Transcript 01 Orders of Magnitude and Units

Orders of Magnitude and
Units
The ‘mole’:
- The amount of a substance can be described
using ‘moles’.
- ‘One mole’ of a substance has 6 x 1023 molecules
in it. (This number is called the Avogadro constant)
- So a chemist may measure out 3 moles of sulphur
and she would know that she has 18 x 1023
molecules of sulphur.
1.1 The Realm of Physics
Q1. How many molecules are there in the Sun?
Info: - Mass of Sun = 1030 kg
- Assume it is 100% Hydrogen
- Avogadro constant = No. of molecules
in one mole of a substance = 6 x 1023
- Mass of one mole of Hydrogen = 2g
A.
Mass of Sun = 1030 x 1000 = 1033 g
No. of moles of Hydrogen in Sun = 1033 / 2 = 5 x 1032
No. of molecules in Sun = ( 6 x 1023 ) x ( 5 x 1032 )
= 3 x 1056 molecules
Orders of Magnitude
Orders of magnitude are numbers on a scale where
each number is rounded to the nearest power of
ten. This allows us to compare measurements,
sizes etc.
E.g. A giraffe is about 6m tall. So to the nearest power of
ten we can say it is 10m = 1x101m = 101m tall.
An ant is about 0.7mm tall. So to the nearest power of
ten we can say it is 1mm = 1x10-3m = 10-3m tall.
So if an ant is 10-3m tall and the giraffe 101m tall, then
the giraffe is bigger by four orders of magnitude.
Orders of magnitude link
Order of Magnitude of some Masses
Order of Magnitude of some Lengths
MASS
LENGTH
grams
meters
electron
10-27
diameter of nucleus
10-15
proton
10-24
diameter of atom
10-10
virus
10-16
radius of virus
10-7
amoeba
10-5
radius of amoeba
10-4
raindrop
10-3
height of human being
100
ant
100
radius of earth
107
human being
105
radius of sun
109
pyramid
1013
earth-sun distance
1011
earth
1027
radius of solar system
1013
sun
1033
distance of sun to nearest
star
1016
milky way galaxy
1044
radius of milky way galaxy
1021
the Universe
1055
radius of visible Universe
1026
Q2. There are about 1x1028 molecules of air in the
lab. So by how many orders of magnitude are
there more molecules in the Sun than in the lab?
A. 1056 / 1028 = 1028 so 28 orders of magnitude
more molecules in the Sun.
Q3. Determine the ratio of the diameter of a
hydrogen atom to the diameter of a hydrogen
nucleus to the nearest order of magnitude.
A. Ratio = 1015 / 1010 = 105
Prefixes
Power
Prefix
Symbol
Power
Prefix
Symbol
1015
peta
P
10-15
femto
f
1012
terra
T
10-12
pico
p
109
giga
G
10-9
nano
n
106
mega
M
10-6
micro
µ
103
kilo
k
10-3
milli
m
Quantities and Units
A physical quantity is a measurable feature of an
item or substance.
A physical quantity will have a value and usually a
unit. (Note: Some quantities such as ‘strain’ are
dimensionless and have no unit).
E.g. A current of 5.3A ;
A mass of 1.5x108kg
Base quantities
The SI system of units starts with seven base
quantities. All other quantities are derived from
these.
Base
Base quantity
quantity
Base
Base Unit
Unit
Abbreviation
Abbreviation
mass (m)
kilogram
kg
Length (l)
metre
m
time (t)
second
s
temperature (T)
Kelvin
K
electric current (I)
Ampere
A
amount of substance (n)
mole
mol
luminous intensity (Iv)
candela
cd
Derived units
The seven base units were defined arbitrarily. The sizes of
all other units are derived from base units.
E.g. Charge in coulombs
This comes from :
Charge = Current x time
so…
coulombs = amps x seconds
or…
C=Axs
so…
C could be written in base
units as As (amp seconds)
Homogeneity
If the units of both side of an equation can be
proved to be the same, we say it is dimensionally
homogeneous.
E.g.
Velocity
ms-1 =
ms-1 =
= Frequency x wavelength
s-1
x
m
ms-1
 homogeneous, therefore this formula is correct.
Dimensional Analysis
The dimensions of a physical quantity show how it
is related to base quantities.
Dimensional homogeneity and a bit of guesswork
can be used to prove simple equations.
E.g. Experimental work suggests that the period of
oscillation of a pendulum moving through small
angles depends upon its length, mass and the
gravitational field strength, g.
So we can write
Period = k mx ly gz
Where k is a dimensionless constant and x,y and z are
unknown numbers.
So…
s = kgx my (ms-2)z
s1 = kgx my+z s-2z
Now equate both sides of the equation:
For s
1 = -2z
For kg
0=x
For m
0 = y+z
so
z = -1/2
so
y = +1/2
So…
Period = k m0 l1/2 g-1/2
Or…
Period =
k
l
g
Q.
Consider a sphere (radius, r) moving through a fluid
of viscosity η at velocity v.
Experimental work suggests that the force acting
upon it is related to these quantities. Use
dimensional analysis to determine the formula.
(Note: the units of viscosity are Nsm-2)
A. You should prove…
F = k ηrv
(F = 6π ηrv)
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