At the endof this chapter, students should be able to:-
❖State basic quantities with their respective SI units: length (m),
time (s), mass (kg), electrical current (A), temperature (K) and
amount of substance (mol).
❖State derive quantities and their respective units and symbols:
velocity (ms-1
), acceleration (ms-2
), work (J), Force (N),
pressure (Pa), energy (J), power (W) and frequency (Hz).
❖Perform dimensional analysis and conversion units.
3.
OUTLINE
1.1 Physical quantities:Basic and derived quantities
1.2 System of unit: SI unit and unit conversion
1.3 Dimension and Dimensional Analysis
4.
Chapter 1
• Physicalquantity is defined as a quantity which can be
measured.
• It can be categorised into two types
1.Basic (base) quantity
2.Derived quantity
• Basic quantity is defined as a quantity which cannot be
derived from any physical quantities.
• Derived quantity is defined as a quantity which can be
expressed in term of base quantity which is a combination of
two or more physical quantities.
UNITS
Chapter 1
• Physicsexperiments involve the measurement of a variety
of quantities.
• These measurements should be accurate and reproducible.
• The first step in ensuring accuracy and reproducibility is
defining the units in which the measurements are made.
8.
Unit is definedas a standard size of measurement of
physical quantities.
Examples:
- 1 second is defined as the time required for 9,192,631,770
vibrations of radiation emitted by a caesium-133 atom.
- 1 kilogram is defined as the mass of a platinum-iridium
cylinder kept at International Bureau of Weights and
Measures Paris.
- 1 meter is defined as the length of the path travelled by
light in vacuum during a time interval of 1/299,792,458 of a
second.
Chapter 1
9.
The common systemof units used today are:-
• S.I unit (International System of Units)
e.g: meter, kilogram, second.
• CGS Unit- UK.
e.g: centimetre, gram, second.
Chapter 1
10.
PREFIXES OF UNIT
•It is used for presenting larger and smaller values.
• Written in the following form (standard scientific
notation)
Example:
An electron mass is about 0.000 000 000 000 000 000
000 000 000 000 910 938 22 kg.
In scientific notation, written as 9.1093822 x 10-31
kg.
Chapter 1
A x 10b
Chapter 1
DIMENSIONAL ANALYSIS
Dimensionis defined as a technique or method which the physical
quantity can be expressed in terms of combination of basic quantities.
• It can be written as [physical quantity or its symbol]
• Table shows the dimension of basic quantities.
13.
Chapter 1
DIMENSIONAL ANALYSIS
•Dimension can be treated as algebraic quantities through the procedure
called dimensional analysis.
• The uses of dimensional analysis are:-
✓ to determine the unit of the physical quantity.
✓ to determine whether a physical equation is correct or not
dimensionally by using the principle of homogeneity.
✓ to derive a physical equation.
Note:
• Dimension of dimensionless constant is 1.
• Dimensions cannot be added or subtracted.
• The validity of an equation cannot determined by dimensional analysis.
• The validity of an equation can only be determined by experiment.
14.
Chapter 1
EXERCISE 1:
Determinea dimension and S.I unit for the following quantities:
a) Velocity
b) Acceleration
c) Linear momentum
d) Force
Answers:
c) MLT-1
, kgms-1
d) MLT-2
, kgms-2
15.
EXERCISE 2:
(a)
(b)
(c)
Checking thehomogeneity/consistency of an equation.
𝑣=𝑢+𝑎𝑡
𝑠=𝑢𝑡+
1
2
𝑎𝑡
2
𝑠=𝑎𝑡 +
1
2
𝑢𝑡
2
Answers:
b) not homogen
c) homogen
16.
Exercise 3 :
Theperiod, T of a simple pendulum depends on its length l, acceleration due
to gravity, g and mass, m. By using dimensional analysis, obtain an equation
for period of the simple pendulum.
Constructing an equation using dimensional analysis
17.
EXERCISE 4
The velocityof wave depends on the wavelength , surface tension and water
density . Determine the wave velocity equation by using dimensional analysis.
Answers:
𝑘
√𝛾
𝜆𝜌
18.
CONVERSION OF UNITS
Chapter1
Length Mass Force
1 mile = 1609 m = 1.609 km 1 metric ton = 10 kg
3 1 lb = 4.448 N
1 ft = 0.3048 m = 30.48 cm 1 slug =14.59 kg 1 N = 0.2248 lb = 10 dyne
5
1 m = 39.37 in = 3.281 ft
1 in. = 0.0254 m = 2.54 cm
19.
CONVERSION FACTORS TOREMEMBER
1 km = 1000 m
1 km = 0.621 mi
1 m = 100 cm
1 cm = 10 mm
1 inch = 2.54 cm
1 kg = 1000 g
1 h = 60 min
1 min = 60 s
1 h = 3600 s
Chapter 1
20.
Reasoning Strategy: ConvertingBetween Units
1. In all calculations, write down the units explicitly.
2. Treat all units as algebraic quantities. When identical units are
divided, they are eliminated algebraically.
3. Use the conversion factors located on the page facing the
inside cover. Be guided by the fact that multiplying or dividing
an equation by a factor of 1 does not alter the equation.
21.
Exercise 5: TheWorld’s Highest Waterfall
The highest waterfall in the world is Angel Falls in Venezuela, with a
total drop of 979.0 m. Express this drop in feet.
22.
Exercise 6:
A solidcube of copper has a density of 8.94 g/cm3
. Convert this value to kg/m3
.
Exercise 7:
Bicyclists in the Tour de France cycle at a speed of 34.0 miles per hour (mi/h) on flat
sections of the road. What is this speed in
a) kilometers per hour (km/h)
b) meters per second (m/s)
(1.609 km = 1 mi, 1 mi = 1609 m, 1 h = 3600s)
Answers:
a) 54.7 km/h b) 15.2 m/s
23.
Exercise 9:
Answer: 6.67X 10-5
mm3
g-1
s-2
.
Given that the gravitational constant, G = 6.673 X 10-11
m3kg-1
s-2
. Determine its value in
mm3
g-1
s-2
.
Exercise 8:
Calculate the area of a rectangle, in km2, where the sides of the rectangle are 7 inches
and 13 inches respectively. Given 1 inch = 2.54 cm.
Answer: 5.87 X 10-8
km2
.
Exercise 10:
Change 17.7 µm/1.6 X 10-3
ps into its SI unit.
Answer: 1.106 X 10-10
m/s
ACCURACY
• In thefields of engineering, industry and statistics, the accuracy of a
measurement system is the degree of closeness of measurements of a
quantity to its actual (true) value.
Low Accuracy HighAccuracy High Accuracy
High Precision Low Precision High
Precision
28.
SIGNIFICANT FIGURE
• Thereare 2 kinds of numbers:
– Exact: the amount of money in your account. Known
with
certainty.
– Approximate: weight, height—anything
MEASURED. No measurement is perfect.
• When a measurement is recorded only those digits that
are
dependable are written down.
• If you measured the width of a paper with your ruler you might
record 21.7 cm.
• To a mathematician 21.70, or 21.700 is the same but,
to a
scientist 21.7 cm and 21.70 cm is NOT the same
• 21.700 cm to a scientist means the measurement is accurate to
within one thousandth of a cm.
29.
• The significantfigures (also called significant digits and
abbreviated sig figs, sign.figs or sig digs) of a number are those
digits that carry meaning contributing to its precision.
• Zeroes are sometime used to locate the decimal point, when the
zeroes are used in that way we say that they are not significant.
30.
SIGNIFICANT FIGURES (SF)Chapter 1
1. All non zero digits in a
number are SF.
2. Zero in between two non
zero digits are SF.
3. For any whole number,
zero at the end of a
number can be a SF or not
a SF It
. depends on the
precision of the reading.
Example
1. (i) 3421 : 4 sf
(ii) 62.5 : 3 sf
2. (i) 503 : 3 sf
(ii) 1.006 : 4 sf
3. (i) 63 000 : 2 sf if the precision
is to the nearest thousand.
(ii) 63 400 : 3 sf if the precision
is to the nearest hundred.
31.
SIGNIFICANT FIGURES (SF)
Chapter1
4. For a decimal number less
than 1, zero placed before any
non-zero digit is not a sf
5. For a decimal number, zero
placed after a non-zero digit is
a sf.
Example
4. (i) 0.0028 : 2 sf
(ii) 0.0902 : 3 sf
5. (i) 7.40 : 3 sf
(ii) 0.020 : 2 sf
32.
SIGNIFICANT FIGURES (SF):
Additionand Subtraction
• The final results of addition or subtraction should no more
precise than the least precise number used
• E.g : Calculate 4.231 + 3.51
• Ans: 7.741= 7.74
Least no of SF
33.
The final resultsof multiplication or division should have
only as many digits as the number with the least number of
significant figures used in the calculation.
E.g : Calculate the area of a rectangle 11 .3 cm by 6.8 cm.
Ans: 76.84 cm = 77 cm
2 2
SIGNIFICANT FIGURES (SF):
Multiplication & Division
Chapter 1
Least no of SF
34.
How many sigfigs?
7
40
0.5
0.00003
7 x 103
7,000,000
3401
2100
2100.0
5.00
0.00412
8,000,050,000