1. Jul 13, 2008

pcfighter

any body know....
question:
true or false??
dimensional analysis can give you the numerical value of constants of propotionality that may appear in a algebra system.....

and can u give the reason too...??

2. Jul 13, 2008

HallsofIvy

Staff Emeritus
False. Dimensional analysis can tell you what units you need and lead you to the general form but cannot tell you what the numerical value of constants must be.

3. Jul 13, 2008

dynamicsolo

An example of this would be finding the area of a plane figure by dimensional analysis. Area has units of square meters (m^2) [or square inches, or whatever length units you prefer]. So the simplest possible expression for the area is A = L^2 .

But what shape is this? As far as dimensional analysis is concerned, it doesn't matter: any plane figure will have an area given by the product of two lengths. A specific shape will have an area which includes a "dimensionless" constant related to the shape. (Also, the shape may have an area involving a product of two distinct lengths.)

So you can have

square: $$A = 1 \cdot x^2$$

rectangle: $$A = 1 \cdot xy$$

triangle: $$A = \frac{1}{2}bh$$

circle: $$A = \pi R^2$$

ellipse: $$A = \pi ab$$

It will be much the same in dealing with any physical quantity by dimensional analysis.

4. Jul 13, 2008

azzkika

how big would the entire matter of the earth be if all space was removed so just matter remained??

there are approximately, 133,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000 atoms that comprise earth, but i dont know enough of atomical dimensions(how much of a proton/neutron etc are comprised of space??), to work this out.

5. Jul 13, 2008

dynamicsolo

First off, a tip: new questions, especially if they're unrelated to the original question starting the thread, should be asked by starting a new thread.

Here's a quick way to find an estimate. You can take the mass of the Earth and divide it by the volume of the Earth (volume of a sphere with Earth's radius) to get an average density for the Earth. The typical density of nuclear matter (just protons and neutrons with "no space between" -- and the electrons hardly count...) is around 10^14 gm/(cm^3). So the volume of the "compressed Earth" would be given by

Compressed volume / Volume of Earth = average density of Earth / density of nuclear matter .

That would give you the volume of the ball of protons and neutrons. If you want to find its radius, solve the equation for the volume of this sphere for the new radius.

I haven't checked your number of atoms, but the approximate diameter of a nucleon (proton or neutron) is about 10^-15 m.