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J. Flesock
II. Unit Conversions
MEASUREMENT
A. SI Units
Quantity Base Unit Abbrev.
Length
Mass
Time
Temp
Amount
Symbol
A. SI Units
Quantity Base Unit Abbrev.
Length
Mass
Time
Temp
meter m
Amount
Symbol
l
A. SI Units
Quantity Base Unit Abbrev.
Length
Mass
Time
Temp
meter
kilogram
m
kg
Amount
Symbol
l
m
A. SI Units
Quantity Base Unit Abbrev.
Length
Mass
Time
Temp
meter
kilogram
second
m
kg
s
Amount
Symbol
l
m
t
A. SI Units
Quantity Base Unit Abbrev.
Length
Mass
Time
Temp
meter
kilogram
second
kelvin
m
kg
s
K
Amount
Symbol
l
m
tT
A. SI Units
Quantity Base Unit Abbrev.
Length
Mass
Time
Temp
meter
kilogram
second
kelvin
m
kg
s
K
Amount mole mol
Symbol
l
m
tTn
A. Derived Units
A. Derived Units
n Combination of base units.
A. Derived Units
n Combination of base units.
n Volume (m3 or cm3)
A. Derived Units
n Combination of base units.
n Volume (m3 or cm3) n length × length × length
1 cm3 = 1 mL1 dm3 = 1 L
A. Derived Units
n Combination of base units.
n Volume (m3 or cm3) n length × length × length
1 cm3 = 1 mL1 dm3 = 1 L
Ø Density (kg/m3 or g/cm3)
A. Derived Units
n Combination of base units.
n Volume (m3 or cm3) n length × length × length
D = MV
1 cm3 = 1 mL1 dm3 = 1 L
Ø Density (kg/m3 or g/cm3)wmass per volume
J. Flesock
A. SI Prefix Conversions
J. Flesock
A. SI Prefix Conversions
1. Find the difference between the exponents of the two prefixes.
J. Flesock
A. SI Prefix Conversions
1. Find the difference between the exponents of the two prefixes.
2. Move the decimal that many places.
J. Flesock
A. SI Prefix Conversions
1. Find the difference between the exponents of the two prefixes.
2. Move the decimal that many places.
To the leftor right?
J. Flesock
A. SI Prefix Conversions
532 m = _______ km
J. Flesock
=
A. SI Prefix Conversions
532 m = _______ km
J. Flesock
=
A. SI Prefix Conversions
NUMBERUNIT
532 m = _______ km
J. Flesock
=
A. SI Prefix Conversions
NUMBERUNIT UNIT
532 m = _______ km
J. Flesock
=
A. SI Prefix Conversions
NUMBERUNIT
NUMBER
UNIT
532 m = _______ km
J. Flesock
=
A. SI Prefix Conversions
NUMBERUNIT
NUMBER
UNIT
532 m = _______ km0.532
J. Flesock
A. SI Prefix Conversions
mega- M 106
deci- d 10-1
centi- c 10-2
milli- m 10-3
Prefix Symbol Factor
micro- µ 10-6
nano- n 10-9
pico- p 10-12
kilo- k 103
BASE UNIT --- 100
J. Flesock
A. SI Prefix Conversions
mega- M 106
deci- d 10-1
centi- c 10-2
milli- m 10-3
Prefix Symbol Factor
micro- µ 10-6
nano- n 10-9
pico- p 10-12
kilo- k 103
BASE UNIT --- 100
J. Flesock
A. SI Prefix Conversions
mega- M 106
deci- d 10-1
centi- c 10-2
milli- m 10-3
Prefix Symbol Factor
micro- µ 10-6
nano- n 10-9
pico- p 10-12
kilo- k 103
mov
e le
ft
BASE UNIT --- 100
J. Flesock
A. SI Prefix Conversions
mega- M 106
deci- d 10-1
centi- c 10-2
milli- m 10-3
Prefix Symbol Factor
micro- µ 10-6
nano- n 10-9
pico- p 10-12
kilo- k 103
mov
e le
ft
BASE UNIT --- 100
J. Flesock
A. SI Prefix Conversions
mega- M 106
deci- d 10-1
centi- c 10-2
milli- m 10-3
Prefix Symbol Factor
micro- µ 10-6
nano- n 10-9
pico- p 10-12
kilo- k 103
mov
e le
ft
mov
e rig
ht
BASE UNIT --- 100
J. Flesock
A. SI Prefix Conversions
1) 20 cm = ______________ m
2) 0.032 L = ______________ mL
3) 45 µm = ______________ nm
4) 805 dm = ______________ km
J. Flesock
A. SI Prefix Conversions
1) 20 cm = ______________ m
2) 0.032 L = ______________ mL
3) 45 µm = ______________ nm
4) 805 dm = ______________ km
0.2
J. Flesock
A. SI Prefix Conversions
1) 20 cm = ______________ m
2) 0.032 L = ______________ mL
3) 45 µm = ______________ nm
4) 805 dm = ______________ km
0.2
32
J. Flesock
A. SI Prefix Conversions
1) 20 cm = ______________ m
2) 0.032 L = ______________ mL
3) 45 µm = ______________ nm
4) 805 dm = ______________ km
0.2
45,000
32
J. Flesock
A. SI Prefix Conversions
1) 20 cm = ______________ m
2) 0.032 L = ______________ mL
3) 45 µm = ______________ nm
4) 805 dm = ______________ km
0.2
0.0805
45,000
32
J. Flesock
B. Dimensional Analysis
n The “Factor-Label” Methodn Units, or “labels” are canceled, or “factored” out
J. Flesock
B. Dimensional Analysis
n The “Factor-Label” Methodn Units, or “labels” are canceled, or “factored” out
J. Flesock
B. Dimensional Analysis
n The “Factor-Label” Methodn Units, or “labels” are canceled, or “factored” out
J. Flesock
B. Dimensional Analysis
n The “Factor-Label” Methodn Units, or “labels” are canceled, or “factored” out
J. Flesock
B. Dimensional Analysis
J. Flesock
B. Dimensional Analysis
n Steps:1. Identify starting & ending units.
2. Line up conversion factors so units cancel.
3. Multiply all top numbers & divide by each bottom number.
4. Check units & answer.
J. Flesock
B. Dimensional Analysis
n Lining up conversion factors:
1 in = 2.54 cm
J. Flesock
B. Dimensional Analysis
n Lining up conversion factors:
1 in = 2.54 cm2.54 cm 2.54 cm
J. Flesock
B. Dimensional Analysis
n Lining up conversion factors:
1 in = 2.54 cm2.54 cm 2.54 cm
= 1
J. Flesock
B. Dimensional Analysis
n Lining up conversion factors:
1 in = 2.54 cm2.54 cm 2.54 cm
1 in = 2.54 cm
= 1
J. Flesock
B. Dimensional Analysis
n Lining up conversion factors:
1 in = 2.54 cm2.54 cm 2.54 cm
1 in = 2.54 cm 1 in 1 in
= 1
J. Flesock
B. Dimensional Analysis
n Lining up conversion factors:
1 in = 2.54 cm2.54 cm 2.54 cm
1 in = 2.54 cm 1 in 1 in
= 1
1 =
J. Flesock
B. Dimensional Analysis
n How many milliliters are in 1.00 quart of milk?
J. Flesock
B. Dimensional Analysis
n How many milliliters are in 1.00 quart of milk?
qt mL
J. Flesock
B. Dimensional Analysis
n How many milliliters are in 1.00 quart of milk?
1.00 qt
qt mL
×
J. Flesock
B. Dimensional Analysis
n How many milliliters are in 1.00 quart of milk?
1.00 qt 1 L1.057 qt
qt mL
×
J. Flesock
B. Dimensional Analysis
n How many milliliters are in 1.00 quart of milk?
1.00 qt 1 L1.057 qt
qt mL
×
J. Flesock
B. Dimensional Analysis
n How many milliliters are in 1.00 quart of milk?
1.00 qt 1 L1.057 qt
qt mL
1000 mL1 L
×
J. Flesock
B. Dimensional Analysis
n How many milliliters are in 1.00 quart of milk?
1.00 qt 1 L1.057 qt
qt mL
1000 mL1 L
×
J. Flesock
B. Dimensional Analysis
n How many milliliters are in 1.00 quart of milk?
1.00 qt 1 L1.057 qt
= 946 mL
qt mL
1000 mL1 L
×
J. Flesock
B. Dimensional Analysis
n You have 1.5 pounds of gold. Find its volume in cm3 if the density of gold is 19.3 g/cm3.
J. Flesock
B. Dimensional Analysis
n You have 1.5 pounds of gold. Find its volume in cm3 if the density of gold is 19.3 g/cm3.
lb cm3
J. Flesock
B. Dimensional Analysis
n You have 1.5 pounds of gold. Find its volume in cm3 if the density of gold is 19.3 g/cm3.
lb cm3
1.5 lb
J. Flesock
B. Dimensional Analysis
n You have 1.5 pounds of gold. Find its volume in cm3 if the density of gold is 19.3 g/cm3.
lb cm3
1.5 lb 1 kg2.2 lb
J. Flesock
B. Dimensional Analysis
n You have 1.5 pounds of gold. Find its volume in cm3 if the density of gold is 19.3 g/cm3.
lb cm3
1.5 lb 1 kg2.2 lb
J. Flesock
B. Dimensional Analysis
n You have 1.5 pounds of gold. Find its volume in cm3 if the density of gold is 19.3 g/cm3.
lb cm3
1.5 lb 1 kg2.2 lb
1000 g1 kg
J. Flesock
B. Dimensional Analysis
n You have 1.5 pounds of gold. Find its volume in cm3 if the density of gold is 19.3 g/cm3.
lb cm3
1.5 lb 1 kg2.2 lb
1000 g1 kg
J. Flesock
B. Dimensional Analysis
n You have 1.5 pounds of gold. Find its volume in cm3 if the density of gold is 19.3 g/cm3.
lb cm3
1.5 lb 1 kg2.2 lb
1000 g1 kg
1 cm3
19.3 g
J. Flesock
B. Dimensional Analysis
n You have 1.5 pounds of gold. Find its volume in cm3 if the density of gold is 19.3 g/cm3.
lb cm3
1.5 lb 1 kg2.2 lb
1000 g1 kg
1 cm3
19.3 g
J. Flesock
B. Dimensional Analysis
n You have 1.5 pounds of gold. Find its volume in cm3 if the density of gold is 19.3 g/cm3.
lb cm3
1.5 lb 1 kg2.2 lb
= 35 cm31000 g1 kg
1 cm3
19.3 g
J. Flesock
B. Dimensional Analysis
n How many liters of water would fill a container that measures 75.0 in3?
J. Flesock
B. Dimensional Analysis
n How many liters of water would fill a container that measures 75.0 in3?
in3 L
J. Flesock
B. Dimensional Analysis
n How many liters of water would fill a container that measures 75.0 in3?
75.0 in3
in3 L
J. Flesock
B. Dimensional Analysis
n How many liters of water would fill a container that measures 75.0 in3?
75.0 in3 (2.54 cm)3
(1 in)3
in3 L
J. Flesock
B. Dimensional Analysis
n How many liters of water would fill a container that measures 75.0 in3?
75.0 in3 (2.54 cm)3
(1 in)3
in3 L
J. Flesock
B. Dimensional Analysis
n How many liters of water would fill a container that measures 75.0 in3?
75.0 in3 (2.54 cm)3
(1 in)3
in3 L
1 L1000 cm3
J. Flesock
B. Dimensional Analysis
n How many liters of water would fill a container that measures 75.0 in3?
75.0 in3 (2.54 cm)3
(1 in)3
in3 L
1 L1000 cm3
J. Flesock
B. Dimensional Analysis
n How many liters of water would fill a container that measures 75.0 in3?
75.0 in3 (2.54 cm)3
(1 in)3 = 1.23 L
in3 L
1 L1000 cm3
J. Flesock
B. Dimensional Analysis
5) Your European hairdresser wants to cut your hair 8.0 cm shorter. How many inches will he be cutting off ?
J. Flesock
B. Dimensional Analysis
5) Your European hairdresser wants to cut your hair 8.0 cm shorter. How many inches will he be cutting off ?
cm in
J. Flesock
B. Dimensional Analysis
5) Your European hairdresser wants to cut your hair 8.0 cm shorter. How many inches will he be cutting off ?
8.0 cm
cm in
J. Flesock
B. Dimensional Analysis
5) Your European hairdresser wants to cut your hair 8.0 cm shorter. How many inches will he be cutting off ?
8.0 cm 1 in2.54 cm
cm in
J. Flesock
B. Dimensional Analysis
5) Your European hairdresser wants to cut your hair 8.0 cm shorter. How many inches will he be cutting off ?
8.0 cm 1 in2.54 cm
cm in
J. Flesock
B. Dimensional Analysis
5) Your European hairdresser wants to cut your hair 8.0 cm shorter. How many inches will he be cutting off ?
8.0 cm 1 in2.54 cm
= 3.2 in
cm in
J. Flesock
B. Dimensional Analysis
6) Taft football needs 550 cm for a 1st down. How many yards is this?
J. Flesock
B. Dimensional Analysis
6) Taft football needs 550 cm for a 1st down. How many yards is this?
cm yd
J. Flesock
B. Dimensional Analysis
6) Taft football needs 550 cm for a 1st down. How many yards is this?
550 cm
cm yd
J. Flesock
B. Dimensional Analysis
6) Taft football needs 550 cm for a 1st down. How many yards is this?
550 cm 1 in2.54 cm
cm yd
J. Flesock
B. Dimensional Analysis
6) Taft football needs 550 cm for a 1st down. How many yards is this?
550 cm 1 in2.54 cm
cm yd
J. Flesock
B. Dimensional Analysis
6) Taft football needs 550 cm for a 1st down. How many yards is this?
550 cm 1 in2.54 cm
cm yd
1 ft12 in
J. Flesock
B. Dimensional Analysis
6) Taft football needs 550 cm for a 1st down. How many yards is this?
550 cm 1 in2.54 cm
cm yd
1 ft12 in
J. Flesock
B. Dimensional Analysis
6) Taft football needs 550 cm for a 1st down. How many yards is this?
550 cm 1 in2.54 cm
cm yd
1 ft12 in
1 yd3 ft
J. Flesock
B. Dimensional Analysis
6) Taft football needs 550 cm for a 1st down. How many yards is this?
550 cm 1 in2.54 cm
cm yd
1 ft12 in
1 yd3 ft
J. Flesock
B. Dimensional Analysis
6) Taft football needs 550 cm for a 1st down. How many yards is this?
550 cm 1 in2.54 cm
= 6.0 yd
cm yd
1 ft12 in
1 yd3 ft
J. Flesock
B. Dimensional Analysis
7) A piece of wire is 1.3 m long. How many 1.5-cm pieces can be cut from this wire?
J. Flesock
B. Dimensional Analysis
7) A piece of wire is 1.3 m long. How many 1.5-cm pieces can be cut from this wire?
cm pieces
J. Flesock
B. Dimensional Analysis
7) A piece of wire is 1.3 m long. How many 1.5-cm pieces can be cut from this wire?
1.3 m
cm pieces
J. Flesock
B. Dimensional Analysis
7) A piece of wire is 1.3 m long. How many 1.5-cm pieces can be cut from this wire?
1.3 m 100 cm1 m
cm pieces
J. Flesock
B. Dimensional Analysis
7) A piece of wire is 1.3 m long. How many 1.5-cm pieces can be cut from this wire?
1.3 m 100 cm1 m
cm pieces
J. Flesock
B. Dimensional Analysis
7) A piece of wire is 1.3 m long. How many 1.5-cm pieces can be cut from this wire?
1.3 m 100 cm1 m
cm pieces
1 piece1.5 cm
J. Flesock
B. Dimensional Analysis
7) A piece of wire is 1.3 m long. How many 1.5-cm pieces can be cut from this wire?
1.3 m 100 cm1 m
cm pieces
1 piece1.5 cm
J. Flesock
B. Dimensional Analysis
7) A piece of wire is 1.3 m long. How many 1.5-cm pieces can be cut from this wire?
1.3 m 100 cm1 m
= 86 pieces
cm pieces
1 piece1.5 cm
C. Temperature
Used to measure the heat energy of a substance or reaction
n Celsius - 25°C
n Fahrenheit =
n Kelvin =
J. Flesock
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