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 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 molecules of sulphur.

1.1 The Realm of Physics Q1. How many molecules are there in the Sun? Info:- Mass of Sun = kg - Assume it is 100% Hydrogen - Avogadro constant = No. of molecules in one mole of a substance = 6 x Mass of one mole of Hydrogen = 2g A. Mass of Sun = x 1000 = g No. of moles of Hydrogen in Sun = / 2 = 5 x No. of molecules in Sun = ( 6 x ) x ( 5 x ) = 3 x 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 = 1x10 1 m = 10 1 m tall. An ant is about 0.7mm tall. So to the nearest power of ten we can say it is 1mm = 1x10 -3 m = m tall. So if an ant is m tall and the giraffe 10 1 m 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 grams LENGTH meters electron diameter of nucleus proton diameter of atom virus radius of virus amoeba10 -5 radius of amoeba raindrop10 -3 height of human being 10 0 ant10 0 radius of earth 10 7 human being10 5 radius of sun 10 9 pyramid10 13 earth-sun distance earth10 27 radius of solar system sun10 33 distance of sun to nearest star milky way galaxy10 44 radius of milky way galaxy the Universe10 55 radius of visible Universe 10 26

Q2. There are about 1x10 28 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 / = 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 = / = 10 5

Prefixes PowerPrefixSymbol petaP terraT 10 9 gigaG 10 6 megaM 10 3 kilok PowerPrefixSymbol femtof picop nanon microµ millim

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 1.5x10 8 kg

Base quantities The SI system of units starts with seven base quantities. All other quantities are derived from these. Base quantityBase UnitAbbreviation Base quantityBase UnitAbbreviation mass (m) Length (l) time (t) temperature (T) electric current (I) amount of substance (n) luminous intensity (I v ) Base quantityBase UnitAbbreviation mass (m)kilogramkg Length (l)metrem time (t)seconds temperature (T)KelvinK electric current (I)AmpereA amount of substance (n)molemol luminous intensity (I v )candelacd

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 = A x s 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 = Frequency x wavelength ms -1 = s -1 x m ms -1 = 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 m x l y g z Where k is a dimensionless constant and x,y and z are unknown numbers. So…s = kg x m y (ms -2 ) z s 1 = kg x m y+z s -2z Now equate both sides of the equation: For s1 = -2z so z = -1/2 For kg0 = x For m0 = y+z so y = +1/2 So… Period = k m 0 l 1/2 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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