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how do i show that is a "unique" product of primes though (again in the whole induction thingum)

ahh and thanks lennon, i'm just feeling a bit miserable atm, everything seems to be going wrong this week! And with my back I dont feel like socialising by going to the bar etc cos all I can do is grin with pain. *Sigh* Ready meal for one tonight!

ahh and thanks lennon, i'm just feeling a bit miserable atm, everything seems to be going wrong this week! And with my back I dont feel like socialising by going to the bar etc cos all I can do is grin with pain. *Sigh* Ready meal for one tonight!

Willa

how do i show that is a "unique" product of primes though (again in the whole induction thingum)

Ah you didn't ask that in your previous question.

That part isn't really amenable to induction.

You need to know of a prime that is p divides a x b then p divides a or b

So if

n = p_1 ..... p_r = P_1 ..... P_S

then p_1 divdes the RHS, and so divides one of them, and so equals one of them by primeness - then continue along these lines.

well that's the bit which was baffling me, because the question asks: Prove by (course of values) induction [which is strong induction in your terminology] that all natural numbers >= 2 can be represented as product of primes

And then as the second part of the question, it says: Prove by course of values induction that all natural numbers >= 2 can be uniquely represented as product of primes.

That's what's throwing me, I dont see how it's related to induction either! How do you show uniqueness based on all smaller natural numbers being uniquely representable?

And then as the second part of the question, it says: Prove by course of values induction that all natural numbers >= 2 can be uniquely represented as product of primes.

That's what's throwing me, I dont see how it's related to induction either! How do you show uniqueness based on all smaller natural numbers being uniquely representable?

Riteeeeeeeeee I have a lovely special relativity question I'm stuck with AGAIN.

====

In collisions between electrons and positrons B+ and B- mesons are produced with equal and opposite momenta in the lab. The B+ and B- undergo decays with mean lives of 1.6x10^-12s in their rest frames. Given that the momentum of each of the B particles is 20GeV/c and their masses are 5279MeV/c^2, calculate the mean distance that a B particle travels before decay as observed in the lab.

====

I was thinking that the way to approach it is with the following method:

1. Fing gamma and the speeds (each travelling in opposite direction, therfefore use addition of velocities formula with the speed of the actual frame.

2. Calculate the mean time from the mean Proper time given and the gamma just calculated

3. Use the speed of each with the mean time to calculate the distance travelled by each particle.

4. Then find the average value.

The problem is I'm not sure how to find the gamma....

====

In collisions between electrons and positrons B+ and B- mesons are produced with equal and opposite momenta in the lab. The B+ and B- undergo decays with mean lives of 1.6x10^-12s in their rest frames. Given that the momentum of each of the B particles is 20GeV/c and their masses are 5279MeV/c^2, calculate the mean distance that a B particle travels before decay as observed in the lab.

====

I was thinking that the way to approach it is with the following method:

1. Fing gamma and the speeds (each travelling in opposite direction, therfefore use addition of velocities formula with the speed of the actual frame.

2. Calculate the mean time from the mean Proper time given and the gamma just calculated

3. Use the speed of each with the mean time to calculate the distance travelled by each particle.

4. Then find the average value.

The problem is I'm not sure how to find the gamma....

well for this one I would start with: momentum = gamma*mv = mv/root(1-v^2/c^2)

then i'd try rearranging that a bit: (m/momentum)^2 = = (1 - v^2/c^2)/v^2 = 1/v^2 - 1/c^2

so v = 1/root((m/momentum)^2 + 1/c^2)

that gives you your v (and hence your gamma). Try going from there. I would do it a slightly different way I think, but since you asked how to find gamma, there you go...i think that's right!

then i'd try rearranging that a bit: (m/momentum)^2 = = (1 - v^2/c^2)/v^2 = 1/v^2 - 1/c^2

so v = 1/root((m/momentum)^2 + 1/c^2)

that gives you your v (and hence your gamma). Try going from there. I would do it a slightly different way I think, but since you asked how to find gamma, there you go...i think that's right!

Willa

The trick, as usual, with these relativity questions, is just always to think: "What do I want to know? And what do I already know!?". You want Gamma, so you essentially want v. You have momentum and you have mass. That should SCREAM at you p = Gamma*mv!

Good point! I was too busy thinking about invariants and energy when it's clearly v.obvious! Thanks!

Hoofbeat

Good point! I was too busy thinking about invariants and energy when it's clearly v.obvious! Thanks!

oh no i shouldn't have told you that! These questions are my only source of mechanics and relativity revision so far! I always lack the motivation to do practice questions myself, but I dont mind helping other people - so it's a good way for me to revise. So errrr.....keep up your struggling with physics for as long as possible, cos it's helping me out at the same time!

Willa

can someone please calculate the moment of inertia of carbon monoxide given carbon RAM = 12, Oxygen RAM = 16, and seperation = 1.128E-10

I get something of the order 10^-46.....that sound correct? (it doesnt match the answers given but the answers given have been wrong in the past!)

I get something of the order 10^-46.....that sound correct? (it doesnt match the answers given but the answers given have been wrong in the past!)

How do you calculate the moment of inertia of an element

Willa

errrrrrrrrrrrrr dont you guys do any rotational mechanics at oxford!? If so then...OH...MY....GOD! (Moment of Intertia = Sum of mr^2 for all particles in a body...and is the equivilent of mass but for rotation)

Lol! No we do do rotational dynamics...but never even thought that you could calculate the MoI of a particle.

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