DIMENSIONAL ANALYSIS

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Transcript 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
1𝑖𝑛
1=
2.54𝑐𝑚
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
1𝑖𝑛
2.85𝑖𝑛
2.85𝑐𝑚 𝑥
=
= 1.12𝑖𝑛
2.54𝑐𝑚
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
2.54𝑐𝑚
1𝑖𝑛
2.54𝑐𝑚
7.00𝑖𝑛 𝑥
= 7𝑥2.54𝑐𝑚 = 17.8𝑐𝑚
1𝑖𝑛
DIMENSIONAL ANALYSIS
O Unit factors can be derived from each
equivalence statement
2.54cm = 1in
O 2 unit factors
2.54𝑐𝑚
1𝑖𝑛
and
1𝑖𝑛
2.54𝑐𝑚
DIMENSIONAL ANALYSIS
2.54𝑐𝑚
1𝑖𝑛
and
1𝑖𝑛
2.54𝑐𝑚
O How to choose – look at direction of required change
O in  cm (need to cancel in – goes in denominator)
2.54𝑐𝑚
1𝑖𝑛
O cm  in (need to cancel cm – goes in denominator)
1𝑖𝑛
2.54𝑐𝑚
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?
2.54𝑐𝑚
25.5𝑖𝑛 𝑥
= 7𝑥2.54𝑐𝑚 = 64.8𝑐𝑚
1𝑖𝑛
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
10.0𝑘𝑚 𝑥
1000𝑚
= 1.00𝑥104 𝑚
1𝑘𝑚
Ex 3) A student entered a 10.0-km run. How long
is the run in miles?
O m  yd
1.00𝑥104 𝑚 𝑥
1.094𝑦𝑑
= 1.094𝑥104 yd
1𝑚
Ex 3) A student entered a 10.0-km run. How long
is the run in miles?
O yd  mi
1.094𝑥104 yd 𝑥
1 𝑚𝑖
= 6.216𝑚𝑖 = 6.22𝑚𝑖
1760𝑦𝑑
O Original 10.0 which has 3 sig figs so you want 3 sig
figs in your answer
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 𝑥
1000𝑚 1.094𝑦𝑑
1𝑚𝑖
𝑥
𝑥
= 6.22 𝑚𝑖
1𝑘𝑚
1𝑚
1760𝑦𝑑
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
435.6𝑔
115𝑙𝑏 𝑥
= 5.22𝑥104 𝑔
1𝑙𝑏
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 𝑐𝑒𝑛𝑡
2.5 𝑑𝑜𝑙𝑙𝑎𝑟𝑠 𝑥
= 250 𝑐𝑒𝑛𝑡𝑠
1 𝑑𝑜𝑙𝑙𝑎𝑟
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
500.0𝑚𝑖 𝑥
1760𝑦𝑑
1𝑚
1𝑘𝑚
𝑥
𝑥
= 804.7𝑘𝑚
1𝑚𝑖
1.094𝑦𝑑 1000𝑚
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
𝑚
𝑠
𝑥
1𝑘𝑚
1𝑚𝑖
60𝑠
60𝑚𝑖𝑛
𝑥
𝑥
𝑥
103 𝑚
1.6093𝑘𝑚
1𝑚𝑖𝑛
1 ℎ𝑟
= 1.15𝑥103 𝑚𝑖/ℎ𝑟
Ex 8) If you are going 50 mi/hr, how many feet/sec
are your going?
O mi  ft ; hr  min  sec
O Equivalence statement
5280 ft=1mi
𝑚𝑖 5280𝑓𝑡
1ℎ𝑟
1𝑚𝑖𝑛
50
𝑥
𝑥
𝑥
= 70𝑓𝑡/𝑠𝑒𝑐
ℎ𝑟
1𝑚𝑖
60𝑚𝑖𝑛 1𝑠𝑒𝑐
Ex 9) Convert 22.50 gal/min to L/s
O gal  L ; min  sec
O Equivalence statement
1 gal = 3.7854L
𝑔𝑎𝑙 3.7854𝐿 1𝑚𝑖𝑛
22.50
𝑥
𝑥
= 1.41953 = 1.420 𝐿/𝑠
𝑚𝑖𝑛
1𝑔𝑎𝑙
60𝑠
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
𝑐𝑚
1 𝑖𝑛
1 𝑑𝑎𝑦
60
𝑥
𝑥
= 0.984 𝑖𝑛/ℎ𝑟
𝑑𝑎𝑦 2.54 𝑐𝑚 24 ℎ𝑟
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 𝑜𝑧 𝑏𝑙𝑠
5 𝑜𝑧 𝑏𝑙
1 𝑞𝑡 𝑏𝑙𝑠 𝑥
𝑥
= 1.6 𝑜𝑧 𝑏𝑙
1 𝑞𝑡 𝑏𝑙𝑠 100 𝑜𝑧 𝑏𝑙𝑠
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
𝑚
1 𝑘𝑚
60 𝑠 60 𝑚𝑖𝑛
15.2
𝑥
𝑥
𝑥
= 54.7 𝑘𝑚/ℎ𝑟
𝑠 1000 𝑚 1 𝑚𝑖𝑛
1 ℎ𝑟
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
𝑖𝑛 2.54𝑐𝑚 1𝑥10−2 𝑚
1𝑘𝑚
1𝑚𝑠
60𝑠 60𝑚𝑖𝑛
. 510
𝑥
𝑥
𝑥
𝑥
𝑥
𝑥
𝑚𝑠
1𝑖𝑛
1𝑐𝑚
1000𝑚 1𝑥10−3 𝑠 1𝑚𝑖𝑛
1ℎ𝑟
= 46.6 𝑘𝑚/ℎ𝑟