physical science & math relationships terra alta/east preston school preston county schools

Physical Science& Math Relationships

Terra Alta/East Preston School

Preston County Schools

2

Physics natural science deals with matter and energy

defines & characterizes interactions between matter and energy

3

Matter a physical substance characteristics of all matter

occupies space has mass

4

Energy capacity for doing work

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

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# 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

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# 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 3 0.039 2 206.1 4 5.90 2

Significant Figures

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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

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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)

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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)

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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)

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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)

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Rounding Example 173.28.062793.57+ 66.296 241.1287

241.1 # decimal places = to least precise

value

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Rounding Example 22.4832

x 30.51

75.762432

75.76 # significant figures =to least accurate

number

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Numerical Relationships direct linear

as x ­ y ­­­(or vice versa) example formula y = k x expressed as proportion y µ x example: x y (for y = 5x) 1 5

2 103 15

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Numerical Relationships direct exponential

direct square (or other exponent) as x ­­y ­­by an exponential value­(or vice

versa)

example formula y = k x2

expressed as proportion y µ x2

example: x y (for y = 5x2) 1 5

2 203 45

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Numerical Relationships (cont.) indirect

as x ­ y ¯ example formula x y = constant

expressed as proportion y ­µ 1/x example: x y (for xy =

100)

1 100

2 50

4 25

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Numerical Relationships (cont.) indirect exponential

inverse square (or other exponent) as x ­ y ¯ by an exponential value­(or vice

versa)

example formula y x2 = constant expressed as proportion y ­µ 1/ x2

example: x y (for x2y = 100)

1 1002 254 6.25

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Graphs used to display

relationships between 2 variables Y-axis (dependent)

measured value

X-axis (independent) controlled

value x-axisy-

axis

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Graphic Relationships ( on linear graph paper)

slope (left to right)

direct = ascending indirect = descending

shape linear = straight exponential = curved

X

Ym

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

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Example #1 Relationship of mA to Intensity

0102030405060708090

100

0 100 200 300 400 500 600

mA

Exp

os

ure

(m

R)

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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)

0

20

40

60

80

100

0 100 200 300 400 500 600

mA

Ex

po

sure

(m

R)

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Example #2 Relationship of the # days before exam to

amount of study time

012345

0 1 2 3 4 5 6

Days before exam

Stu

dy T

ime

(HR

S)

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Quantities & Units quantity = measurable property

quantity definition (what is measured) length distance between two points

mass amount of matter (not weight)

time duration of an event

unit = standard used to express a measurementquantity unit other units length meter

mass kilogram

time second

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Unit SystemsSystem length mass time

English foot slug (pound) second

metric SI** meterkilogram second

** also ampere, Kelvin, mole, candela

metric MKS meterkilogram second

metric CGS centimeter gram second

Do not mix unit systems when doing calculations!!

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Converting Units convert 3825 seconds to hours

identify conversion factor(s) neededfactors needed: 60 sec = 1 min & 60 min = 1 hour

arrange factors in logical progressionFor seconds hours

sec min/sec hour/min set up calculation

60min

1hour

60sec

1minsec3825 hour1.0625 hour1.063

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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 prefix symbol factor numerical equivalent

tera- T 1012 1 000 000 000 000 giga- G 109 1 000 000 000 mega- M 106 1 000 000 kilo- k 103 1 000 centi- c 10-2 .01 milli m 10-3 .001 micro- m 10-6 .000 001 nano- n 10-9 .000 000 001 pico- p 10-12 .000 000 000 001

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Rules for Using Prefixes To use a prefix divide by prefix value &

include the prefix with the unit kmmm km

m 45104500045000 3

lmlml mll 85.10850850 3

To remove a prefix multiply by prefix value & delete prefix notation from the unit

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Base Quantities & Units (SI) describes a fundamental property of matter cannot be broken down further quantity SI unit definition for quantity

length meter distance between two points

mass kilogram amount of matter (not weight)

time second duration of an event

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Derived Quantities & Units properties which arrived at by combining base

quantities

quantity units definition for quantityarea m x m m2 surface measure

volume m x m x m m3 capacity

velocity m/s m/s distance traveled per unit time

acceleration m/s/s m/s2 rate of change of velocity

ms-2

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Derived Quantities with Named Units quantities with complex SI units

quantity units definition

frequency Hertz Hz # of ?? per second

force Newton N "push or pull"

energy Joule J ability to do work

absorbed dose Gray Gy radiation energy

deposited (rad) in matter

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Solving Problems1. Determine unknown quantity2. Identify known quantities3. 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

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Mechanics study of motion & forces motion = change in position or orientation types of motion

translationone place to another

rotation around axis of object's mass

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Measuring Quantities in Mechanics all have magnitude & unit

scalar vs. vector quantities Scalar -- magnitude & unit

Vector -- magnitude, unit & direction

run 2 kmvs

run 2 km east

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Vector Addition/Subtraction requires use of graphs, trigonometry or

special mathematical rules to solve example:

F1

F2F1 + F2 =Net force

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Quantities in Mechanics speed

rate at which an object covers distance rate

indicates a relationship between 2 quantities $/hour exams/tech # of people/sq. mile

speed = distance/time

speed is a scalar quantity

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Speed (cont.)v =

d t

d in mt in sv = m/s

same at all times

total distancetotal time

General Formula:

Variations:

instantaneous uniform average

dist

ance

time

v at 1 point in time

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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 = .25x10-9s

= 6 x 10-2 m / .25x10-9s

= 2.40000 x 108 m/s (raw answer)

= 2.4 x 108 m/s (sig. fig. answer)

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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?

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Acceleration rate of change of velocity with time

if velocity changes there is acceleration includes: ­­v ¯­v ­D direction formula:

­­D v = vf - vi

units v in m/s t in s a = m/s2

a = ­ ­­D v ­ ­­D t

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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 = vi 12 s = D t 32 m/s = vf

a = D v / D t

D v = vf - vi = 32m/s - 48 m/s = -16 m/s

a = -16m/s / 12 s = -1.3333333333 m/s2

= -1.3 m/s2 [ -sign designates slowing down]

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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

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Newton's Laws of Motion

1. Inertia

2. Force

3. Recoil

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Newton's First Law defined -- in notes inertia: resistance to a D in motion

property of all matter mass = a measure of inertia

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InertiaSemi-trailer truck

large mass large inertia

Bicycle small mass small inertia

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Newton's 2nd Law (Force) Force

anything that can ­D object's motion Fundamental forces

Nuclear forces "strong" & "weak"

Gravitational force Electromagnetic force

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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

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2nd Law (Force) defined -- in notes formula for the quantity “force”

force = mass x acceleration

F = m x aa =

v t

kg ms2

Newton N

units kg x m/s2

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Example Problem for 2nd LawWhat is the net force needed to accelerate a 5.1 kg laundry cart to 3.2 m/s2?

F =?? 5.1 kg = mass 3.2 m/s2 = acceleration

F = m a

= 5.1 kg x 3.2 m/s2

= 16.32 kg m/s2

= 16 N

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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/s2] / 110kg

= 2.5 m/s2

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Example 3An 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 = ­D t 15 m/s = ­D v

F = m a ­\ m = F/a

a = Dv/Dt

= 15 m/s / 2 s = 7.5 m/s2

m = 376 [kg m/s2] / 7.5 m/s2

= 50.13333333333 kg = 50 kg

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Weight adaptation of Newton's 2nd law weight = force caused by the pull of gravitation

weight ¹ massgravitational force inertia of the object

varies with gravity always constant

unit = N [pound] unit = kg [slug]

when g is a constant then weight proportional mass

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Weight (cont.) formula for quantity “weight”

modified from force formulaF = m x a

Wt. = m x g gearth = 9.8m/s2

kg ms2

Newton N

units kg x m/s2

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Weight ProblemWhat is the weight (on earth) of a 42 kg person?

Wt. = ?? 42 kg = mass [9.8m/s2 = gravity]

Wt. = m x g

= 42 kg x 9.8m/s2

= 411.6 kg m/s2

= 410 N

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Weight Problem #2What is the mass of a 2287N mobile x-ray unit?

mass = ?? 2287N = Wt [9.8m/s2 = gravity]

Wt. = m x g

m = Wt./g

= 2287N / 9.8m/s2

= 233.3673469388 kg

= 233.4 kg

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3rd Law (Recoil) Defined -- in notes

no single force in nature all forces act in pairs

action vs. reaction formula

FAB = -FBA

A

B

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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 = m x v units

= kg x m/s= kg m

s

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Momentum vs. Mass (Inertia)p = m x v

p µ m

¯­m = ¯­p­­m = ­ pDirect proportional relationship

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Momentum vs. Velocityp = m x

v

p µ v50 km/hr

¯­v = ¯­p

100 km/hr

­­v = ­ pDirect proportional relationship

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Momentum ProblemWhat 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 = m x v

= 8.8 kg x 1.24 m/s

= 10.912 kg m/s

= 11 kg m/s

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Momentum Problem #2What is the speed of a 3.5x104 kg car that has a momentum of 1.4x105 kg m/s?

velocity = ?? 3.5x104 kg = mass 1.4x105 kg m/s = momentum

p = m x v

v = p / m

= 1.4x105 kg m/s / 3.5x104 kg

= 4.0 x 100 m/s

= 4.0 m/s

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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

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Conservation of Linear Momentum

momentum after a collision will equal momentum before collision

results in a redistribution momentum among the objects

p1 = p2

m1v1 = m2v2

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Example

m1v1= 1kg m/s mv = 0

mv = 0 m2v2= 1kg m/s

before collision

collision occurs

after collision

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Example #2

m1v1= 5kg m/s mv = 0

m2v2= 5kg m/s

before collision

collision occurs

after collisionm2 = mA + mB v2 = vA + vB

A B

A B

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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

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Work (cont.) formula

Work = force x distance W = F x d

units = N x m=kg m

s2 x m

kg m2

s2 = Joule J=

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ExampleHow 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 = distance 130 N = force

W = F x d

= 130 N x 8.2 m

= 1066 N m

= 1100 J (1.1 kJ)

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 = F x d

= 162 N x 32 m

= 5184 N m

= 5200 J or 5.2 kJ

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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

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Kinetic Energy work done by the motion of an object

translation, rotation, or vibration formula

KE = ½ mass x velocity squared = ½ m v2

units = kg x [m/s]2

kg m2

s2 = Joule J=

73

ExampleFind 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 v2

= ½ x 450 kg x [6 m/s]2

= 8100 kg m2 /s2

= 8000 J or 8 kJ

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Potential Energy capacity to do work because of the object's

position in a force field fields

nuclear electromagnetic gravitational

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Gravitational Potential Energy barbell with PE formula

PEg = mass x gravity x height

= m x g x h

units= kg x m/s2 x m

=

hg

m

kg m2

s2

= Joule J

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ExampleHow 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/s2 = gravity]

Peg = m x g x h

= 460 kg x 9.8 m/s2 x 42 m

= 189 336 kg m2 /s2

= 190 000 J or 1.9x105 J or 190 kJ

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Rest Mass Energy energy due to mass Einstein's Theory formula (variation of KE formula)

Em = mass x speed of light squared

= m c2 [c = 3x108 m/s] units = kg x [m/s]2

kg m2

s2 = Joule J=

78

ExampleWhat is the energy equivalent of a 2.2 kg object? Em = ?? 2.2 kg = mass [3x108 m/s = speed of light]

Em = m c2

= 2.2 kg x [3x108 m/s ]2

= 1.98 x 1017 kg m2 /s2

= 2.0 x 1017 J [trailing 0 is significant]

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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

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Power Rate at which work is done

Faster work = more power Rate at which energy changes

Large E D = more power

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Power (cont.) formula

power = work / time or ­D energy / time

P = W / t or DE / t units = J / s

kg m2

s3 = Watt W=

kg m2

s2 = s

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ExampleHow 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

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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.

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Heat Transfer movement of heat energy from the hotter to

cooler object (or portion of object) 3 methods of transfer

conduction convection radiation

85

conduction primary means in solid objects classification of matter by heat transfer

conductors--rapid transfer insulator--very slow to transfer

86

convection primary means in gasses and liquids convection current--continuing rise of

heated g/l and sinking of cool g/l

87

radiation transfer without the use of a medium

(i.e. no solid, liquid or gas) occurs in a vacuum

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

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

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Temperature ScalesScales Boil (steam) Freeze (ice) No KE

Fahrenheit 212° 32° -460°

Celsius 100° 0° -273°

Kelvin (SI) 373 273 0

1K = 1°C = 1.8°FConversion formulae

°F = 32 + (1.8 °C)

°C = (°F - 32) ¸ 1.8

K = °C + 273