Transcript Measurements - walker2015

Standards of Measurements
Accuracy and Precision


Accuracy – how close a measured value is to
the actual value
Precision – how close the measured values
are to each other
Significant Figures

All nonzero digits are significant.


1, 2, 3, 4, 5, 6, 7, 8, 9
Zeros within a number are always significant.

Both 4308 and 40.05 have four significant figures.
Significant Figures

Zeros that set the decimal point are not
significant.
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470,000 has two significant figures.
0.000084 has two significant figures.
Trailing zeros that aren't needed to hold the
decimal point are significant.

4.00 has three significant figures.
Significant Figures
If the least precise measurement in a calculation has
three significant figures, then the calculated answer
can have at most three significant figures.
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Mass = 34.73 grams
Volume = 4.42 cubic centimeters.

Rounding to three significant figures, the density is
7.86 grams per cubic centimeter.
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Scientific Notation

For large numbers, moving the decimal to the
left will result in a positive number


For small numbers, moving the decimal to the
right will result in a negative number


346500 = 3.46 x 105
0.000145 = 1.45 x 10-4
For numbers less than 1 that are written in
scientific notation, the exponent is negative.
Scientific Notation

Before numbers in scientific notation can be
added or subtracted, the exponents must be
equal.
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5.32 x 105 + 9.22 x 104
5.32 x 105 + 0.922 x 105
5.32 + 0.922 x 105
6.24 x 105
Scientific Notation

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When numbers in scientific notation are
multiplied, only the number is multiplied. The
exponents are added.
(3.33 x 102) (2.71 x 104)
(3.33) (2.71) x 102+4
9.02 x 106
Scientific Notation


When numbers in scientific notation are
divided, only the number is divided. The
exponents are subtracted.
4.01 x 109
1.09 x 102

4.01 x 109-2
1.09

3.67 x 107
Scientific Notation

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A rectangular parking lot has a length of 1.1 ×
103 meters and a width of 2.4 × 103 meters.
What is the area of the parking lot?
(1.1 x 103 m) (2.4 x 103 m)
(1.1 x 2.4) (10 3+3) (m x m)
2.6 x 106 m2
SI Units
Kilo- (k)
1000
Hecto- (h)
100 Deka- (da)
10 Base Unit
m, L, g
Deci- (d)
0.1 Centi- (c)
0.01 Milli- (m)
Mnemonic device:
King Henry Died By Drinking Chocolate Milk
0.001
Metric System


Meter (m) – The basic
unit of length in the
metric system
Length – the distance
from one point to
another

A meter is slightly longer
than a yard
Metric System

Liter (L) – the basic unit of
volume in the metric system

A liter is almost equal to a
quart
Metric System

Gram (g) – The basic unit of
mass
Derived Units

Combination of base units

Volume – length  width  height
1 cm3 = 1 mL

Density – mass per unit volume (g/cm3)
M
D=
V
M
D V
Density
1) An object has a volume of 825 cm3 and a
density of 13.6 g/cm3. Find its mass.
GIVEN:
WORK:
V = 825 cm3
D = 13.6 g/cm3
M=?
M = DV
M
D V
M = 13.6 g x 825 cm3
cm3
M = 11,220 g
1
Density
2) A liquid has a density of 0.87 g/mL. What
volume is occupied by 25 g of the liquid?
GIVEN:
WORK:
D = 0.87 g/mL
V=?
M = 25 g
V=M
D
M
D V
V=
25 g
0.87 g/mL
V = 28.7 mL
Density
3) You have a sample with a mass of 620 g & a
volume of 753 cm3. Find the density.
GIVEN:
WORK:
M = 620 g
V = 753 cm3
D=?
D=M
V
M
D V
D=
620 g
753 cm3
D = 0.82 g/cm3
Dimensional Analysis / Unit Factors

Dimensional analysis – a problem-solving
method that use any number and can be
multiplied by one without changing its value
Dimensional Analysis / Unit Factors

How many hours are there in a year?
24 hr x 365 days =
1 day
1 year

8760 hr
1 year
There’s 8,760 hours in a year.
Dimensional Analysis / Unit Factors

The distance from Grove Hill to Thomasville
is 15 miles. How many feet is that?
15 mi
1

x 5280 ft
1 mi
=
79200 ft
1
There’s 79,200 feet in 15 miles.
Dimensional Analysis / Unit Factors

Convert 36 cm/s to mi/hr
36 cm
sec
3600 sec x 1 in
x
x
1 hr
2.54 cm
1 ft
12 in
1 mi
5280 ft
= 129600 mi = 0.805 mi/hr
160934.4 hr
Temperature


A degree Celsius is almost twice as large as
a degree Fahrenheit.
You can convert from one scale to the other
by using one of the following formulas:
Temperature

Convert 90 degrees Fahrenheit to Celsius

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o
o
C = 5/9 ( F - 32)
o
C = 5/9 (90 - 32)
o
C = 0.55555555555555556 (58)
o
C = 32.2
Temperature
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Convert 50 degrees Celsius to Fahrenheit

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o
o
F = 9/5 ( C ) + 32
o
F = 9/5 (50 ) + 32
o
F = 1.8 (50) + 32
o
F = 90 + 32
o
F = 122
Temperature
The SI base unit for temperature is the kelvin
(K).
•
•
A temperature of 0 K, or 0 kelvin, refers to the
lowest possible temperature that can be reached.
In degrees Celsius, this temperature is
–273.15°C. To convert between kelvins and
degrees Celsius, use the formula:
Temperature