### DIMENSIONAL ANALYSIS

```DIMENSIONAL
ANALYSIS
- Convert a given result from one system of
units to another
- Unit factor method
Ex 1) A pin measuring 2.85 cm in
length. What is its length in inches?
O Need an equivalence statement
2.54cm = 1in
O Divide both sides by 2.54cm
O Unit Factor
1
1=
2.54
O Multiply any expression by this unit factor
and it will not change its value
Ex 1) A pin measuring 2.85 cm in
length. What is its length in inches?
O Pin is 2.85cm need to multiply by the unit
factor
1
2.85
2.85
=
= 1.12
2.54
2.54
Ex 2) A pencil is 7.00 in long. What is
the length in cm?
O Convert in  cm
O Need equivalence statement 2.54cm = 1in
O Unit Factor
2.54
1
2.54
7.00
= 72.54 = 17.8
1
DIMENSIONAL ANALYSIS
O Unit factors can be derived from each
equivalence statement
2.54cm = 1in
O 2 unit factors
2.54
1
and
1
2.54
DIMENSIONAL ANALYSIS
2.54
1
and
1
2.54
O How to choose – look at direction of required change
O in  cm (need to cancel in – goes in denominator)
2.54
1
O cm  in (need to cancel cm – goes in denominator)
1
2.54
Ex 3) You want to order a bicycle with a 25.5in
frame, but the sizes in the catalog are given only
in cm. What size should you order?
2.54
25.5
= 72.54 = 64.8
1
Ex 3) A student entered a 10.0-km run. How long
is the run in miles?
O km  mi
O Equivalence statement 1m = 1.094yd
O Strategy first
km  m  yards  mi
O Equivalence statements:
1km = 1000m
1m = 1.094 yd
1760yd = 1 mi
Ex 3) A student entered a 10.0-km run. How long
is the run in miles?
O km  m
10.0
1000
= 1.00104
1
Ex 3) A student entered a 10.0-km run. How long
is the run in miles?
O m  yd
1.00104
1.094
= 1.094104 yd
1
Ex 3) A student entered a 10.0-km run. How long
is the run in miles?
O yd  mi
1.094104 yd
1
= 6.216 = 6.22
1760
O Original 10.0 which has 3 sig figs so you want 3 sig
Ex 3) A student entered a 10.0-km run. How long
is the run in miles?
O Can combine all conversions into one step
10.0km
1000 1.094
1

= 6.22
1
1
1760
DO NOW
Ex 4) If a woman has a mass 115lb, what is her
mass
in
grams?
O lb  g
O Equivalence Statement
1lb = 453.6g
435.6
115
= 5.22104
1
Ex 5) If a woman has a mass 115lb, what is her
mass in grams?
O dollar  cents
O Equivalence Statement
1 dollar = 100 cents
100
2.5
= 250
1
Ex 6) Determine the length of a 500.0 mi
automobile race in km
O mi  yd  m  km
O Equivalence statements –
1mi = 1760yd
1m = 1.094yd
1km = 1000m
500.0
1760
1
1

= 804.7
1
1.094 1000
Ex 7) The average speed of a nitrogen molecule in
air at 25C is 515m/s. Convert this speed to miles
per hour
O m/s  km/s  mi/s  mi/min  mi/hr
O Equivalence statements –
1km = 103
1mi = 1.6093km
60s = 1min
60min = 1hr
15

1
1
60
60

103
1.6093
1
1 ℎ
= 1.15103 /ℎ
Ex 8) If you are going 50 mi/hr, how many feet/sec
O mi  ft ; hr  min  sec
O Equivalence statement
5280 ft=1mi
5280
1ℎ
1
50

= 70/
ℎ
1
60 1
Ex 9) Convert 22.50 gal/min to L/s
O gal  L ; min  sec
O Equivalence statement
1 gal = 3.7854L
3.7854 1
22.50

= 1.41953 = 1.420 /

1
60
Ex 10)Bamboo can grow up to 60.0 cm/day.
Calculate this growth rate in inches per hour
O cm  in ; day  hr
O Equivalence statement
1 in = 2.54 cm

1
1
60

= 0.984 /ℎ
2.54  24 ℎ
Ex 11)How much bleach (bl) would you need to
make a quart of 5% bleach solution (bls)? – (refer to
brightstorm youtube video for this question if you get confused)
O qt  oz ; need 5% bls
O Equivalence statement
1 qt = 32 oz
5oz bs = 5% Bleach
100 oz bls = 100 % bleach solution
32
5
1

= 1.6
1   100
Ex 12) A bumblebee flies with a ground speed of
15.2 m/s. Calculate its speed in km/hr
O m  km ; s  min  hr
O Equivalence statement

1
60  60
15.2

= 54.7 /ℎ
1000  1
1 ℎ
Ex 13) Convert 0.510 in/ms to km/hr
O in  cm  m  km; ms  s  min  hr
O Equivalence statement
1 in = 2.54cm
2.54 110−2
1
1
60 60
. 510

1
1
1000 110−3  1
1ℎ
= 46.6 /ℎ
```