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

1. (Original post by kappleberry)
^ isn't it just simple trig?

I havent yet looked through it, so may be wrong
Yeah the calculation is simple trig it's just the explanation of it that I wasn't too clear about
2. Could someone please try and quickly explain the life cycle of stars?... All my books say slightly different things, and its confusing me..

Thanks!
3. I need a LOT of help =(
http://cid-5432f573d3fc3f5a.office.l...hysics%20Q.doc Going through those Qs atm. Have been through all of them once. But some just aren't clear to me.
For starters, could someone please explain Q1 to me =#

the gravitational field strength on the surface of the earth is g. The gravitational field strength on the surface of a planet of twice the radius and same density is....

Thank youu!
There is also a mark scheme, I think you've checked it.
Which part didnt you understand?
there is...but i still dont get it. We haven't done anything to do with light years in class. I looked through the cgp textbook and saw that 1 light year = 9.46 x 10^15, but do we need to know this? =/
5. (Original post by kappleberry)
there is...but i still dont get it. We haven't done anything to do with light years in class. I looked through the cgp textbook and saw that 1 light year = 9.46 x 10^15, but do we need to know this? =/
Well a light year is basically the distance travelled by a ray of light during the course of 1 year. You can just used distance = speed x time to work out the distance of 1 light year:

speed = 3 x 10^8 m/s
time = 1 year = 365 x 24 x 60 x 60 seconds

Just multiply the speed and time together and you should get the value they've given in the book.
6. ..and also part iii of that Q..where did they get the value for the energy from?
I know power = energy/time. but we dont have a value for t =/
7. ^^thank you so much! Gonna have a read now! =]
8. Sorry...i know youve gone through it in the notes, but could you just exaplain what youve done in q2 of the spec paper?
9. Legend! You got any more of these? Quite good for a quick read.
10. http://cid-5432f573d3fc3f5a.office.l...hysics%20A.doc

Q19 (ii)....whats =/

11. Okay.
(Original post by kappleberry)
I need a LOT of help =(
http://cid-5432f573d3fc3f5a.office.l...hysics%20Q.doc Going through those Qs atm. Have been through all of them once. But some just aren't clear to me.
For starters, could someone please explain Q1 to me =#

the gravitational field strength on the surface of the earth is g. The gravitational field strength on the surface of a planet of twice the radius and same density is....

Thank youu!
You know that g = GM/r2.
We need gp (planet) in terms of ge (Earth).
We know that the rp = 2re, and that density of that planet pp = pe = p that of the Earth.
p = M/V (Density = Mass/Volume) => M = pV.
Volume of a sphere is 4пr3/3.
Hence the gravitational field on Earth is:
ge = G*(p*4пre3/3)/re2 = 4пGpre3/3re2 = 4пGpre/3
The gravitational field on the other planet would be:
gp = G*(p*4пrp3/3)/rp2 = G*(p*4п(2re)3/3)/(2re)2 = 4пGp(8*re3)/3(4*re2) = 4пGp(2*re)/3 = 2*(4пGpre/3) = 2ge

(Original post by kappleberry)
..and also part iii of that Q..where did they get the value for the energy from?
I know power = energy/time. but we dont have a value for t =/
L (luminosity) = P (total power output) = ΔE/Δt.
We need Δm/Δt.
ΔE = Δmc2 = ΔE/Δt = Δmc2/Δt => Δm/Δt = c2*ΔE/Δt = c2*L.
12. (Original post by kappleberry)
http://cid-5432f573d3fc3f5a.office.l...hysics%20A.doc

Q19 (ii)....whats =/

Is it b) (ii)?
Well, you know the ratio of distances, you know the total distance.
R + r = 1.5*1011
As part b) (i) askes to show, R/r = 600 => R = 600r.
600r + r = 1.5*1011
r = 1.5*1011/601 = 2.5*108m

Using R/r = 577 or so, as you would calculate in b) (i), would result in r = 2.6*108m, which is still within the acceptable range.

Did I miss any questions asked?
13. I dont know how to use laTeX so i've just done it on paper (might be abit hard to follow but il try help if you get stuck).

I remember going through this question at school and for 1 mark its way too much to do, probably why i remember this question so well lol.

Is it b) (ii)?
Well, you know the ratio of distances, you know the total distance.
R + r = 1.5*1011
As part b) (i) askes to show, R/r = 600 => R = 600r.
600r + r = 1.5*1011
r = 1.5*1011/601 = 2.5*108m

Using R/r = 577 or so, as you would calculate in b) (i), would result in r = 2.6*108m, which is still within the acceptable range.

Did I miss any questions asked?
I get itt!! Thank you so much fr yur helpp!! =]

Would you mind helping with another one?

So sorry!

http://cid-5432f573d3fc3f5a.office.l...hysics%20Q.doc
Q 24(b) I do understand why they got 712 as the wavelength. Only why would they add 56 to the known wavelength when the Q asks where the line on the spectra would appear? I thought youd add it to 656..?

thank yuuu
15. (Original post by Uberuvy)
I dont know how to use laTeX so i've just done it on paper (might be abit hard to follow but il try help if you get stuck).

I remember going through this question at school and for 1 mark its way too much to do, probably why i remember this question so well lol.

Thank you Uberuvy! =]
16. (Original post by kappleberry)
I do understand why they got 712 as the wavelength. Only why would they add 56 to the known wavelength when the Q asks where the line on the spectra would appear? I thought youd add it to 656..?

thank yuuu
656 + 56 = 712
I don't know if im understanding your question right but it seems that they did add it to 656.
17. (Original post by kappleberry)
I get itt!! Thank you so much fr yur helpp!! =]

Would you mind helping with another one?

So sorry!

http://cid-5432f573d3fc3f5a.office.l...hysics%20Q.doc
Q 24(b) I do understand why they got 712 as the wavelength. Only why would they add 56 to the known wavelength when the Q asks where the line on the spectra would appear? I thought youd add it to 656..?

thank yuuu

I will use
ƛ = original (656nm)
ƛ' = received (684nm from X)
Δƛ = change (ƛ' - ƛ )
Red shift, z = Δƛ/ƛ = v/c. When the distance, d, is doubled, from Hubble's law, v = 2dH0 is also doubled. So from Y it would be:
Δƛy/ƛ = 2v/c
Δƛy = 2vƛ/c = 2Δƛx = 2*(684-656) = 56.
From this you calculate ƛ'y by Δƛy = ƛ'y - ƛ => ƛ'y = Δƛy + ƛ.

I will use
ƛ = original (656nm)
ƛ' = received (684nm from X)
Δƛ = change (ƛ' - ƛ )
Red shift, z = Δƛ/ƛ = v/c. When the distance, d, is doubled, from Hubble's law, v = 2dH0 is also doubled. So from Y it would be:
Δƛy/ƛ = 2v/c
Δƛy = 2vƛ/c = 2Δƛx = 2*(684-656) = 56.
From this you calculate ƛ'y by Δƛy = ƛ'y - ƛ => ƛ'y = Δƛy + ƛ.
So you always add the new change to the original?

=] I cant thank yu enoughh.

yur a staaar
19. I have a strong feeling that this may be a stupid question, but I'm going over the nuclear decay topic and I was just wondering, how does a Nuclide differ from a Nuclei?

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