Hey there! Sign in to join this conversationNew here? Join for free

OCR Physics B G495 Field and Particle Pictures June 21st 2011 Exam Thread Watch

    Offline

    0
    ReputationRep:
    (Original post by hannahbeex)
    Anyone want to explain why the Earth bulges at the equator? Other than just 'Cos it's rotating'...just in case it comes up in section C
    Large-scale mantle convection due to axial rotation?
    Offline

    2
    ReputationRep:
    (Original post by Summerdays)
    A better way to think of it, after realising that pretty much all of the mass lies in the nucleus is that the atomic radius equals 4.6x10^-14/6x10-5 = 7.67x10^-10m

    The MASS of the gold atom thus equals to 7.67x10^-10 x 1.9 x10^4 =3.6x10^-23 (m=Vxrho)

    This is the same mas that is contained in the NUCLEUS because of my previous assumption.

    So p = M/v = 3.6x10^-23 x (4/3pi(4.6x10-14)^3) = 8.8x10^16

    V = 4/3pi(4.6x10-14)^3
    Cheers for that, makes more sense now.

    Fingers crossed ratios and uncertainties are kept to a minimum tomorrow though.
    Offline

    0
    ReputationRep:
    found this on yahoo answers:

    the centripetal acceleration of the earth's rotation slightly cancels out gravity, points on the equator are the furthest from the earth's axis so they experience the greatest acceleration

    not entirely sure why the acceleration cancels out gravity, any explanations?
    Offline

    2
    ReputationRep:
    (Original post by Mikkels88)
    found this on yahoo answers:

    the centripetal acceleration of the earth's rotation slightly cancels out gravity, points on the equator are the furthest from the earth's axis so they experience the greatest acceleration

    not entirely sure why the acceleration cancels out gravity, any explanations?
    Maybe because some of the gravitational force is used to stop the mass flying off at a tangent making it weaker or something but not sure - chances are they wouldn't ask a 5-mark question to explain it anyway and that little description would do fine.
    Offline

    2
    ReputationRep:
    OCR B are idiots if they ask such a question. That concept isn't intuitive and it isnt something that can be adequately be explained without it being taught!
    Offline

    13
    ReputationRep:
    (Original post by Mikkels88)
    found this on yahoo answers:

    the centripetal acceleration of the earth's rotation slightly cancels out gravity, points on the equator are the furthest from the earth's axis so they experience the greatest acceleration

    not entirely sure why the acceleration cancels out gravity, any explanations?
    Well, the acceleration is to keep the direction of velocity at a tangent to the earth's surface, so if you increase that acceleration needed then it could be considered to cancel out gravity.
    Offline

    2
    ReputationRep:
    (Original post by Summerdays)
    Sure. His thesis says that the momentum of an object detremines their wavelength, the larger the momentum the smaller the wavelength, but it's really the MASS that determines if this effect is noticable or not.

    An electron ha a definite wavelength, depending on its speed. If the wavelength is as large a the diameter of the nucleus, diffreaction occurs, sintheta = 1.2lamda/d where d is the diameter of the nucleus. Each nucleus acts as an obstacle (like an aperture) - interference results between the phasors of the paths available to the electron either side of the nucleus. The theta from this equation determine the angle where a MINIMUM occurs. The electron travel all paths. The smaller the wavelength of an electron, the greater it's resolving abilities (based on rayleigh's criterion, th = lambda/d.) If the energy is large enough an electron can cause the gluon field to stretch so much that mesons are released. An electron is used because it is not effected by the strong force, in the nucleus because it is a lepton
    Cracking stuff Can you also whiz through the bit on the wave model? I know about energy levels, electrons having wave-like character and how the standing wave patterns change as n increases, but I might have a couple o' other gaps in my understanding. xD
    Offline

    13
    ReputationRep:
    (Original post by Summerdays)
    OCR B are idiots if they ask such a question. That concept isn't intuitive and it isnt something that can be adequately be explained without it being taught!
    I think explaining why points on the equator would experience the greatest acceleration would be well within the syllabus though.
    Offline

    0
    ReputationRep:
    (Original post by Mikkels88)
    found this on yahoo answers:

    the centripetal acceleration of the earth's rotation slightly cancels out gravity, points on the equator are the furthest from the earth's axis so they experience the greatest acceleration

    not entirely sure why the acceleration cancels out gravity, any explanations?
    Thanks, still don't really understand it but I'm just going to memorise that and hope it'll be good enough if it comes up tomorrow :/
    Offline

    2
    ReputationRep:
    (Original post by Unkempt_One)
    I think explaining why points on the equator would experience the greatest acceleration would be well within the syllabus though.

    Not really, to be honest. The only other concept that even remotely explains that is an elliptical orbit...
    Offline

    0
    ReputationRep:
    (Original post by Unkempt_One)
    I think explaining why points on the equator would experience the greatest acceleration would be well within the syllabus though.
    can you simply say they are furthest from the axis of rotation (centre of the circle)?

    but in my mind it doesnt work out, cus if F = mv^2 / r , surely being further is in some ways increasing r and so reducing the force? the only way i can think is because they are furthest out, they are travelling furthest and so have the greatest velocity and thus increasing the force / acceleration... arrrr mind****
    Offline

    2
    ReputationRep:
    (Original post by Unkempt_One)
    Well, the acceleration is to keep the direction of velocity at a tangent to the earth's surface, so if you increase that acceleration needed then it could be considered to cancel out gravity.
    I don't see why it would though...
    Offline

    13
    ReputationRep:
    (Original post by Mikkels88)
    can you simply say they are furthest from the axis of rotation (centre of the circle)?

    but in my mind it doesnt work out, cus if F = mv^2 / r , surely being further is in some ways increasing r and so reducing the force? the only way i can think is because they are furthest out, they are travelling furthest and so have the greatestvelocity and thus increasing the force / acceleration... arrrr mind****
    Centripetal force can also be expressed as mrXomega^2. Omega, the angular speed, will intuitively be the same at any point on the earth's surface, so omega doesn't vary, not does m, and r increases, hence the centripetal force required increases.

    EDIT: Stated force then talked about acceleration for a bit. Corrected.
    Offline

    13
    ReputationRep:
    (Original post by Summerdays)
    I don't see why it would though...
    Nope, I don't think it cancels out gravity. I think that Yahoo answers thing doesn't entirely appreciate the meaning of centripetal acceleration.
    Offline

    2
    ReputationRep:
    (Original post by arianex)
    Cracking stuff Can you also whiz through the bit on the wave model? I know about energy levels, electrons having wave-like character and how the standing wave patterns change as n increases, but I might have a couple o' other gaps in my understanding. xD
    Hmm, the electron vibrates at specific wavelengths where its wavelength is equal to an integer multiple of the circumference of the nucleus. The electron's wavelength forms standing waves "at the limits" of the nuclues. The wavelength of the electron determines the enrgy of the elctron (E = hc/lambda.) The ground state of the electron corresponds to the first harmonic of its standing wave.

    This link might also help http://www.sparknotes.com/testprep/b...section3.rhtml
    Offline

    2
    ReputationRep:
    (Original post by Unkempt_One)
    Centripetal force can also be expressed as rXomega^2. Omega, the angular speed, will intuitively be the same at any point on the earth's surface, so omega doesn't vary, and r increases, hence the centripetal force required increases.
    This answer is a good answer though But it still doesn't explain that statement.
    Offline

    0
    ReputationRep:
    i'll be so glad when ocr b is out of my life.
    Offline

    13
    ReputationRep:
    (Original post by Summerdays)
    This answer is a good answer though But it still doesn't explain that statement.
    Do you mean why the earth bulges at the equator?
    Offline

    2
    ReputationRep:
    (Original post by Unkempt_One)
    Do you mean why the earth bulges at the equator?
    Hmm, well I guess because the equator is further away from the centre of mass thus the elements at the equator eperience the greatest centripetal force (is this correct?)
    I am really talking about gravity cancelling.
    Offline

    0
    ReputationRep:
    I would say that at the equator its rotating much faster than at the poles, so has a larger acceleration and a larger centripedal force pulling it outwards, causing the bulge, non?
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Would you rather give up salt or pepper?
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Write a reply...
    Reply
    Hide
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.