# anyone know physics?



## Lachlan (Jul 3, 2008)

Hi. does anyone know anything about basic physics. im trying to learn some stuff


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## mind_games (Nov 30, 2008)

:lol. Post the specific topic or problem you're having trouble with.


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## mohammed (Feb 12, 2010)

I have a BS degree in physics.


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## mazer (Feb 12, 2010)

Yep, something you need help with?


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## C 13 (Feb 21, 2010)

mind_games said:


> :lol. Post the specific topic or problem you're having trouble with.


Hahaha! :b


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## Catlover4100 (Feb 10, 2009)

I took two semesters of physics last year and got a 91%. It wasn't my favorite class, and it was pretty hard, but I got through it!


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## marenubium87 (Jan 11, 2009)

I'm a physics major, but I'd defer to the guy above that actually has the degree.


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## zookeeper (Jun 3, 2009)

I know some psychics. They might be able to give you the answer too.


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## Lachlan (Jul 3, 2008)

ok, thanks for the replies. 
so does anyone know if the centripedal force due to the earths spin, which a person is effected by at the equator, makes the net force acting on him smaller than a person standing at the north pole. because at the pole gravity effects a person, but the centripedal force does not?


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## Lachlan (Jul 3, 2008)

thanks


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## splattice (Sep 7, 2009)

The centripetal force is the net force acting on a body that's moving in circular motion. The body is moving in circular motion due to actual forces like gravity, the normal reaction from the ground, and friction keeping the body fixed relative to the surface of the Earth.


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## marenubium87 (Jan 11, 2009)

The short answer - yes. We had to calculate this in class once. I forget the exact answer, but it's tiny. You *may* weigh about a percent less at the equator, all other things being equal.

edit - was wrong. It's more like 1/5 of one percent.

http://en.wikipedia.org/wiki/Equatorial_bulge


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## Lachlan (Jul 3, 2008)

*nuclear question i cant figure out*

so i dont think ive got the right answer, but im gona ask my tuto tommorow. anybody know?

*1. The problem statement, all variables and given/known data*
60Co, half life = 5.2 years, decays by emittion of a beta particle (0.31 MeV) and two gamma particles (1.71MeV and 1.33MeV). what is the minimum initial mass needed of 60Co that will have an activity of at least 10Ci after 30 months?
*2. Relevant equations*
half life = ln(2)/decasy constant(λ).
Amount of nuclei (N) = rate of decay(R)/decay constant(λ).
N=N(starting) x e^(-λ)(t)
in this equation t is the time between the starting amount and the amount, N, left after the decay.

*3. The attempt at a solution*
I feel reasonably competent at the simpler problems involving these formulas, just plug the number in basically. But for this one, i am a little unsure how to include the energy numbers. or if they are the right formulas to use. they dont normally put numbers in if they are not needed. !
i can calculate a number which is the amount of nuclei needed to start with, when you have the decay rate they given, and time (i got 1.222x10^20). but this is not the mass. 
so im thinking i must have to use the energy in decay with the nuclei number to calculate the mass maybe. or is there a way i can go straight from the number of nuclei to the mass that i am missing?


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## marenubium87 (Jan 11, 2009)

I'm assuming you've asked your tutor by now. But, just in case.

So you're almost there. What you calculated is the number of 60Co atoms needed, but they need a mass. So at this point it's stoichiometry. i.e.

# of Co atoms you have / # of Co atoms per mole (this is avogadro's number) = # of moles of Co atoms you have

take that result , and multiply by the molar mass of 60Co (59.934 g/mol) to get the number of grams of 60Co you need.


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## Lachlan (Jul 3, 2008)

Thanks. yeah, i realized that i just needed the mass in units, so i could just multiply the number of nuclei by the weight of 1 atom (in units). I think im geting it but its taking me quite a bit of time.


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