AQA Physics A - PHYA5 (18/06/12) - Exam thread

Reply 380

username570910

2

Original post by number23

under charles' low it should be V s proportional to T (not P)

Yes, now that is a mistake. I apologise.

Reply 381

Amirrryy

1

in old questions, they would give us the mass of a nucleus and ask us to work out the energy right? i mean they wont ask us something in this form right? cos i dont know the mass of the nucleus :L much appreciated

The iron isotope 56
26 Fe has a very high binding energy per nucleon.
Calculate its value in MeV.

Reply 382

username570910

2

Updated version (there was a mistake with Charles' law). Sorry :s-smilie:

Original post by Augmented hippo

Okay absolute magnitude ranks stars based on how bright they would appear from 10 parsecs away. It's a funny scale as going down by 5 on the scale means a star is 100 times brighter. If you know how bright a star is from 10 parsecs (it's absolute magnitude) compared to how bright it appears from earth
(apparent magnitude) you can work out how far away it is. The formula is derived using inverse square rule but you don't really need to know that. The formula is m-M=5log(d/10) where m is apparent magnitude, M is absolute magnitude and d is distance in parsecs. You can rearrange the formula so if you know m and M you can work out the distance to a star. Hope this helps but I think the best way to get to grips with it is by doing a few questions.

Cheeers dude, but whats with the minus numbers in magnitudes? :s-smilie:

(i.e, you get -2 as a magnitude)

and you said that going down by 5 (i.e from 8 to 3) makes a star brighter by 100times, so how many times would a star be brighter if it went down by say for example by 1, or 10? - if it went down by 10, would it be 200 times (2*100)?

Reply 384

soldiersixteen

Original post by FrightBright

Well it says the useful power out put is 300 watts...

That means that it generated more energy, so you x by 10.

As for 0.238 bit even I dont really know :biggrin:

, I've done no chemistry, all I know is that the nucleon number is 238, so you divide that by 1000 to get it to kg... I think it's relative mass or something.

So, that would work for any other element as well?
Thanks.

Reply 385

helpme12345

Original post by leeandrewarmstrong

Updated version (there was a mistake with Charles' law). Sorry :s-smilie:

Thanks! looks really helpful :smile:

If you're doing medical do have any notes for that? :P

Reply 386

FrightBright

Original post by soldiersixteen

So, that would work for any other element as well?
Thanks.

Pretty confident. Only trick I see is Oxygen is 02 so it's 32, otherwise it's never failed me.

Reply 387

Hulksmash

Original post by don'tTRIP.

I reckon the 6-marker will be on Hertz's discovery of radio waves :dice:

You may well be right, electron microscopes, photoelectric effect, wave/particle duality or a em wave question which could be about Hertz's discovery of radio waves.

Reply 388

username570910

2

Original post by helpme12345

Thanks! looks really helpful :smile:

If you're doing medical do have any notes for that? :P

Sorry, i'm an astro guy haha :smile:

Reply 389

Augmented hippo

2

Original post by Red Richie

Cheeers dude, but whats with the minus numbers in magnitudes? :s-smilie:

(i.e, you get -2 as a magnitude)

and you said that going down by 5 (i.e from 8 to 3) makes a star brighter by 100times, so how many times would a star be brighter if it went down by say for example by 1, or 10? - if it went down by 10, would it be 200 times (2*100)?

Okay, when they first sorted out this system they said Vega, a star, is going to have an apparent magnitude of 0. Therefore any source of light that appears brighter than Vega from earth will have a negative apparent magnitude. Also, absolute magnitudes can be less than 0. If you were ten parsecs from a light source and it appeared brighter than Vega appears from earth it would have a negative absolute magnitude.

Because going up 5 in the scale means 100 times dimmer going up one in the scale means the star is 100^(1/5) time dimmer. Going down ten in the scale would mean the light source is 100^2 times brighter. Say you move by amount x up or down the scale that will mean the star is 100^(x/5) times brighter/dimmer depending on if you move up or down the scale.

Reply 390

H2SO4

Original post by Augmented hippo

Okay absolute magnitude ranks stars based on how bright they would appear from 10 parsecs away. It's a funny scale as going down by 5 on the scale means a star is 100 times brighter. If you know how bright a star is from 10 parsecs (it's absolute magnitude) compared to how bright it appears from earth
(apparent magnitude) you can work out how far away it is. The formula is derived using inverse square rule but you don't really need to know that. The formula is m-M=5log(d/10) where m is apparent magnitude, M is absolute magnitude and d is distance in parsecs. You can rearrange the formula so if you know m and M you can work out the distance to a star. Hope this helps but I think the best way to get to grips with it is by doing a few questions.

In June 10 paper,
They ask us to prove the difference of brightness is nearly 8000 times if the difference of absolute magnitude is 9.8

The marking scheme said is 2.5^9.8 = 8000
How about if I use 100^9.8/5 = 8300Which will give a more accurate ans,
Will the examiner know this and accept my alternative method.

Reply 391

aim28

Original post by helpme12345

They mention the formula on page 218 of the nelson thorne's book

I've attached the question I tried to do, the first formula (in the formula book) didn't work, and the solution said to use the other formula but I don't know why :s-smilie:

Also I don't think it's old spec because the equation is in the book

Okay, i have done it one way not using that equation and got 9.3x10-14 :/ not sure that's right, what was the answer??

Reply 392

number23

15

Original post by don'tTRIP.

I reckon the 6-marker will be on Hertz's discovery of radio waves :dice:

thanks, so what points would get us the marks:

•

set up inclues transmitter and detector loop

•

a high voltage is connected to the trnsmitter

•

this causes sparks to jump across terminals

•

radio waves were produced and detcted by the detctor

•

because radio waves are transverse, the detctor should be adjusted at correct angle due to polarisation

does anyone want to go over plans for possible 6 markers in turning points?

Reply 393

jonnny

14

Do we need to derive the pressure of an ideal gas?

This was posted from The Student Room's iPhone/iPad App

Reply 394

Laurence94

Original post by helpme12345

is the formula for the kinetic energy of a gas molecule 1.5RT/Na
or 1.5nRT ??? :confused:

only the first one is in the formula booklet but I saw the second one used as a solution in a question...

can someone please explain when you'd use either formula?

Thanks :smile:

It's 3RT/2Na, (i.e. 1.5(RT/Na) and should be on the formula sheet....page 3, 3rd line from bottom, right column :smile:

Reply 395

helpme12345

Original post by Laurence94

It's 3RT/2Na, (i.e. 1.5(RT/Na) and should be on the formula sheet....page 3, 3rd line from bottom, right column :smile:

Hey,

I meant the other formula, isn't on the formula sheet
KE = 1.5nRT where n=mols

it's on pg 218 of the textbook

Do you have any idea when to use each formula? :smile:

Reply 396

helpme12345

Original post by leeandrewarmstrong

Sorry, i'm an astro guy haha :smile:

haha ok, well thanks anyway for the other notes :smile:

Reply 397

soldiersixteen

Original post by number23

thanks, so what points would get us the marks:

•

set up inclues transmitter and detector loop

•

a high voltage is connected to the trnsmitter

•

this causes sparks to jump across terminals

•

radio waves were produced and detcted by the detctor

•

because radio waves are transverse, the detctor should be adjusted at correct angle due to polarisation

does anyone want to go over plans for possible 6 markers in turning points?

sure

Reply 398

number23

15

photoelectric effect:

•

classical wave theory predicted energy proportional to intensity over enough time all electrons would leave metal

•

disproved by photelectric effect where emissions occured over certain frequency and instantaneously with range of KE

•

einstein explained it by saying light consisted of dicrete energy wave packets of enrgy=hf which interact with one electron at a time

•

the minimum frequency is such that hf=work function (minimum energy for photelectron to be released, depends on metal)

stopping potential:

•

make cathode increasingly positive so electrons have to do more work to get away from it

•

so maximum kinetic energy is reduced by eV and given by KE=hf-workfunction-eV

•

stopping potential is the voltage required to reduce KE to 0

•

rearrange: $eV_s=hf-work function$

•

stopping potential ranges with frequency

•

the stopping potential for different frequencied of light can be measured using photocell(collects light) ammeter potential divider voltmeter and cell. when stopping potential is reached, current is 0