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    i know u cant really do this for physics but any predictions??
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    (Original post by SuruthiG)
    i know u cant really do this for physics but any predictions??
    No clue :lol: ! Predicting Edexcel is hard as hell.
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    How do you do this question?
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    (Original post by target21859)
    How do you do this question?
    It is B because Period = 1 / frequency and the equation for period of a spring oscillation = 2 pi x root (m/k) which means it doesn't matter whether the object is on earth or the moon as gravitational field strength (g) doesn't affect the period and hence doesn't affect the frequency
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    (Original post by Moe21)
    It is B because Period = 1 / frequency and the equation for period of a spring oscillation = 2 pi x root (m/k) which means it doesn't matter whether the object is on earth or the moon as gravitational field strength (g) doesn't affect the period and hence doesn't affect the frequency
    ah thanks makes perfect sense
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    Can someone please explain question 3 to me I just don't get it at all?
    https://a4942901ab27cf2817f7a4f7497d...%20Physics.pdf
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    (Original post by Moe21)
    Can someone please explain question 3 to me I just don't get it at all?
    https://a4942901ab27cf2817f7a4f7497d...%20Physics.pdf
    Well you know that the change would be a decrease, so you can immediately discount C and D.
    I did it by saying g mars = GM/r^2. Since the height of Olympus Mons is 0.6% of r, then the height is 0.006r, hence g mountain = GM/(r+0.006r)^2 which is GM/(503r/500)^2
    Since GM and r are constants, you can simply approximate g Mars to be 1 and g Mountain to be 1/(503/500)^2 which is approximately 0.9881
    0.9881-1 is approximately - 0.0119 which is equivalent to - 1.19% so the answer is B
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    (Original post by Moe21)
    It is B because Period = 1 / frequency and the equation for period of a spring oscillation = 2 pi x root (m/k) which means it doesn't matter whether the object is on earth or the moon as gravitational field strength (g) doesn't affect the period and hence doesn't affect the frequency
    But if the question was about a pendulum then the answer would be 1Hz because
    ?
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    Name:  Screen Shot 2016-06-26 at 12.16.46.png
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Size:  8.3 KB?
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    (Original post by PhysicsIP2016)
    Well you know that the change would be a decrease, so you can immediately discount C and D.
    I did it by saying g mars = GM/r^2. Since the height of Olympus Mons is 0.6% of r, then the height is 0.006r, hence g mountain = GM/(r+0.006r)^2 which is GM/(503r/500)^2
    Since GM and r are constants, you can simply approximate g Mars to be 1 and g Mountain to be 1/(503/500)^2 which is approximately 0.9881
    0.9881-1 is approximately - 0.0119 which is equivalent to - 1.19% so the answer is B
    I get it now thank you!


    (Original post by mrbeady9)
    But if the question was about a pendulum then the answer would be 1Hz because
    ?
    Yea that's right
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    (Original post by candycake)
    Sorry, there isn't one - unit 5 was first examined in June 2010.
    Thats really strange because there is a unit 4 jan 2010 :')
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    (Original post by CasioGamer98)
    Name:  Screen Shot 2016-06-26 at 12.16.46.png
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    Flux= Luminosity/ 4pid^2 so we know that Luminosity has the unit W and d has the unit m^2 so

    F= W/m^2. Try simplifying the unit W as this is a derived unit into J/s and then J into Nm. You will get the answer as A.
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    Not sure what you're meant to do here.
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    (Original post by target21859)
    Not sure what you're meant to do here.
    Name:  IMG_5114.jpg
Views: 75
Size:  382.9 KBis 6.67x10^24 molecules correct?
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    (Original post by CasioGamer98)
    Name:  IMG_5114.jpg
Views: 75
Size:  382.9 KBis 6.67x10^24 molecules correct?
    Yes thanks I think I understand now.
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    Name:  Screen Shot 2016-06-26 at 9.23.23 PM.png
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    (Original post by apocolyptic)
    Name:  Screen Shot 2016-06-26 at 9.23.23 PM.png
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Size:  22.7 KBAnswer and explanation please
    The answer is B. The density of the universe is not accurately known to us due to dark matter being undiscovered till now. If Dark Matter is present, then the actual density would be less than the critical density so this means that the universe will keep on expanding i.e it is an open universe because there is continuous expansion taking place.
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    (Original post by sabahshahed294)
    The answer is B. The density of the universe is not accurately known to us due to dark matter being undiscovered till now. If Dark Matter is present, then the actual density would be less than the critical density so this means that the universe will keep on expanding i.e it is an open universe because there is continuous expansion taking place.
    How will the actual density be less if dark matter is present?
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    (Original post by apocolyptic)
    How will the actual density be less if dark matter is present?
    I'm not exactly sure myself but I recall reading somewhere this.(Idk whether my logic is right though :/ )
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    (Original post by sabahshahed294)
    I'm not exactly sure myself but I recall reading somewhere this.(Idk whether my logic is right though :/ )
    but dark matter increases the mass of the universe? Surely it would increase the density by that logic?
 
 
 
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