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# General Physics I PHYS 1410

UNT

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This 96 page Class Notes was uploaded by Annalise Morissette on Sunday October 25, 2015. The Class Notes belongs to PHYS 1410 at University of North Texas taught by Arup Neogi in Fall. Since its upload, it has received 110 views. For similar materials see /class/229205/phys-1410-university-of-north-texas in Physics 2 at University of North Texas.

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Date Created: 10/25/15

Phys 1410 PhysicsI Prerequisite Math 1100 or equivalent College Algebra Arup Neogi Professor of Nanotechnology Photonics Materials LaboratOry Department Qf PhysicSS University of North Texas Textbook Physics Volume 1 9th Edition Cuttnell amp Johnson John Wiley and Son s Prerequisite in Mathematics Math 1100 or equivalent Simple fractions algebraic equations unit conversions multiplication division Grading Attendance 6 Quiz 4 Homework 10 Midterm I 20 Midterm II 20 Midterm III 20 Final Exam 20 Cosmic View the Milky Way at 10 million light years from the Earth Then move through space towards the Earth in successive orders of magnitude Macroscopic A tall oak tree just outside the buildings of the National High Magnetic Field Laboratory in Tallahassee Florida After that begin to move from the actual size of a leaf into a microscopic world that reveals leaf Microscopic cell walls the cell nucleus chromatin DNA Subatomic Electrons Neutrons Protons h p quotimam mwicl Ila edupmnemau 1 Irr 1cv mln39 LA39quotltHfzl i ll l lllLli X him Chapter 1 1 l T he Nature of Physics Physics has developed out of the efforts of men and women to explain our physical environment Physics encompasses a remarkable variety of phenomena planetary orbits radio and TV waves magnetism lasers many more 10 1 l T he Nature of Physics Physics predicts how nature will behave in one situation based on the results of experimental data obtained in another situation Newton s Laws gt Rocketry Maxwell s Equations gt Telecommunications 11 1 2 Physics experiments involve the measurement of a variety of quantities These measurements should be accurate and reproducible The first step in ensuring accuracy and reproducibility is defining the units in which the measurements are made 12 12 SI units meter m unit of length kilogram kg unit of mass second 3 unit of time 13 12 Units 14 12 Units 15 Table M Units of Measurement Syetem SI CGS BE Length Meter 111 Centimeter em Poet ft Mass Kilogram kg Gram g Slug 51 Time Second 8 Second 5 Second 8 16 I 2 The units for length mass and time as well as a few others are regarded as base SI units These units are used in combination to define additional units for other important physical quantities such as force and energy 17 13 T he Role of Units in Problem Solving THE CONVERSION OF UNITS 1ft 03048 m 1mi 1609 km 1 hp 746 w 1 liter 10393 m3 18 13 T he Role of Units in Problem Solving Example 1 The World s Highest Waterfall The highest waterfall in the world is Angel Falls in Venezuela with a total drop of 9790 m Express this drop in feet Since 3281 feet 1 meter it follows that 3281 feet1 meter 1 3281feet 3212 feet lmeter Length 9790 meters 19 Table 12 Standard Pre xes Used to Denote Multiples of Ten Pre x Symbol Fawn i tera T IIII2 gigaquot G 109 mega M 10 kilo k 03 hecto h 102 deka da 10L deci d 10 1 cermi c 10392 milli m 1 0 3 micro u 10quot6 name n 10 9 pico p ID 2 femto f 10 5 Apptndix A cuntajms a discuussiml of powers 0f ten and scienli c notation l1Proncunced jig39a 20 13 T he Rfo le of Units in Problem Solving Reasoning Strategy Converting Between Units 391 In all calculations write down the units explicitly 2 Treat all units as algebraic quantities When identical units are divided they are eliminated algebraically 3 Use the conversion factors located on the page 39facing the inside cover Be guided by the fact that multiplying or dividing an equation by a factor of 1 does not alter the equation 21 13 The Role of Units in Problem Solving Example 2 Interstate Speed Limit Express the speed limit of 65 mileshour in terms of meterssecond Use 5280 feet 1 mile and 3600 seconds 1 hour and 3281 feet 1 meter Speed 2 65 mm Speed 2 95 feet feet second miles 5280 feet lhour 95 hour mile 3600 5 second feet lmeter meters second 3281feet second 22 13 T he Role of Units in Problem Solving DIMENSIONAL ANALYSIS L length M mass T time Is the following equation dimensionally correct 23 13 The Role of Units in Problem Solving Is the following equation dimensionally correct x 2 v2 1 L HT L 24 14 Trigar mmetiy h hypotenuse he length of side opposite the 900 an le 6 r 8 ha length of side adjacent to the angle 9 25 14 Trigonometry h 8111 6 0 h h hypotenuse ho length of side h opposite the GOSH a 90 angle 9 h ha length of side adjacent to the angle 9 h tan 6 0 h 26 tan 9 4 tan 50quot 672m 2 1 ha 62 m ho tan 500 672m 800m 27 14 Trigonometry 0 6 sin 1 h h h hypotenuse ho length of side 1 h opposite the 6 COS f 900 an la 6 r g 9 ha length of side adjacent to the angle 9 6 tan1 ha 28 i faw r a w 1 0 29 14 Trigonometry Pythagorean theorem 12 12 12 0 Cl h hypotenuse ho length of side opposite the 9039 an le 9 r g ha length of side adjacent to the angle 9 3O 15 Scalars and Vectors A scalar quantity is one that can be described by a single number temperature speed mass A vector quantity deals inherently with both magnitude and direction velocity force displacement 31 15 Swims and Vectors Arrows are used to represent vectors The direction of the arrow gives the direction of the vector By convention the length of a vector arrow is proportional to the magnitude of the vector 3 32 33 L6 VectarAddition and Subtraction Often it is necessary to add one vector to another Tail to head 0 gt1 ii Start V V 0 Finish at v 34 Tai to head K J I Start 39 Finish ii 5m 3m 8m 35 Finish Tai l to head 36 zion Wl 600 m X I 200 m 37 1 6 VectorAddition and Subtraction Ifzc omf00mf RzJQDOmY00mY63mn 200 m 600 m 38 16 VectorAddition and Subtraction tan 200600 6 tan 1200600184 632 m 200 m 600 m 39 L6 VectarAddition and Subtraction When a vector is multiplied by 1 the magnitude of the vector remains the same but the direction of the vector is reversed I J Tailto head 1 Ii 40 1 6 VectorAddition and Subtraction wl A1 41 17 The Components of a Vector Finish i and 37 are called the x vector component and the y vector component of f 42 1 7 The Components of a Vector y gt gtl X The vector components of A are two perpendicular vectors Ax and Ky that are parallel to the x and y axes and add together vectorially so that A Ax Ky 43 17 The Components of a Vector It is often easier to work with the scalar components rather than the vector components Ax and AV are the scalar components of A n A y i and 37 are unit vectors with magnitude 1 A Ax X Ayy Ax r Agt 44 17 The Components of a Vector Example A displacement vector has a magnitude of 175 m and points at an angle of 500 degrees relative to the X axis Find the X and y components of this vector sin6 yr y rsine 175 msin500 134 m coslt9xr x rcos 175 mXcosSODquot 112m F112m 134m7 45 7 m u l Ax Ayy 3x5 By Ax 3365 By CxAxBx CyZAyBy 47 Chapter 2 48 Motion Basics What is Motion changing position How do we know that something has motion How can we change its motion What controls motion Why do things stop amp go they way they do Aristotle vs Galileo Technical terms used to describe motion I Speed and Velocity l Acceleration 49 quotGalileo Formulated the Concepts of Speed amp Velocity How does its position change with timequot How fast is it goingquot Speed Average Speed distance time Ex miles per39 hour39 per39 means divided byquot Compare snail to race car39 driver39 and what about an airplane pilot 50 How fast is it going Speed Velocifyl What direction is it going 0 Velocity is a Vector quantit A vector quantity specn les dlr39ec Ion as well as magnitude 0 Motion is relative Constant Motion vs Changing Motion Is the earth really movingquot 51 Kinematics deals with the concepts that are needed to describe motion Dynamics deals with the effect that forces have on motion Together kinematics and dynamics form the branch of physics known as Mechanics 52 21 Displacement Origin 0 f it Displacement A X b 120 2 initial position i 2 nal position A12 2 i i0 displacement 53 21 Displacement 54 21 Displacement i20m Aiz i m i0TOm Ai2i i0 20m 70m 50m 55 21 Displacement i02 2Om X50m Aii i0 50m 20m70m 56 22 Speed and Velocity Average speed is the distance traveled divided by the time required to cover the distance Distance Average speed Elapsed tlme SI units for speed meters per second ms 57 22 Speed and Velocity Example 1 Distance Run by a Jogger How far does a jogger run in 15 hours 5400 s if his average speed is 222 ms Distance Average speed 2 Elapsed time Distance 2 Average speedXElapsed time 222 ms 5400 s 12000 m 58 22 Speed and Velocity Average velocity is the displacement divided by the elapsed time D39 l t Average velocrry w Elapsed time i i Ai t t At 59 22 Speedfand Velocity Example 2 The World s Fastest JetEngine Car Andy Green in the car ThrustSSC set a world record of 3411 ms in 1997 To establish such a record the driver makes two runs through the course one in each direction to nullify wind effects From the data determine the average velocity for each run Start Finish Ea A 1609 m u 146955 r0Os Finish Start d A 1609 m I 6O 22 Speed and Velocity 05 147405 A 1609m 039 V 3395ms if E At 4740 S Start Finish t E ASE 1609 m a a i4 i95 s 00 5 33427m t 4695 5 Finish Start 4a2 r A Ai 1609 m I7 61 22 Speed and Velocity The instantaneous velocity indicates how fast the car moves and the direction of motion at each instant of time 62 The notiOn of acceleration emerges when a change in velocity is combined with the time during which the quot change occurs 63 M 39 n 64 2 3 Acceleration Example 3 Acceleration and Increasing Velocity Determine the average acceleration of the plane V70Oms x72260kmh t0OS 1295 all 6 60 260kmh Okmh 9 Okmh t to 29s Os I s 65 gt W 18 kmh 2 Acceleration Example 3 Acceleration and Decreasing Velocity 13n s 28ny ml 39 II 50ms2 12s 9s 5 5 0 mm V 28 ms 1 x 13 ms gt ampgt x2 gt m 24 Equations of Kinematics for Constant Acceleration ltl SNI X0 l 0 N941 V VO t IO It is customary to dispense with the use of boldface symbols overdrawn with arrows for the displacement velocity and acceleration vectors We will however continue to convey the directions with a plus or minus sign 69 24 Equatibns 0f Kinematics for Constant Acceleration Let the object be at the origin when the clock starts x020 100 x x x 0 V v tt0 gt t x217tv0 vt 7O 71 24 ofKinematicsfor Constant Acceleration Five kinematic variables 1 displacement X 2 acceleration constant a 3 final velocity at time t v 4 initial velocity v0 5 elapsed time t 72 24 Equations 0f Kinematics for Constant Acceleration vv0at J v0vtv04 v04 aft 1 2 x VJ3at i x2 73 24 Equations of Kinematics for Constant Acceleration a 20 mfs u x vat at2 60 ms80 s2Om32X8O s2 2 110 m 74 40fKinematicsf0r Constant Acceleration 39 110 0 ms 1 31 mfs v I I Hr E I n r b1 Example 6 Catapulting a Jet Find its displacement V00mS czz31ms2 x v 62 ms 75 24 Equations of Kinematics for Constant Acceleration azv tvo gt tzv avo 1 1 VVo x3v0 vt 2310 v 76 24 Equatibns 0f Kinematics for Constant Acceleration L 1 I if 7 v 100 ml39s a3J mfs2 r 1v262mr39s 39 ilt39 uj gt 3971Vquot 3 Tl L Ni 3 W i 7 7 H 7 77 713 k 2 1 I r h 77 24 Equatibns 0f Kinematics for Constant Acceleration Equations of Kinematics for ConstantAcceleration vv0at v0 vt NIH x 2 2 v v0 2ax x v0tat2 78 2 5 Applications of the Equations of Kinematics Reasoning Strategy 1 Make a drawing 2 Decide which directions are to be called positive and negative 3 Write down the values that are given for any of the five kinematic variables 4 Verify that the information contains values for at least three of the five kinematic variables Select the appropriate equation 5 When the motion is divided into segments rememberthat the final velocity of one segment is the initial velocity for the next 6 Keep in mind that there may be two possible answers to a kinematics problem 79 2 5 Applications of the Equations of Kinematics Example 8 An Accelerating Spacecraft A spacecraft is traveling with a velocity of 3250 ms Suddenly the retrorockets are fired and the spacecraft begins to slow down with an acceleration whose magnitude is 100 ms2 What is the velocity of the spacecraft when the displacement of the craft is 215 km relative to the point where the retrorockets began nng x a v v0 t 215000 m 100 III82 3250 ms 80 3939 1215 km quot g5 39 3 a L hit H a l I 1 v3 3250 mfs I 25U mfs N x 215 kl39h Eu 8 1 2 5 Applications of the Equations of Kinematics x a v 10 t 215000 m 100 III82 3250 ms 2 2 V v02ax gtv1v022ax v i3250ms2 2100ms2 215000 m i2500 ms 2 6 Freebz Falling Bodies In the absence of air resistance it is found that all bodies at the same location above the Earth fall vertically with the same acceleration If the distance of the fall is small compared to the radius of the Earth then the acceleration remains essentially constant throughout the descent This idealized motion is called freefall and the acceleration of a freely falling body is called the acceleration due to gravity g980ms2 or 322fts2 83 4 1 Tng Bvdiess g 98011182 Airefillecl Evacuated tu be tu be a I 84 Bodies Example 10 A Falling Stone Astone is dropped from the top of a tall building After 3003 of free fall what is the displacement y of the stone i i 85 980 ms2 0 ms 300 s 86 2 6 Freely Falling Bodies y 980 ms2 0 ms 300 s y 2 var azt2 0 ms300 s 441 m i 2 980ms2300 s2 87 2 6 Freely Falling Bodies Example 12 How High Does it Go The referee tosses the coin up with an initial speed of 500ms In the absence if air resistance how high does the coin go above its point of release v 0 ms glI h v0 500 ms 88 2 6 Freebz Falling Bodies v 0 ms f I4 gt v0 500 ms y a V V0 980 1 1 182 0 ms 500 ms 89 2 6 Freely Falling Bodies y a 1 V0 t 980 ms2 0 ms 500 ms v2 v2 12 v022ay gt yz O 2a y 2 v2 v Oms 500ms 2a 2 980ms2 2128111 2 6 Frieer Falling Bodies Conceptual Example 14 Acceleration Versus Velocity There are three parts to the motion of the coin On the way up the coin has a vector velocity that is directed upward and has decreasing magnitude At the top of its path the coin momentarily has zero velocity On the way down the coin has downwardpointing velocity with an increasing magnitude In the absence of air resistance does the acceleration of the coin like the velocity change from one part to another 91 v Bodies Gonceptual Example 15 Taking Advantage of Symmetry JDOes the pellet in part b strike the ground beneath the cliff C With a smaller greater or the same speed as the pellet 139 i part a KW l l l l 92 2 7 Graphical Analysis of Velocity and Acceleration 16 12 8 Position xm 4 I 93 Position 1139 m 1200 800 400 I IiAnabsis 0f Velocity and Acceleration Positive gt velocity Zero velocity i 4 Negative velocity E 2 39 A z 400 s apom m4005 quot 39I l I l I IA 3x5400m 1 400m I 3 1 i At 200 s I I l I I I 200 400 600 800 1000 1200 1400 1600 1800 Time 5 94 2 7 Graphical Analysis of Velocity and Acceleration Position x m 800 600 400 200 Tangent line Ax 26 m 100 150 200 250 Time I s 50 95 2 7 Graphical Analysis of Velocity and Acceleration 36 E E 24 51 3 12 D I O O l 2 3 4 5 Time I 3 Av 12ms Slope 6ms2 At 23 96

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