1. Physics Applied to Radiology Chapter 3 Fundamentals of Physics

2. 2 Physics natural science deals with matter and energy defines & characterizes interactions between matter and energy

3. 3 Matter a physical substance characteristics of all matter occupies space has mass

4. 4 Energy capacity for doing work

5. 5 Math exact vs. approximate numbers exact -- defined or counted approximate -- measured examples your height # of chairs in room # of seconds in a minute # seconds to run 100 m dash

6. 6 # of digits in a value when... leading & trailing zeros are ignored trailing 0 may be designated as significant the decimal place is disregarded How many significant figures? Value:significant figures 3.47 0.039 206.1 5.90 Significant Figures

7. 7 # of digits in a value when... leading & trailing zeros are ignored trailing 0 may be designated as significant the decimal place is disregarded How many significant figures? Value:significant figures 3.473 0.0392 206.14 5.902 Significant Figures

8. 8 Accuracy vs. Precision accuracy -- # of significant figures 3.47 is more accurate than 0.039 precision -- decimal position of the last significant figure 0.039 is more precise than 3.47

9. 9 Example Describe the accuracy and precision of the following information. 2.5 cm metal sheet with a.025 cm coat of paint accuracy is same for both (2 sig. fig.) precision is > for paint (1/1000 vs. 1/10)

10. 10 Rounded Numbers all approximate # are rounded last digit of approx. number is rounded last sig. fig. of an approx. # is never an accurate # error of last number is ½ of the last digit's place value (if place value is.1 then error =.05)

11. 11 Rounded Number example: if a measured value = 32.63 error is.005 (½ of.01) actual # is between 32.635 (32.63 +.005) 32.625 (32.63 -.005)

12. 12 Rounding Rules round at the end of the total calculation do not round after each step in complex calculations when - or + use least precise # (same # of decimal places) when x or ÷ use least accurate # (same # of sig. figures)

13. 13 Rounding Example 1 73.2 8.0627 93.57 + 66.296 241.1287 241.1# decimal places = to least precise value

14. 14 Rounding Example 2 2.4832 x 30.51 75.762432 75.76# significant figures =to least accurate number

15. 15 Numerical Relationships direct linear as x  y  (or vice versa) example formula y = k x expressed as proportion y  x example:xy(for y = 5x) 15 210 315

16. 16 Numerical Relationships direct exponential direct square (or other exponent) as x  y  by an exponential value  (or vice versa) example formula y = k x 2 expressed as proportion y  x 2 example:xy(for y = 5x 2 ) 15 220 345

17. 17 Numerical Relationships (cont.) indirect as x  y  example formula x y = constant expressed as proportion y  1/x example:xy(for xy = 100) 1100 250 425

18. 18 Numerical Relationships (cont.) indirect exponential inverse square (or other exponent) as x  y  by an exponential value  (or vice versa) example formula y x 2 = constant expressed as proportion y  1/ x 2 example:xy(for x 2 y = 100) 1100 225 46.25

19. 19 Graphs used to display relationships between 2 variables Y-axis (dependent) measured value X-axis (independent) controlled value x-axis y-axis

20. 20 Graphic Relationships ( on linear graph paper) slope (left to right) direct = ascending indirect = descending shape linear = straight exponential = curved

21. 21 Evaluating Graphed Information identify variables describe shape & slope of line correlate information to theory

22. 22 Example #1 Relationship of mA to Intensity

23. 23 Example #1 (evaluated) Relationship of mA to Intensity variables independent = mA dependent = Exposure shape & slope slope = ascending (=direct) shape = straight line (=linear) correlate to theory mA has a direct linear relationship to exposure; as mA increases exposure increases in a similar fashion; the graph demonstrates that if you double the mA (200 to 400) you also double the exposure (30 mR to 60 mR )

24. 24 Example #2 Relationship of the # days before exam to amount of study time

25. 25 Quantities & Units quantity = measurable property quantitydefinition (what is measured) lengthdistance between two points massamount of matter (not weight) timeduration of an event unit = standard used to express a measurement quantityunitother units length meter mass kilogram time second

26. 26 Unit Systems Systemlengthmasstime Englishfootslug (pound) second metric SI**meterkilogramsecond ** also ampere, Kelvin, mole, candela metric MKSmeterkilogramsecond metric CGScentimetergramsecond Do not mix unit systems when doing calculations!!

27. 27 Converting Units convert 3825 seconds to hours identify conversion factor(s) needed factors needed: 60 sec = 1 min & 60 min = 1 hour arrange factors in logical progression For seconds  hours sec  min/sec  hour/min set up calculation

28. 28 Dimensional Prefixes Bushong, table 2-3 (pg 23) used with metric unit systems modifiers used with unit a power of 10 to express the magnitude prefixsymbolfactor numerical equivalent tera- T10 12 1 000 000 000 000 giga- G10 9 1 000 000 000 mega- M10 6 1 000 000 kilo- k10 3 1 000 centi- c10 -2.01 milli m10 -3.001 micro-  10 -6.000 001 nano- n10 -9.000 000 001 pico- p10 -12.000 000 000 001

29. 29 Rules for Using Prefixes To use a prefix divide by prefix value & include the prefix with the unit To remove a prefix multiply by prefix value & delete prefix notation from the unit

30. 30 Base Quantities & Units (SI) describes a fundamental property of matter cannot be broken down further quantity SI unitdefinition for quantity length meter distance between two points mass kilogram amount of matter (not weight) time second duration of an event

31. 31 Derived Quantities & Units properties which arrived at by combining base quantities quantityunitsdefinition for quantity area m x mm 2 surface measure volume m x m x mm 3 capacity velocity m/sm/sdistance traveled per unit time acceleration m/s/sm/s 2 rate of change of velocity ms -2

32. 32 Derived Quantities with Named Units quantities with complex SI units quantityunitsdefinition frequencyHertzHz # of ?? per second forceNewtonN "push or pull" energyJouleJ ability to do work absorbed dose Gray Gy radiation energy deposited (rad) in matter

33. 33 Solving Problems 1. Determine unknown quantity 2. Identify known quantities 3. Select an equation (fits known & unknown quantities) 4. Set up numerical values in equation same unit or unit system 5. Solve for the unknown write answer with magnitude & units raw answer vs. answer in significant figures

34. 34 Mechanics study of motion & forces motion = change in position or orientation types of motion translation one place to another rotation around axis of object's mass

35. 35 Measuring Quantities in Mechanics all have magnitude & unit scalar vs. vector quantities Scalar -- magnitude & unit Vector -- magnitude, unit & direction run 2 km vs run 2 km east

36. 36 Vector Addition/Subtraction requires use of graphs, trigonometry or special mathematical rules to solve example: F1F1 F2F2 F 1 + F 2 = Net force

37. 37 Quantities in Mechanics speed rate at which an object covers distance rate indicates a relationship between 2 quantities $/hourexams/tech# of people/sq. mile speed = distance/time speed is a scalar quantity

38. 38 Speed (cont.) v = d t d in m t in s v = m/s same at all times total distance total time General Formula: Variations: instantaneousuniformaverage distance time v at 1 point in time

39. 39 Speed Example An e - travels the 6.0 cm distance between the anode & the cathode in.25 ns. What is the e - speed? [Assume 0 in 6.0 is significant] v = ?? 6.0 cm = distance.25 ns = time v= d/t (units: m/s  need to convert) 6.0 cm = 6.0 x 10 -2 m.25 ns =.25x10 -9 s = 6 x 10 -2 m /.25x10 -9 s = 2.40000 x 10 8 m/s(raw answer) = 2.4 x 10 8 m/s (sig. fig. answer)

40. 40 Velocity speed + the direction of the motion vector quantity A boat is traveling east at 15 km/hr and must pass through a current that is moving northeast at 10 km/hr. What will be the true velocity of the boat?

41. 41 Acceleration rate of change of velocity with time if velocity changes there is acceleration includes:  v  v  direction formula:  v = v f - v i units v in m/s t in s a = m/s 2 a =  v  t

42. 42 Acceleration Example A car is traveling at 48 m/s. After 12 seconds it is traveling at 32 m/s. What is the car’s acceleration? a = ?48 m/s = v i 12 s =  t32 m/s = v f a =  v /  t  v = v f - v i = 32m/s - 48 m/s = -16 m/s a = -16m/s / 12 s = -1.3333333333 m/s 2 = -1.3 m/s 2 [ -sign designates slowing down]

43. 43 Application of v and a in Radiology KE (motion) of e- used to produce x rays controlling the v of e- enables the control of the photon energies Brems photons are produced when e - undergo a -a close to the nucleus of an atom

44. 44 Newton's Laws of Motion 1. Inertia 2. Force 3. Recoil

45. 45 Newton's First Law defined -- in notes inertia:resistance to a  in motion property of all matter mass = a measure of inertia

46. 46 Inertia Semi-trailer truck  large mass  large inertia Bicycle  small mass  small inertia

47. 47 Newton's 2nd Law (Force) Force anything that can  object's motion Fundamental forces Nuclear forces "strong" & "weak" Gravitational force Electromagnetic force

48. 48 Mechanical Force push or pull vector quantity net force = vector sum of all forces push on box + friction from floor equilibrium -- net force = 0 Vector sum

49. 49 2nd Law (Force) defined -- in notes formula for the quantity “force” force = mass xacceleration F = m x a a =  v  t kg m s 2 Newton N units kgxm/s 2

50. 50 Example Problem for 2nd Law What is the net force needed to accelerate a 5.1 kg laundry cart to 3.2 m/s 2 ? F =?? 5.1 kg = mass 3.2 m/s 2 = acceleration F= m a = 5.1 kg x 3.2 m/s 2 = 16.32 kg m/s 2 = 16 N

51. 51 Example 2: A net force of 275 N is applied to a 110 kilogram mobile unit. What is the unit's acceleration? acceleration =?? 275 N = F 110 kg = mass F= m a a= F/m = 275[kg m/s 2 ] / 110kg = 2.5 m/s 2

52. 52 Example 3 An object experiences a net force of 376N. After 2 seconds the change in the object's velocity 15m/s. What is the object's mass? mass =?? 376 N = F 2 s =  t 15 m/s =  v F = m a  m = F/a a=  v/  t = 15 m / s / 2 s = 7.5 m/s 2 m = 376 [kg m/s 2 ] / 7.5 m/s 2 = 50.13333333333 kg = 50 kg

53. 53 Weight adaptation of Newton's 2nd law weight = force caused by the pull of gravitation weight  mass gravitational force inertia of the object varies with gravityalways constant unit = N [pound]unit = kg [slug] when g is a constant then weight proportional mass

54. 54 Weight (cont.) formula for quantity “weight” modified from force formula F = m xa Wt.=mx gg earth = 9.8m/s 2 kg m s 2 Newton N units kg x m/s 2

55. 55 Weight Problem What is the weight (on earth) of a 42 kg person? Wt. = ?? 42 kg = mass [9.8m/s 2 = gravity] Wt.= mx g = 42 kg x 9.8m/s 2 = 411.6 kg m/s 2 = 410 N

56. 56 Weight Problem #2 What is the mass of a 2287N mobile x-ray unit? mass = ?? 2287N = Wt [9.8m/s 2 = gravity] Wt.=mx g m = Wt./g = 2287N / 9.8m/s 2 = 233.3673469388 kg =233.4 kg

57. 57 3rd Law (Recoil) Defined -- in notes no single force in nature all forces act in pairs action vs. reaction formula F AB = -F BA A B

58. 58 Momentum (Linear) measures the amount of motion of an object tendency of an object to go in straight line when at a constant velocity formula p=mx v units =kgxm/s = kg m s

59. 59 Momentum vs. Mass (Inertia) p=mx v p  m  m =  p  m =  p Direct proportional relationship

60. 60 Momentum vs. Velocity p=mx v p  v 50 km/hr  v =  p 100 km/hr  v =  p Direct proportional relationship

61. 61 Momentum Problem What is the momentum of a 8.8 kg cart that has a speed of 1.24 m/s? p = ?? 8.8 kg = mass 1.24 m/s = velocity p= mx v = 8.8 kgx 1.24 m/s = 10.912 kg m/s = 11 kg m/s

62. 62 Momentum Problem #2 What is the speed of a 3.5x10 4 kg car that has a momentum of 1.4x10 5 kg m/s? velocity = ?? 3.5x10 4 kg = mass 1.4x10 5 kg m/s = momentum p=m x v v = p / m = 1.4x10 5 kg m/s / 3.5x10 4 kg = 4.0 x 10 0 m/s = 4.0 m/s

63. 63 Conservation Laws Statements about quantities which remain the same under specified conditions. Most Notable Conservation Laws Conservation of Energy Conservation of Matter Conservation of Linear Momentum

64. 64 Conservation of Linear Momentum momentum after a collision will equal momentum before collision results in a redistribution momentum among the objects p 1 = p 2 m 1 v 1 = m 2 v 2

65. 65 Example m 1 v 1 = 1 kg m/s mv = 0 m 2 v 2 = 1 kg m/s before collision collision occurs after collision

66. 66 Example #2 m 1 v 1 = 5 kg m/s mv = 0 m 2 v 2 = 5 kg m/s before collision collision occurs after collision m 2 = m A + m B v 2 = v A + v B A B A B

67. 67 Work defined -- in notes measures the change a force has on an object's position or motion If there is NO change in position or motion, NO mechanical work is done. F d

68. 68 Work (cont.) formula Work= force x distance W= F x d units= N x m = kg m s 2 x m kg m 2 s 2 = Joule J =

69. 69 Example How much mechanical work is done to lift a 12 kg mass 8.2 m off of the floor if a force of 130 N is applied? work = ??12 kg = mass 8.2 m = distance130 N = force W= Fx d = 130 N x 8.2 m = 1066 N m = 1100 J(1.1 kJ)

70. 70 Example #2 A 162 N force is used to move a 45 kg box 32 m. What is the work that is done moving the box? work = ?? 162 N = force 45 kg = mass 32 m = distance W= Fxd = 162 N x 32 m = 5184 N m = 5200 Jor 5.2 kJ

71. 71 Energy property of matter enables matter to perform work broad categories Kinetic Energy: due to motion Potential Energy: due to position in a force field Rest Energy: due to mass

72. 72 Kinetic Energy work done by the motion of an object translation, rotation, or vibration formula KE= ½ mass x velocity squared = ½ m v 2 units= kg x [m/s] 2 kg m 2 s 2 = Joule J =

73. 73 Example Find the kinetic energy of a 450 kg mobile unit moving at 6 m/s. kinetic energy = ?? 450 kg = mass 6 m/s = velocity KE= ½ m v 2 = ½ x 450 kg x [6 m/s] 2 = 8100 kg m 2 /s 2 = 8000 J or 8 kJ

74. 74 Potential Energy capacity to do work because of the object's position in a force field fields nuclear electromagnetic gravitational

75. 75 Gravitational Potential Energy barbell with PE formula PE g = mass x gravity x height = m x g x h units = kg x m/s 2 x m = h g m kg m 2 s 2 = Joule J

76. 76 Example How much energy does a 460 kg mobile unit possess when it is stationed on the 3rd floor of the hospital? (42m above ground) PE = ?? 460 kg = mass 42 m = height [9.8 m/s 2 = gravity] Pe g = m x g x h = 460 kg x 9.8 m/s 2 x 42 m = 189 336 kg m 2 /s 2 = 190 000 J or 1.9x10 5 J or 190 kJ

77. 77 Rest Mass Energy energy due to mass Einstein's Theory formula (variation of KE formula) E m = mass x speed of light squared = m c 2 [c = 3x10 8 m/s] units= kg x [m/s] 2 kg m 2 s 2 = Joule J =

78. 78 Example What is the energy equivalent of a 2.2 kg object? E m = ?? 2.2 kg = mass [3x10 8 m/s = speed of light] E m = m c 2 = 2.2 kg x [3x10 8 m/s ] 2 = 1.98 x 10 17 kg m 2 /s 2 = 2.0 x 10 17 J [trailing 0 is significant]

79. 79 Conservation Of Energy (Matter) Energy is neither created nor destroyed but can be interchanged (Matter is neither created nor destroyed but can be interchanged) Because mass has rest energy, conservation of matter & energy can be combined

80. 80 Power Rate at which work is done Faster work = more power Rate at which energy changes Large E  = more power

81. 81 Power (cont.) formula power= work / time or  energy / time P= W / t or  E / t units = J / s kg m 2 s 3 = Watt W = kg m 2 s 2 = s

82. 82 Example How much power is used when an 80N force moves a box 15 m during a 12 s period of time? (hint: solve for work first) P = ?? 80 N = force 15 m = distance 12 s = time P= W/t& W = Fd P= (F d) / t = (80 N x 15 m) / 12 s = 100 Nm/s = 100 W

83. 83 Heat energy internal kinetic energy of matter from the random motion of molecules or atoms KE & PE of molecules heat E in matter moves from area of higher E in object to area of lower internal E Unit -- Calorie (a form of the joule) amount of heat required to raise one gram of water one degree Celsius.

84. 84 Heat Transfer movement of heat energy from the hotter to cooler object (or portion of object) 3 methods of transfer conduction convection radiation

85. 85 conduction primary means in solid objects classification of matter by heat transfer conductors--rapid transfer insulator--very slow to transfer

86. 86 convection primary means in gasses and liquids convection current--continuing rise of heated g/l and sinking of cool g/l

87. 87 radiation transfer without the use of a medium (i.e. no solid, liquid or gas) occurs in a vacuum

88. 88 Heat Radiation term “radiation” may simply refer to heat energy and not the transfer of heat infra-red radiation, part of EM spectrum, is heat energy

89. 89 Effects of Heat Transfer change in physical state of matter solid  liquid  gas melt boil change in temperature measure of the average KE of an object relative measure of sensible heat or cold

90. 90 Temperature Scales Scales Boil (steam)Freeze (ice)No KE Fahrenheit 212°32°-460° Celsius 100°0°-273° Kelvin (SI) 3732730 1K = 1°C = 1.8°F Conversion formulae °F = 32 + (1.8 °C) °C = (°F - 32)  1.8 K = °C + 273