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AQA Physics Unit 1 PHYA1 20th May 2013

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My default expression is 'An isotope of an element is made up from...'

That's a good one too, I just found this past paper that had "an isotope is a variation of an element with..." so I think I'm good for that now

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Reply 841

x-Sophie-x

Original post by Cheesegrater4

i thought it varies the resistance so changes voltage. ??

Ah great. But surely that would vary the current too?

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Reply 842

chrishy2012

Its probably a stupid question, but how do you know what type of interaction a reaction is? It always comes up and i always tend guess and put weak which tends to be right most of the time but is their an actual way of telling which one it is!

Thankss :smile:

Reply 843

x-Sophie-x

Original post by Surf

Well, see ya guys, good luck.... I've got to leave now...
Remember to relax - take it one question at a time!
:cool:

Best of luck! xD

You're leaving early!

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Reply 844

Micheal123456

Can some one go through jan 2013 question 7biii) please . And also is it only hadrons that experience strong interaction and leptons don't?

Reply 845

FrankB3

Original post by x-Sophie-x

Did they use 2sf in the question?
That's why they would have rounded to 120.

But don't worry, unless they ask you for a certain number of Sig figs, you won't lose marks if you give your answer to something appropriate :smile:

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Ah, they probably had. Cheers anyway :wink:

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Reply 846

ria_devil

hows everyone feeling bout the exam today?

Reply 847

ria_devil

any suggestions on what the 6 mark might be on?

Reply 848

BayHarborButcher

Original post by x-Sophie-x

Does a variable resistor vary the current initially? Or resistance? Or both?

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I'm pretty sure a variable resistor alters the current that flows through the circuit, and as V is constant and V=IR then that affects the resistance.

Reply 849

NtainS

Can someone confirm that the W+ boson mediates beta+ decay and that the W- boson mediates beta- decay. Cheers

Reply 850

x-Sophie-x

Original post by Micheal123456

Can some one go through jan 2013 question 7biii) please . And also is it only hadrons that experience strong interaction and leptons don't?

The emf is the total voltage in the circuit.
So, 12v + (2x2.04v) as there are two wires which equals 16.1V.

Ans yep, that's correct :smile:

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Reply 851

tetra-s

Original post by FrankB3

Guys can someone explain to me in Jun 12, Q7.a.ii, it's asked us to calculate the resistance, and the correct equation is: R=(8/0.067) it even backs this up in the mark scheme. The answer to that is 119.4, so 119 right? But in the mark scheme it has rounded up to 120. I don't get why they've done this and it's really annoying me because I'll lose marks this way. Can someone please explain when I'm supposed to round up all of a sudden?

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They like to put their answer to 2 sig figs I've noticed, don't worry as long as you've done it right you'll get the marks. It's only on the question where it specifically asks you to put your answer to the appropriate sig figs, then you put your answer to the amount of sig figs indicated in the question.

Reply 852

x-Sophie-x

In this interaction, can someone please explain why strangeness is not conserved?

Sigma plus--> pi plus + n

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Reply 853

Lelcats

I'm okay with Particles and Quantum Phenomena but Electricity is going to be the end of me. Good luck guys!

Reply 854

username1099191

Original post by x-Sophie-x

Thank you, that was a good explanation.

I understand why for excitation, although one thing that still confuses me is that in ionisation the electron completely leaves the atom in one go, so doesn't need exact energy values incident on it.. and the 'extra' energy is the KE on the electron.

Or have I completely missed the point here?

This one is a bit weird and perhaps a bit tricky to explain, and I'd just like to point out that you are right in terms of Ek of the electron there - but this is a longer explanation as to why that happens:

Using the stair analogy again (I just made that analogy up on the spot btw :cool:

), this time think of the "ionisation shell" as the infinite shell, n=∞ - you will most likely have seen this on those energy level diagrams in past papers, maybe without realising. Also on these diagrams, have you ever noticed that the energy levels are labelled as negative numbers? This is also important.
The ground state, usually n=1, will be labelled with a negative number - let's say that an atom has an n=1 of -11eV. This means that in order to leave the atom entirely, the atomic electron has to take at least 11eV from a bombarding electron in order to "get back to 0eV" (shoddy terminology there). From there, any extra energy will be converted to Ek of this electron - just as you said at the beginning! So in a sense, that discrete 11eV is still required, but this time excess energy can still be transferred. :smile:

Reply 855

Val9090

Original post by Micheal123456

Can some one go through jan 2013 question 7biii) please . And also is it only hadrons that experience strong interaction and leptons don't?

Hadrons INTERACT through the strong interaction, but charged hadrons DECAY through the WEAK interaction. So whenever there is an equation with a hadron becoming something else, it is through the WEAK interaction. The only charged hadron that doesn't decay (afaik) is the the proton, because it is stable.

Reply 856

BayHarborButcher

Original post by Micheal123456

Can some one go through jan 2013 question 7biii) please . And also is it only hadrons that experience strong interaction and leptons don't?

7biii it tells you in the question that the bulb is 12v and it tells you there are two wires (which you previously calculated the p.d. across). Hence 12 + 2.1 + 2.1 = 16.2v. The strong nuclear force only acts on quarks where as the weak acts on quarks and leptons :smile:

Reply 857

username1099191

Original post by NtainS

Can someone confirm that the W+ boson mediates beta+ decay and that the W- boson mediates beta- decay. Cheers

Confirmed! :biggrin:

Reply 858

NtainS

Original post by jazzynutter

Confirmed! :biggrin:

Excellent

- thanks

Reply 859

x-Sophie-x

Original post by BayHarborButcher

I'm pretty sure a variable resistor alters the current that flows through the circuit, and as V is constant and V=IR then that affects the resistance.