a nonturbulent, steady-state manner Bernoulli’s Equation, cont. n States that the sum of the pressure, kinetic energy per unit volume, and the potential energy per unit volume has the same value at all points along a streamline 1 2 P + rv + rgy = constant 2 Bernoulli’s Equation Slide 13-36 Atmospheric Pressure patmos = 1 atm = 103,000 Pa Slide 13-16 Constrained Flow: Continuity Slide 13-33 Acceleration of Fluids Slide 13-34 Pressure Gradient in a Fluid Slide 13-35 Applications of Bernoulli’s Principle: Venturi Tube n Shows fluid flowing through a horizontal constricted pipe n Speed changes as diameter changes n Can be used to measure the speed of the fluid flow n Swiftly moving fluids exert less pressure than do slowly moving fluids An Object Moving Through a Fluid n Many common phenomena can be explained by Bernoulli’s equation n At least partially n In general, an object moving through a fluid is acted upon by a net upward force as the result of any effect that causes the fluid to change its direction as it flows past the object Application – Golf Ball n The dimples in the golf ball help move air along its surface n The ball pushes the air down n Newton’s Third Law tells us the air must push up on the ball n The spinning ball travels farther than if it were not spinning  Application – Airplane Wing n The air speed above the wing is greater than the speed below n The air pressure above the wing is less than the air pressure below n There is a net upward force n Called lift n Other factors are also involved TEMPERATURE AND HEAT TRANSFER TEMPERATURE ¢ Physical quantity which measures the degree of coldness or hotness of a body ¢ Definition of temperature is based on the concepts of thermal equilibrium ¢ Quantity that determines when objects are in thermal equilibrium.(the object do not necessarily have the same energy when in the thermal equilibrium) ¢ The SI units of temperature is the kelvin (symbol K) HEAT ¢ Is the energy which is transferred from one object to another due to a difference in temperature ZEROTH LAW OF THERMODYNAMICS ¢ If two objects each in thermal equilibrium with a third object, then all three objects are thermal equilibrium THERMAL EQUILIBRIUM ¢ When two objects in contact have the same temperature, these two object are in thermal equilibrium THERMAL EQUILIBRIUM Energy flows as heat from a region at a higher temperature to one at a lower temperature if the two are in contact through a diathermic wall, as in (a) and (c). However, if the two regions have identical temperatures, there is no net transfer of energy as heat even though the two regions are separated by a diathermic wall (b). The latter condition corresponds to the two regions being at thermal equilibrium. ZEROTH LAW OF THERMODYNAMICS The experience summarized by the Zeroth Law of thermodynamics is that, if an object A is in thermal equilibrium with B and if B is in thermal equilibrium with C, then B is in thermal equilibrium with A. FORMS OF HEAT TRANSFER ¢Three forms of heat transfer: ¢Conduction ¢Convection ¢Radiation CONDUCTION ¢Conduction involves the transfer of heat through direct contact ¢Heat conductors conduct heat well, insulators do not CONVECTION ¢Takes place in liquids and gases as molecules move in currents ¢Heat rises and cold settles to the bottom RADIATION ¢ Heat is transferred through space ¢Energy from the sun being transferred to the Earth TEMPERATURE SCALES THERMAL EXPANSION OF SOLIDS AND LIQUID ¢ Most objects expand as their temperature increases ¢ For example, the cooper(barrel maker) heated irons hoop red hot to make them expand before fitting them around the wooden staves of a barrel. (∆L/Lo)= (α ∆T) ∆L=L-Lo and ∆L=L-Lo. The length at temperature T is L= Lo + ∆L = ( 1 + α ∆T) Lo α= coefficient of thermal expansion ¢ The fractional change in volume of a solid or liquid is also proportional to the temperature change as long temperature change is not too large ¢ (∆V/Vo)= (β∆T) ¢ For solid, the coefficient of volume expansion is three times the coefficient of linear expansion EXAMPLE ¢ Two motor rods, one aluminum and one brass, are each clamped at one. At 0.0oC, the rods are each 50.0cm long and separated by 0.024cm at their unfastened ends. At what temperature will the rods just come into contact? (Assume that the base to which the rods are clamped undergoes a negligibly small thermal expansion) ¢ Lo = 50.0cm, To = 0.0oC for both ¢ Look up αbr= 1.9 x 10-6 K-1 ; αA1= 22.5 x 10-6 K-1 ¢ Requirement : ∆L br + ∆LA1 = 0.024cm ¢ Tf= To + ∆T The brass rod expands by ¢ ∆Lbr = (αbr ∆T)Lo The brass aluminum rod by ¢ ∆LA1= (αA1 ∆T)Lo ¢ the sum of the two expansion is known ¢ ∆Lbr + ∆LA1 = 0.024cm ¢ Since both the initial lengths and the temperatures changes are the same (αA1 + αbr ∆T)x Lo = 0.024cm ¢ ∆T = (0.024cm)/(αA1 + αbr)Lo ¢ = (0.024cm)/[(1.9 x 10-6 K-1 + 22.5 x 10-6 K-1 ) x 50.0cm] ¢ = 11.6 oC The temperature at which the two touch is ¢ Tf = To + ∆T= 0.0 oC + 11.6 oC à12 oC ¢ ∆LA1= (αA1 ∆T)Lo ¢ = 22.5 x 10-6 K-1 x 11.6K x 50.0cm = 0.013cm ¢ ∆Lbr = (αbr ∆T)Lo ¢ = 1.9 x 10-6 K-1 x 11.6K x 50.0cm = 0.011cm Total expansion = 0.013cm+ 0.011cm = 0.024cm KINETIC THEORY OF GASES 20 PRESSURE OF A GAS DEPENDS ON: ¢Number of molecules ¢Volume in which they are contained ¢Average kinetic energy of molecules 21 BOYLE’S LAW If the amount and temperature of a gas remain constant, the pressure exerted by the gas varies inversely with the volume. P µ 1/V or PV = k 22 BOYLE’S LAW (VALID AT LOW PRESSURES) The pressure-volume dependence of a fixed amount of perfect gas at different temperatures. Each curve is a hyperbola (pV = constant) and is called an isotherm. APPLYING BOYLE’S LAW P1V1 = P2V2 If some oxygen gas collected at 760 mm Hg is allowed to expand from 5.0 L to 10.0 L without changing the temperature, what pressure will the oxygen gas exert? 380 mm Hg 24 The relationship between temperature and volume CHARLES’ LAW The volume of a quantity of gas, held at a constant pressure, varies directly with the Kelvin temperature. VµT or V = kT 26 CHARLES LAW The variation of the volume of a fixed amount of gas with the temperature constant. Note that in each case they extrapolate to zero volume at -273.15° C. The lines are isotherms. APPLYING CHARLES’ LAW V1/T1 = V2/T2 A 225 mL volume of gas is collected at 58.0 oC. What volume would this sample of gas occupy at standard temperature, assuming pressure remains constant. 186 mL 28 Charles’s Law: as the temperature of a gas increases, the volume increases proportionally, provided that the pressure and amount of gas remain constant, (V1/T1 ) = (V2/T2 ) ¢ Example A sample of gas occupies 3.5 L at 300 K. What volume will it occupy at 200 K? V1 = 3.5 L, T1 = 300K, V2 = ?, T2 = 200K Using Charles’ law: V1/T1 = V2/T2 3.5 L / 300 K = V2 / 200 K V2 = (3.5 L/300 K) x (200 K) = 2.3 L COMBINING BOYLE’S AND CHARLES’ LAWS The volume of P1V1 = P2V2 a gas measured T1 T2 at 755 mm Hg and 23.0 C is o 455 mL. What 417 mL is the volume of the gas at STP? 30 SAMPLE PROBLEM At STP, the volume of a gas is 325 mL. What volume does it occupy at 20.0 oC and 756 mm Hg? 351 mL 31 SAMPLE PROBLEM A gas occupies a volume of 245 mL at 30.0 C and 746 mm o Hg. Assuming no change in pressure, at what temperature will the gas occupy a volume of 490. mL? 606 K 32 THE IDEAL GAS LAW PV = nRT THE IDEAL GAS LAW An ideal gas is an idealized model for real gases that have sufficiently low densities. The condition of low density means that the molecules of the gas are so far apart that they do not interact (except during collisions that are effectively elastic). The ideal gas law expresses the relationship between the absolute pressure (P), the Kelvin temperature (T), the volume (V), and the number of moles (n) of the gas. PV = nRT Where R is the universal gas constant. R = 8.31 J/(mol · K). THE IDEAL GAS LAW The constant term R/NA is referred to as Boltzmann's constant, in honor of the Austrian physicist Ludwig Boltzmann (1844–1906), and is represented by the symbol k: PV = NkT ¢ If a volume vs. temperature graph is plotted for gases, most lines can be interpolated so that when volume is 0 the temperature is -273 °C. ¢ Naturally, gases don’t really reach a 0 volume, but the spaces between molecules approach 0. ¢ At this point all molecular movement stops. ¢ –273°C is known as “absolute zero” (no EK) ¢ Lord Kelvin suggested that a reasonable temp-erature scale should start at a true zero value. ¢ He kept the convenient units of °C, but started at absolute zero. Thus, K = °C + 273. 62°C = ? K: K=°C+273 = 62 + 273 = 335 K ¢ Notice that kelvin is represented as K not °K. THE IDEAL GAS LAW PV = nRT P = Pressure (in kPa) V = Volume (in L) T = Temperature (in K) n = moles R = 8.31 kPa L K mol R is constant. If we are given three of P, V, n, or T, we can solve for the unknown value. Recall, From Boyle’s Law: P1V1 = P2V2 or PV = constant From combined gas law: P1V1/T1 = P2V2/T2 or PV/T = constant IDEAL GAS LAW QUESTIONS 1. How many moles of CO2(g) is in a 5.6 L sample of CO2 measured at STP? 2. a) Calculate the volume of 4.50 mol of SO2(g) measured at STP. b) What volume would this occupy at 25°C and 150 kPa? (solve this 2 ways) 3. How many grams of Cl2(g) can be stored in a 10.0 L container at 1000 kPa and 30°C? 4. At 150°C and 100 kPa, 1.00 L of a compound has a mass of 2.506 g. Calculate its molar mass. 5. 98 mL of an unknown gas weighs 0.087 g at SATP. Calculate the molar mass of the gas. Can you determine the identity of this unknown gas? 1. Moles of CO2 is in a 5.6 L at STP? P=101.325 kPa, V=5.6 L, T=273 K PV = nRT (101.3 kPa)(5.6 L) = n (8.31 kPa L/K mol)(273 K) (101.325 kPa)(5.6 L) = n = 0.25 mol (8.31 kPa L/K mol)(273 K) 2. a) Volume of 4.50 mol of SO2 at STP. P= 101.3 kPa, n= 4.50 mol, T= 273 K PV=nRT (101.3 kPa)(V)=(4.5 mol)(8.31 kPa L/K mol)(273 K) (4.50 mol)(8.31 kPa L/K mol)(273 K) V= = 100.8 L (101.3 kPa) 2. b) Volume at 25°C and 150 kPa (two ways)? Given: P = 150 kPa, n = 4.50 mol, T = 298 K (4.50 mol)(8.31 kPa L/K mol)(298 K) V= = 74.3 L (150 kPa) From a): P = 101.3 kPa, V = 100.8 L, T = 273 K Now P = 150 kPa, V = ?, T = 298 K P1V1 P2V2 = T1 T2 (101.3 kPa)(100 L) (150 kPa)(V2) = (273 K) (298 K) (101.3 kPa)(100.8 L)(298 K) = 74.3 L (V2) = (273 K)(150 kPa) 3. How many grams of Cl2(g) can be stored in a 10.0 L container at 1000 kPa and 30°C? PV = nRT P= 1000 kPa, V= 10.0 L, T= 303 K (1000 kPa)(10.0 L) = n = 3.97 mol (8.31 kPa L/K mol)(303 K) 3.97 mol x 70.9 g/mol = 282 g 4. At 150°C and 100 kPa, 1.00 L of a compound has a mass of 2.506 g. Calculate molar mass. PV = nRT P= 100 kPa, V= 1.00 L, T= 423 K (100 kPa)(1.00 L) = n = 0.02845 mol (8.31 kPa L/K mol)(423 K) g/mol = 2.506 g / 0.02845 mol = 88.1 g/mol 5. 98 mL of an unknown gas weighs 0.081 g at SATP. Calculate the molar mass. PV = nRT P= 100 kPa, V= 0.098 L, T= 298 K (100 kPa)(0.098 L) = n = 0.00396 mol (8.31 kPa L/K mol)(298 K) g/mol = 0.081 g / 0.00396 mol = 20.47 g/mol It’s probably neon (neon has a molar mass of 20.18 g/mol) KINETIC THEORY OF GASES The pressure that a gas exerts is caused by the impact of its molecules on the walls of the container.  KINETIC THEORY OF G ASES The pressure that a gas exerts is caused by the impact of its molecules on the walls of the container. It can be shown that the average translational kinetic energy of a molecule of an ideal gas is given by, where k is Boltzmann's constant and T is the Kelvin temperature. DERIVATION OF, Consider a gas molecule colliding elastically with the right wall of the container and rebounding from it. The force on the molecule is obtained using Newton’s second law as follows, DP åF = Dt , The force on one of the molecule, According to Newton's law of action–reaction, the force on the wall is equal in magnitude to this value, but oppositely directed. The force exerted on the wall by one molecule, mv 2 = L If N is the total number of molecules, since these particles move randomly in three dimensions, one-third of them on the average strike the right wall. Therefore, the total force is: Vrms = root-mean-square velocity. Pressure is force per unit area, so the pressure P acting on a wall of area L2 is Pressure is force per unit area, so the pressure P acting on a wall of area L2 is Since the volume of the box is V = L3, the equation above can be written as, PV = NkT KINETIC THEORY OF THE IDEAL GAS Microscopic basis of Pressure -the forces a gas exerts on a surface is due to collision that gas molecules make with that surfaces -the pressure depends on three things a) How many molecules there are b) How often each one collides with the wall c) Momentum transfer due to each collision - ASSUMPTION KINETIC THEORY OF GASES 1. There are a large number of molecules, N, each of mass m, moving in random directions with a variety of speeds. 2. The molecules are, on the average, far apart from one another (i.e., low pressure). 3. The molecules are assumed to obey the laws of classical mechanics (vice quantum mechanics), and are assumed to interact with one another only when they collide. Potential energy is small enough to be ignored. 4. Collisions with another molecule of the wall of the container are assumed to be completely elastic, like the collisions of perfectly elastic billiard balls. Assume the collisions are of very short duration compared to the time between collisions. Then we can ignore the potential energy associated with collisions in comparison to the kinetic energy between collisions. TEMPERATURE AND KINETIC ENERGY TRANSLATION OF MOLECULE GAS ¢ The average translational kinetic energy of molecules in an ideal gas is directly proportional to the absolute temperature. ¢ Temperature is a measure of energy K=(1/2)(mv2)=(3/2)kT Root-Mean-Square Velocity Vrms=√v2 = √(3kT/M) The mean speed, v, is the average of the magnitudes of the speeds of molecules. The root-mean-square velocity, is the square root of the sum of the squares of the molecular velocities. We use rms velocity because this is how the above equation was derived. ¢ According to the kinetic theory, the pressure of a gas is proportional to the average translational kinetic energy of the molecules ¢ P= (2/3)(N/V) ¢ The average translational kinetic energy of the molecules is proportional to the absolute temperature ¢ = (3/2)(kT) ¢ The speed of a gas molecule that has the average kinetic energy is called the rms speed ¢ =(1/2)(mvrms2) THE INTERNAL ENERGY OF A MONATOMIC IDEAL GAS