Transcript Physical Science and Math Relationships

Physical Science
& Math Relationships
Terra Alta/East Preston School
Preston County Schools
Physics


natural science
deals with matter and energy


defines & characterizes
interactions between matter and energy
2
Matter


a physical substance
characteristics of all matter


occupies space
has mass
3
Energy

capacity for doing work
4
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
5
Significant Figures

# 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:
 3.47
 0.039
 206.1
 5.90
significant figures
6
Significant Figures

# 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:
 3.47
 0.039
 206.1
 5.90
significant figures
3
2
4
2
7
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
8
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)

9
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)
10
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)
11
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)
12
Rounding Example 1
73.2
8.0627
93.57
+ 66.296
241.1287
241.1
# decimal places =
to least precise value
13
Rounding Example 2
2.4832
x 30.51
75.762432
75.76
# significant figures =
to least accurate number
14
Numerical Relationships

direct linear

as x y (or vice versa)
y = kx
expressed as proportion y  x
example: x
y
(for y = 5x)
1
5
2
10
3
15
example formula

15
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  x2
example: x
y
(for y = 5x2)
1
5
2
20
3
45
16
Numerical Relationships (cont.)

indirect

as x
y
x y = constant
expressed as proportion y 1/x
example: x
y
(for xy = 100)
1 100
2
50
4
25
example formula

17
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 1/ x2
example: x
y
(for x2y = 100)
1 100
2 25
4
6.25
18
Graphs

used to display
relationships between
2 variables

Y-axis (dependent)
measured value

X-axis (independent)
controlled value
x-axis
19
Graphic Relationships ( on linear graph paper)
 slope
(left to right)

Y
X
direct = ascending

indirect = descending
m
 shape

linear = straight

exponential = curved
20
Evaluating Graphed Information



identify variables
describe shape & slope of line
correlate information to theory
21
Example #1
Relationship of mA to Intensity
Exposure (m R)

100
90
80
70
60
50
40
30
20
10
0
0
100 200 300 400 500 600
mA
22
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
Exposure (mR)

100
80
60
40
20
0
0 100 200 300 400 500 600
mA
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)
23
Example #2
Relationship of the # days before exam to
amount of study time
Study Time (HRS)

5
4
3
2
1
0
0
1
2
3
4
5
6
Days before exam
24
Quantities & Units

quantity = measurable property
quantity
length
mass
time

definition (what is measured)
distance between two points
amount of matter (not weight)
duration of an event
unit = standard used to express a measurement
quantity
length
mass
time
unit
other units
meter
kilogram
second
25
Unit Systems
System
length
mass
time
English
foot
slug (pound) second
metric SI**
meter
kilogram second
** also ampere, Kelvin, mole, candela
metric MKS meter
kilogram second
metric CGS
centimeter gram
second
Do not mix unit systems when doing calculations!!
26
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
1min 1hour
3825 sec 

60sec 60min
1.0625hour 1.063hour
27
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
tera-
gigamegakilocentimilli
micronanopico-
symbol
T
G
M
k
c
m
m
n
p
factor
numerical equivalent
1012 1 000 000 000 000
109
106
103
10-2
10-3
10-6
10-9
10-12
1 000 000 000
1 000 000
1 000
.01
.001
.000 001
.000 000 001
.000 000 000 001
28
Rules for Using Prefixes

To use a prefix divide by prefix value &
include the prefix with the unit
45000m  45000m  10

3 m
km
 45km
To remove a prefix multiply by prefix value &
delete prefix notation from the unit
850ml  850ml 10
3 l
ml
 .85l
29
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
30
Derived Quantities & Units

properties which arrived at by combining base
quantities
quantity
units
area
mxm
m2
surface measure
volume
mxmxm
m3
capacity
velocity
m/s
m/s
distance traveled per unit time
m/s2
rate of change of velocity
acceleration m/s/s
definition for quantity
ms-2
31
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
(rad)
Gy
radiation energy deposited
in matter
32
Solving Problems
1.
2.
3.
4.
Determine unknown quantity
Identify known quantities
Select an equation (fits known & unknown quantities)
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
33
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
34
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
35
Vector Addition/Subtraction


requires use of graphs, trigonometry or
special mathematical rules to solve
example:
F2
F1
F1 + F2 =
Net force
36
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
37
Speed (cont.)
General Formula:
d
v =
t
d in m
t in s
v = m/s
Variations:
instantaneous
uniform
average
v at 1 point in time
same at
all times
total distance
total time
time
38
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 = ??
v = d/t
6.0 cm = distance
.25 ns = time
(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)
39
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?
40
Acceleration

rate of change of velocity with time



if velocity changes there is acceleration
includes: v
formula:
v
v
a =
t
v = vf - vi
direction
units
v in m/s
t in s
a = m/s2
41
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 =  t
32 m/s = vf
a=v/t
 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]
42
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 eundergo a -a close to the nucleus of an atom
43
Newton's Laws of Motion
1. Inertia
2. Force
3. Recoil
44
Newton's First Law


defined -- in notes
inertia: resistance to a  in motion


property of all matter
mass = a measure of inertia
45
Inertia
Semi-trailer truck
 large mass
 large inertia
Bicycle
 small mass
 small inertia
46
Newton's 2nd Law (Force)

Force


anything that can object's motion
Fundamental forces

Nuclear forces

"strong" & "weak"
Gravitational force
 Electromagnetic force

47
Mechanical Force


push or pull
vector quantity

net force = vector sum of all forces
push on box + friction from floor
Vector sum

equilibrium -- net force = 0
48
2nd Law (Force)


defined -- in notes
formula for the quantity “force”
force
F
=
=
mass x
m x
acceleration
a
v
a= t

units
kg
x
kg m
s2
m/s2
Newton
N
49
Example Problem for 2nd Law
What 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 = ma
= 5.1 kg x 3.2 m/s2
= 16.32 kg m/s2
= 16 N
50
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 =ma
a = F/m
= 275[kg m/s2] / 110kg
= 2.5 m/s2
51
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/s2
m = 376 [kg m/s2] / 7.5 m/s2
= 50.13333333333 kg = 50 kg
52
Weight

adaptation of Newton's 2nd law

weight = force caused by the pull of gravitation
weight
gravitational force
varies with gravity
unit = N [pound]


mass
inertia of the object
always constant
unit = kg [slug]
when g is a constant then weight proportional mass
53
Weight (cont.)

formula for quantity “weight”
modified from force formula
F =
Wt. =
units
m
m
x
x
kg
x m/s2
kg m
s2
a
g
gearth = 9.8m/s2
Newton
N
54
Weight Problem
What is the weight (on earth) of a 42 kg person?
Wt. = ?? 42 kg = mass
Wt. =
=
=
=
[9.8m/s2 = gravity]
m x g
42 kg x 9.8m/s2
411.6 kg m/s2
410 N
55
Weight Problem #2
What is the mass of a 2287N mobile x-ray unit?
2287N = Wt [9.8m/s2 = gravity]
mass = ??
Wt.
m
=
=
=
=
=
m
x g
Wt./g
2287N / 9.8m/s2
233.3673469388 kg
233.4 kg
56
3rd Law (Recoil)

Defined -- in notes



no single force in nature
all forces act in pairs
 action vs. reaction
A
formula
FAB = -FBA
B
57
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=

units
=
=
m
x
v
kg x m/s
kg m
s
58
Momentum vs. Mass (Inertia)
p = m x
p  m
v
m= p
m = p
Direct proportional relationship
59
Momentum vs. Velocity
p = m x
p  v
v
50 km/hr
100 km/hr
v= p
v = p
Direct proportional relationship
60
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 =
=
=
=
m
x
v
8.8 kg x 1.24 m/s
10.912 kg m/s
11 kg m/s
61
Momentum Problem #2
What 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
62
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
63
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
64
Example
before collision
m1v1= 1kg m/s
mv = 0
collision occurs
after collision
mv = 0
m2v2= 1kg m/s
65
Example #2
A
B
m1v1= 5kg m/s
before collision
mv = 0
collision occurs
A
B
after collision
m2v2= 5kg m/s
m2 = mA + mB
v2 = vA + vB
66
Work

defined -- in notes


measures the change
a force has on an
object's position or
motion
F
If there is NO
change in position or
motion, NO
mechanical work is
done.
d
67
Work (cont.)

formula
Work = force
W = F

units
= N
= kg m
=
s2
kg m2
s2
x distance
x
d
x
m
x m
= Joule
J
68
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 = distance
130 N = force
W= F
x d
= 130 N x 8.2 m
= 1066 N m
= 1100 J
(1.1 kJ)
69
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
70
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
71
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
72
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 v2
= ½ x 450 kg x [6 m/s]2
= 8100 kg m2 /s2
= 8000 J or 8 kJ
73
Potential Energy


capacity to do work because of the object's
position in a force field
fields



nuclear
electromagnetic
gravitational
74
Gravitational Potential Energy


barbell with PE
formula
m
PEg = mass x gravity x height
= m x g
x h

h
units
= kg x
m/s2
=
kg m2
s2
=
Joule J
x m
g
75
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
Peg
42 m = height [9.8 m/s2 = gravity]
= 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
76
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
77
Example
What is the energy equivalent of a 2.2 kg object?
Em = ??
2.2 kg = mass
[3x108 m/s = speed of light]
Em = m c 2
= 2.2 kg x [3x108 m/s ]2
= 1.98 x 1017 kg m2 /s2
= 2.0 x 1017 J [trailing 0 is significant]
78
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
79
Power

Rate at which work is done


Faster work = more power
Rate at which energy changes

Large E  = more power
80
Power (cont.)

formula
power = work / time
P = W /t

or
or
energy / time
E / t
units = J / s
=
=
kg m2
s2
s
kg m2
s3
= Watt
W
81
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
82
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.
83
Heat Transfer


movement of heat energy from the hotter to
cooler object (or portion of object)
3 methods of transfer



conduction
convection
radiation
84
conduction


primary means in solid objects
classification of matter by heat transfer


conductors--rapid transfer
insulator--very slow to transfer
85
convection


primary means in gasses and liquids
convection current--continuing rise of
heated g/l and sinking of cool g/l
86
radiation

transfer without the use of a medium


(i.e. no solid, liquid or gas)
occurs in a vacuum
87
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
88
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
89
Temperature Scales
Scales
Boil (steam)
212°
Celsius
100°
Kelvin (SI) 373
1K = 1°C =
Fahrenheit
Freeze (ice)
No KE
32°
0°
273
1.8°F
-460°
-273°
0
Conversion formulae
°F = 32 + (1.8 °C)
°C = (°F - 32)  1.8
K = °C + 273
90