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# Edexcel A2 Physics Unit 5 'Physics from Creation to Collapse' watch

1. (Original post by lefneosan)
so the other definitions are wrong?
I cant say they are just wrong..
'You can directly calculate their luminosity' - yes, otehrwise where would we get it from?
'A standard candle is an object whose luminosity is constant' - for some period of time it is constant (and for supernova explosions we believe the maximum luminosity is also constant).

(Original post by The Magnificent KoloToure)
But where has it come from?

Potential energy is mgh and Elastic Strain Energy is 1/2Fx, which is why I (still) don't get the equation. Don't think we need to learn it though, it's not even in the CGP book.
E = ½Fx is just a formula for elastic potential energy, just like gravitational potential or kinetic, we have to leave with it.
Yeah, there might be some derivation of it, but I dont want to go into details.
Substituting F = kx into E = ½Fx gives E = ½kx2.

(Original post by OL1V3R)
Okay, how about I do 2 revision sessions on Saturday:

1:00pm to 2:30pm - Astrophysics
5:00pm to 6:30pm - General 6PH05 FAQ (i.e. problem session)
2. Economics done. Now I have 4 days to master Unit 5 physics
3. (Original post by yadas)
I cant say they are just wrong..
'You can directly calculate their luminosity' - yes, otehrwise where would we get it from?
'A standard candle is an object whose luminosity is constant' - for some period of time it is constant (and for supernova explosions we believe the maximum luminosity is also constant).

E = ½Fx is just a formula for elastic potential energy, just like gravitational potential or kinetic, we have to leave with it.
Yeah, there might be some derivation of it, but I dont want to go into details.
Substituting F = kx into E = ½Fx gives E = ½kx2.

Well I'm a bit pressed for time because I have to meet up with a friend in town and then get ready for my Year 13 prom this evening.

What topic shall we do today? How about 1:00pm until 2:30pm today?
4. (Original post by lefneosan)
I have two questions though: Why do you need to use two different positions of the earth over a six month period and also in the third worked example given, they say that the the angle measured after a six month period was 1.52 arcsenconds different from that originally measured. so doesnt that mean that the angle isnt 1.52 divided by 2 but actually 1.52 as the questions suggests that there was originally 1.52 and then another 1.52 was measured?
Because we can get maximum displacement of the Earth exactly in a period of 6 months. (Is this sentence grammatically correct?)
Greater displacement means greater parallax angle (the one that is very small), so this increases the accuracy of the measurements.
Also the Earth is on the same line, where the Sun is at the centre, after 6 months.
If we use different time, the Sun would not be at the centre.
And it is the distance from the Sun that we determine, bacause it's very99 far.
And for the worked exmaple, I think they mean the angle 2θ on fig. 7.3.2..

(Original post by lefneosan)
Also yadas do you agree that the miles hudson book disagrees with this wiki definition http://en.wikipedia.org/wiki/Arcsecond because wiki gives arc measurements as part of a circle but the hudson book as part of a degree. Are they consistent?
Yeah, of course they are consistent.
As Oliver mentioned in this thread earlier, $s=r\theta$.

(Original post by OL1V3R)
Well I'm a bit pressed for time because I have to meet up with a friend in town and then get ready for my Year 13 prom this evening.

What topic shall we do today? How about 1:00pm until 2:30pm today?
Oh, I think we could skip today then..
1:00pm in the UK will be less than in half an hour? Not many people will come anyway..

(Original post by JayAyy)
Economics done. Now I have 4 days to master Unit 5 physics
Be there tomorrow.
5. (Original post by yadas)
Oh, I think we could skip today then..
1:00pm in the UK will be less than in half an hour? Not many people will come anyway..
Okay let's skip the session today.
6. (Original post by yadas)
Because we can get maximum displacement of the Earth exactly in a period of 6 months. (Is this sentence grammatically correct?)
Greater displacement means greater parallax angle (the one that is very small), so this increases the accuracy of the measurements.
Also the Earth is on the same line, where the Sun is at the centre, after 6 months.
If we use different time, the Sun would not be at the centre.
And it is the distance from the Sun that we determine, bacause it's very99 far.
And for the worked exmaple, I think they mean the angle 2θ on fig. 7.3.2..

Yeah, of course they are consistent.
As Oliver mentioned in this thread earlier, $s=r\theta$.

Oh, I think we could skip today then..
1:00pm in the UK will be less than in half an hour? Not many people will come anyway..

Be there tomorrow.
Thanks dont worry about the worked example. COuld you perhaps reexplain the reason why we take parallax from two places cos Im not understanding. Thanks!
7. (Original post by lefneosan)
Could you perhaps reexplain the reason why we take parallax from two places cos Im not understanding. Thanks!
Alright, havinghoops said what I was going to say.
Having just one position will not show any difference related to the background.
Having two positions, the difference gives us the angle
8. Lefneosan: When the observer moves, the object being observed appears to move relative to a fixed background. You see this everyday. When you're out walking, look at the objects in the distance. The objects closer to you appear to move more than the ones further from you.

Yadas will probably explain fully.
Attached Images

9. (Original post by havinghoops)
Lefneosan: When the observer moves, the object being observed appears to move relative to a fixed background. You see this everyday. When you're out walking, look at the objects in the distance. The objects closer to you appear to move more than the ones further from you.

Yadas will probably explain fully.
10. Do we need to know the trends across the star classes OBAFGKM?
11. (Original post by yadas)
E = ½Fx is just a formula for elastic potential energy, just like gravitational potential or kinetic, we have to leave with it.
Yeah, there might be some derivation of it, but I dont want to go into details.
Substituting F = kx into E = ½Fx gives E = ½kx2.
That's great, thanks yadas.

And another quick question for you, or anyone else. Can someone explain to me how the mass and stiffness constant of a structure are related to their natural frequency? They say they are in one of the revision books without saying why, it came up in the spec paper too.
12. (Original post by The Magnificent KoloToure)
That's great, thanks yadas.

And another quick question for you, or anyone else. Can someone explain to me how the mass and stiffness constant of a structure are related to their natural frequency? They say they are in one of the revision books without saying why, it came up in the spec paper too.
Well using the formula for the period of oscillations of a spring mass oscillator:

$\newline T = 2\pi\sqrt{\frac{m}{k}} \newline \newline f_{0}=1/T \newline \newline \therefore f_{0}=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

The period of the oscillations is its natural period (i.e. when the system freely oscillates), which means that the natural frequency is the reciprocal of the period.
13. (Original post by OL1V3R)
Do we need to know the trends across the star classes OBAFGKM?
I guess knowing that mass/radius/luminosity/temperature deacreses, and the life time increases from O to M would be sufficient.
14. Ah yeah, of course.
15. (Original post by yadas)
I guess knowing that mass/radius/luminosity/temperature deacreses, and the life time increases from O to M would be sufficient.
Do we need to be able to explain the trends?

As far as I can remember, hotter stars have more mass and a greater luminosity. That means that they use up energy at a quicker rate and as a result they spend less time on the main sequence. However, I don't understand this because surely if white dwarfs are much hotter than red giants but they have less mass, why do they have a smaller luminosity?

Do we not need to be able to explain these trends, or do we just have to use the HR diagram to explain the life of a star?
16. (Original post by OL1V3R)
Do we need to be able to explain the trends?

As far as I can remember, hotter stars have more mass and a greater luminosity. That means that they use up energy at a quicker rate and as a result they spend less time on the main sequence. However, I don't understand this because surely if white dwarfs are much hotter than red giants but they have less mass, why do they have a smaller luminosity?

Do we not need to be able to explain these trends, or do we just have to use the HR diagram to explain the life of a star?
It's only the bold part that is in the syllabus..
But who knows, they might still ask.

As you know, Stefan-Boltzmann law states that L = 4пσr2T4.
Looking at H-R diagram, the difference in T is 2-10 times.
To keep the same L while decreasing r, r can be 22-102 = 4-100 times lower.
Indeed the radius of a red giant is much greater than 100 times the radius of a white dwarf.
I mean the increase in r2 is much greater than decrease in T4.
17. Blimey you guys are KEEN.
You know I'm just learning it all now :P
18. (Original post by JayAyy)
Economics done. Now I have 4 days to master Unit 5 physics
same here!

I think I might fail this unit.
19. Does anyone have the link to the sample paper?
20. (Original post by unamed)
same here!

I think I might fail this unit.
I can't fail this unit. I want to get 120, because I don't trust my marks in any of my other units

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