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Thread: HSC Physics Marathon 2013-2015 Archive

  1. #51
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    Re: 2013 HSC physics marathon

    Quote Originally Posted by bleakarcher View Post
    Yeah, I just meant applying the mass dilation formula to F=ma:
    a=F/(m/sqrt(1-v^2/c^2))=F*sqrt(1-v^2/c^2)/m
    As v->c, a->0.
    thank you for that.

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    Junior Member omgiloverice's Avatar
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    Re: 2013 HSC physics marathon

    If our galaxy, the Milky Way, is 20 kiloparsecs or 65 000 light-years in radius, calculate how fast a spacecraft would need to travel so that its occupants could travel right across it in 45 years.
    Interesting question.
    Hehe cheap textbooks here :3


    http://community.boredofstudies.org/...d.php?t=314250


    and freebies :3

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    Re: 2013 HSC physics marathon

    Fuuuuuar! Just noticed some of my tutor students pwning this thread. Well done you three
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    Executive Member iBibah's Avatar
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    Re: 2013 HSC physics marathon

    QUESTION: Explain why all low earth orbit satellites will eventually fall to the Earth’s surface.

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    Re: 2013 HSC physics marathon

    Quote Originally Posted by iBibah View Post
    QUESTION: Explain why all low earth orbit satellites will eventually fall to the Earth’s surface.
    All Low Earth Orbit (LEO) satellites will eventually fall to the Earth's surface due to atmospheric drag. Because LEOs are typically 160-2000km above the surface of the Earth, the satellite tends to collide with air molecules, this creates friction and thus kinetic energy is converted into heat and sound energy, thereby further losing kinetic energy and in time, altitude.

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    This too shall pass Sy123's Avatar
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    Re: 2013 HSC physics marathon




  7. #57
    This too shall pass Sy123's Avatar
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    Re: 2013 HSC physics marathon

    Quote Originally Posted by Sy123 View Post


    Just a note, am I allowed to ask questions which need students to solve simultaneous equations? Surely general maths students can do that.

    =======================



    Last edited by Sy123; 14 Feb 2013 at 9:45 PM.

  8. #58
    what is that?It is Cowpea RealiseNothing's Avatar
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    Re: 2013 HSC physics marathon

    Quote Originally Posted by Sy123 View Post
    Just a note, am I allowed to ask questions which need students to solve simultaneous equations? Surely general maths students can do that.

    =======================



    You may want to fix up the given equation, you've done a typo I think.

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    Re: 2013 HSC physics marathon

    Quote Originally Posted by RealiseNothing View Post
    You may want to fix up the given equation, you've done a typo I think.
    haha oops thanks for the heads up

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    Re: 2013 HSC physics marathon

    Q: Calculate the effective value of 'g' at the equator. (Given g=9.81ms^-2)

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    Re: 2013 HSC physics marathon

    Quote Originally Posted by iBibah View Post
    Q: Calculate the effective value of 'g' at the equator. (Given g=9.81ms^-2)
    A. The equation to find the value of 'g' is ge= GM/r^2. The radius of the Earth is only slightly bulged at the equator (and flattened at the poles), resulting in only a slight increase in radius and variation from the given "g=9.81ms^-2". The resulting acceleration due to gravity at the equator is about 9.782ms^-2 (sea level).

    Q. Describe how the motor effect is used in a loudspeaker. [3 marks]

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    Retired Nov '14 someth1ng's Avatar
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    Re: 2013 HSC physics marathon

    Quote Originally Posted by Anvaeon View Post
    A. The equation to find the value of 'g' is ge= GM/r^2. The radius of the Earth is only slightly bulged at the equator (and flattened at the poles), resulting in only a slight increase in radius and variation from the given "g=9.81ms^-2". The resulting acceleration due to gravity at the equator is about 9.782ms^-2 (sea level).

    Q. Describe how the motor effect is used in a loudspeaker. [3 marks]
    It clearly said "calculate" - an response like that would get 0.
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    Re: 2013 HSC physics marathon

    Quote Originally Posted by someth1ng View Post
    It clearly said "calculate" - an response like that would get 0.
    This.

    Radius of earth = 6378km (probably should have given this)

    Therefore velocity at equator: v = (2*pi*6378000)/24*60*60

    Centripetal acceleration = v^2/r = 462.82^2/6378000=0.03373 ms^-2

    Therefore effective value of 'g' at the equator is reduced by 0.033 ms^-2, and the resultant effective value of 'g' is 9.777 ms^-2.

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    re: HSC Physics Marathon Archive

    Nice solution Habibi!

    New Question:

    A satellite is launched from the equator to a stable orbit of altitude 600km. Determine the work done on the satellite.
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    re: HSC Physics Marathon Archive

    Quote Originally Posted by Fizzy_Cyst View Post
    Nice solution Habibi!

    New Question:

    A satellite is launched from the equator to a stable orbit of altitude 600km. Determine the work done on the satellite.
    That's just calculating the change in GPE, isn't it?
    Last edited by someth1ng; 24 Feb 2013 at 8:29 PM.
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  16. #66
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    re: HSC Physics Marathon Archive

    Quote Originally Posted by someth1ng View Post
    That's just calculating the change in GPE, isn't it?
    Nope
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    re: HSC Physics Marathon Archive




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    re: HSC Physics Marathon Archive

    If Power loss = v^2/r, then why do we transmit electricity at high voltage. Should this not increase power loss? Explain.
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    re: HSC Physics Marathon Archive

    Quote Originally Posted by Fizzy_Cyst View Post
    Nope
    I'm pretty sure it is, isn't it?
    If we define work properly, let point A be on the equator and point B be the point 600km away:



    Where F is force, and x is the distance.

    By Newton's Law of Universal Gravitation:









    Where E_p is the gravitational potential energy.

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    re: HSC Physics Marathon Archive

    You should write it as P=I^2 R. Thus since power is proportional to the square of current, there is less power loss when there is low current. Since V=P/I, for a low current, there is a high voltage. Thus, a step-up transformer is used to increase the voltage for minimal power/energy loss in transmission lines.

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    re: HSC Physics Marathon Archive

    Quote Originally Posted by Sy123 View Post
    I'm pretty sure it is, isn't it?
    If we define work properly, let point A be on the equator and point B be the point 600km away:



    Where F is force, and x is the distance.

    By Newton's Law of Universal Gravitation:









    Where E_p is the gravitational potential energy.
    You are making an assumption
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    Executive Member bleakarcher's Avatar
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    re: HSC Physics Marathon Archive

    Quote Originally Posted by Fizzy_Cyst View Post
    Nice solution Habibi!

    New Question:

    A satellite is launched from the equator to a stable orbit of altitude 600km. Determine the work done on the satellite.
    Is it the change in the mechanical energy of the satellite?
    Physics is to mathematics like sex is to masturbation.” —Richard Feynman

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    re: HSC Physics Marathon Archive

    Quote Originally Posted by bleakarcher View Post
    Is it the change in the mechanical energy of the satellite?
    Yes

    Not only does work need to be done to lift it from the Earth to 600km altitude, work must also be done to make its speed increase to the orbital velocity of an orbit of alitutude 600km.

    Think about it, change in Ep is just saying 'lifting it from the surface of Earth and placing at a point 600km away', whereas this satellite is ORBITING, not just sitting there, hence it needs more energy and the amount of extra energy it needs is equal to the change in Ek of the satellite
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    re: HSC Physics Marathon Archive

    Ahh, forgot about that - that's why I'm not gonna be majoring in Physics :P
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    re: HSC Physics Marathon Archive






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