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    (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
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    Could someone please try and quickly explain the life cycle of stars?... All my books say slightly different things, and its confusing me..

    Thanks!
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    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....

    the answers 2g...but why?

    Thank youu!
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    (Original post by yadas)
    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? =/
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    (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.
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    ..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 =/
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    Here's my Astrophysics notes I made earlier this afternoon.
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    ^^thank you so much! Gonna have a read now! =]
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    Sorry...i know youve gone through it in the notes, but could you just exaplain what youve done in q2 of the spec paper?
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    Legend! You got any more of these? Quite good for a quick read.
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    http://cid-5432f573d3fc3f5a.office.l...hysics%20A.doc

    Q19 (ii)....whats =/

    help please?
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    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....

    the answers 2g...but why?

    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.
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    (Original post by kappleberry)
    http://cid-5432f573d3fc3f5a.office.l...hysics%20A.doc

    Q19 (ii)....whats =/

    help please?
    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?
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    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.

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    (Original post by yadas)
    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!! =]
    :grin:

    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
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    (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! =]
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    (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.
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    (Original post by kappleberry)
    I get itt!! Thank you so much fr yur helpp!! =]
    :grin:

    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
    Thank you for asking.

    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.
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    (Original post by yadas)
    Thank you for asking.

    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
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    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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