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AQA Physics PHYA5 - Thursday 18th June 2015 [Exam Discussion Thread] Watch

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    (Original post by CD223)
    Absolute magnitude was reading off of a graph so everyone got slightly different values. Try and plug into your calculator slightly different values (1.981, 1.982, 1.983 etc) and you'll see they vary dramatically.


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    Are you going to make an unofficial MS?
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    (Original post by ubisoft)
    Are you going to make an unofficial MS?
    I will do, considering everyone's contributions and including my own input, but I will do that on Tuesday after my M2 and Computing exams are over.


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    (Original post by CD223)
    Absolute magnitude was reading off of a graph so everyone got slightly different values. Try and plug into your calculator slightly different values (1.981, 1.982, 1.983 etc) and you'll see they vary dramatically.


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    None of them vary as much as his though. His is essentially -9.7x10^-6 which is way out. It should be to power of -3
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    (Original post by ImNormal)
    None of them vary as much as his though. His is essentially -9.7x10^-6 which is way out. It should be to power of -3
    Oh sorry, well observed haha, didn't realise it was a factor of 1000 out. I'm sure it was just a case of not remembering.


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    Hi guys, thanks for all your feedback, so here is an updated version. I would be grateful for answers to the six marker and the Schwarzschild radius.
    1) (i) Closest distance was sqrt 2.57^2 + 1^2 x 1.5x10 11 = 4x10 11(ii) Show that theta = 3.12x10 6 radians(iii)Draw a Casegrain telescope (the correct diagram is shown in the June 2011Mark scheme)(iv)The crater can be resolved in detail because the angle subtended is largerthan the theoretical min. resolving limit2) (i) Six marker, Don't ask.(ii) m - M = 5log25/10 = -0.00973)(i) T = 16111K(ii) The star was actually a white dwarf but I messed up and put main sequence(iii) H balmer lines because the star's atmosphere was hot enough to havehydrogen in its n = 2 state and the deep prominent absorption lines in theBalmer series in the absorption spectra of stars with this surface temperature.4)(i) A black hole is an object whose escape velocity is greater than c(ii) Schwarzschild Radius, can't remember.(iii) 4.96x10 17 s
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    Hi guys, thanks for all your feedback, so here is an updated version. I would be grateful for answers to the six marker and the Schwarzschild radius.
    1) (i) Closest distance was sqrt 2.57^2 + 1^2 x 1.5x10 11 = 4x10 11
    (ii) Show that theta = 3.12x10 6 radians
    (iii)Draw a Casegrain telescope (the correct diagram is shown in the June 2011Mark scheme)
    (iv)The crater can be resolved in detail because the angle subtended is largerthan the theoretical min. resolving limit

    2) (i) Six marker, Don't ask.
    (ii) m - M = 5log25/10 = -0.0097

    3)(i) T = 16111K
    (ii) The star was actually a white dwarf but I messed up and put main sequence
    (iii) H balmer lines because the star's atmosphere was hot enough to have hydrogen in it's n=2 state andthe deep prominent absorption lines in the Balmer series in the absorption spectra of stars with this surface temperature.

    4)(i) A black hole is an object whose escape velocity is greater than c
    (ii) Schwarzschild Radius, can't remember.
    (iii) 4.96x10 17 s
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    (Original post by TheRAG)
    Hi guys, thanks for all your feedback, so here is an updated version. I would be grateful for answers to the six marker and the Schwarzschild radius.
    1) (i) Closest distance was sqrt 2.57^2 + 1^2 x 1.5x10 11 = 4x10 11
    (ii) Show that theta = 3.12x10 6 radians
    (iii)Draw a Casegrain telescope (the correct diagram is shown in the June 2011Mark scheme)
    (iv)The crater can be resolved in detail because the angle subtended is largerthan the theoretical min. resolving limit

    2) (i) Six marker, Don't ask.
    (ii) m - M = 5log25/10 = -0.0097

    3)(i) T = 16111K
    (ii) The star was actually a white dwarf but I messed up and put main sequence
    (iii) H balmer lines because the star's atmosphere was hot enough to have hydrogen in it's n=2 state andthe deep prominent absorption lines in the Balmer series in the absorption spectra of stars with this surface temperature.

    4)(i) A black hole is an object whose escape velocity is greater than c
    (ii) Schwarzschild Radius, can't remember.
    (iii) 4.96x10 17 s
    I cannot remember the question well enough to know I read it correctly but can you explain your working out for 1i) please?
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    (Original post by adamjhodg)
    I cannot remember the question well enough to know I read it correctly but can you explain your working out for 1i) please?
    I used Pythagoras and square rooted 2.75^2 and added 1 (these were the valued in Au) then multiplied by 1.5x10 11 to convert to meters
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    (Original post by TheRAG)
    Hi guys, thanks for all your feedback, so here is an updated version. I would be grateful for answers to the six marker and the Schwarzschild radius.
    1) (i) Closest distance was sqrt 2.57^2 + 1^2 x 1.5x10 11 = 4x10 11(ii) Show that theta = 3.12x10 6 radians(iii)Draw a Casegrain telescope (the correct diagram is shown in the June 2011Mark scheme)(iv)The crater can be resolved in detail because the angle subtended is largerthan the theoretical min. resolving limit2) (i) Six marker, Don't ask.(ii) m - M = 5log25/10 = -0.00973)(i) T = 16111K(ii) The star was actually a white dwarf but I messed up and put main sequence(iii) H balmer lines because the star's atmosphere was hot enough to havehydrogen in its n = 2 state and the deep prominent absorption lines in theBalmer series in the absorption spectra of stars with this surface temperature.4)(i) A black hole is an object whose escape velocity is greater than c(ii) Schwarzschild Radius, can't remember.(iii) 4.96x10 17 s
    Summised 6 marker.

    ~Apparent magnitude graph shows the variation in apparent magnitude of the eclipsing binaries.
    ~The deeper troughs show that the less luminous star is eclipsing the more luminous star and visa versa for the shallower troughs.
    ~The maximum height of the graph occurs when the two star are equidistant from the earth ergo one quarter further in there orbit from an eclipse.
    ~Period can be calculated as the time between two consecutive troughs of equal depth. Therefore 5-1=4 days.
    ~Wavelength analysis graph shows the variation in wavelength of the Balmer line centered on ~628 nm.
    ~Change in wavelength is a consequence of the Doppler effect when the star is moving towards or away from the earth in its orbit.
    ~Red shift occurs when the star is moving away from us therefore an increase in the wavelength of the light and visa versa for blue shift.
    ~Orbital speed may be calculated by z=-delta.lamda/lamda and then feeding that answer back into z=v/c.
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    (Original post by TheRAG)
    I used Pythagoras and square rooted 2.75^2 and added 1 (these were the valued in Au) then multiplied by 1.5x10 11 to convert to meters
    Just realised i used the parsec->metre coversion for it...damn!
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    1) (i) Closest distance was sqrt 2.57^2 +1^2 x 1.5x10 11 = 4x10 11(ii) Show that theta = 3.12x10 -6 radians(iii)Draw a Casegrain telescope (the correct diagram is shown in the June2011Mark scheme)(iv)The crater can be resolved in detail because the angle subtended is larger thanthe theoretical min. resolving limit2) (i) ~Apparent magnitude graph shows the variation in apparent magnitude of theeclipsing binaries.~The deeper troughs show that the less luminous star is eclipsing the moreluminous star and visa versa for the shallower troughs.~The maximum height of the graph occurs when the two star are equidistant fromthe earth ergo one quarter further in there orbit from an eclipse.~Period can be calculated as the time between two consecutive troughs of equaldepth. Therefore 5-1=4 days.~Wavelength analysis graph shows the variation in wavelength of the Balmer linecentered on ~628 nm. ~Change in wavelength is a consequence of the Doppler effect when the star ismoving towards or away from the earth in its orbit.~Red shift occurs when the star is moving away us therefore an increase in thewavelength of the light and visa versa for blue shift.~Orbital speed may be calculated by z=delta.lamda/lamda and then feeding thatanswer back into z=v/c.(ii) m - M = 5log25/10 = -0.00973)(i) T = 16111K(ii) The star was a white dwarf (iii) H balmer lines because the star's atmosphere was hot enough to havehydrogen in it's n=2 state andthe deep prominent absorption lines in the Balmerseries in the absorption spectra of stars with this surface temperature.4)(i) A black hole is an object whose escape velocity is greater than c(ii) Schwarzschild Radius, can't remember.(iii) 4.96x10 17 s
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    What are the sig fig for astro I did 2 sig fig for the majority :/ I'm guessing I've lost a lot of marks if it's meant to be 3 sig figs
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    (Original post by adamjhodg)
    Just realised i used the parsec->metre coversion for it...damn!
    What did you get?
    It did state that the distances were in AU.
    But if you got 2.92 AU, you should get 1 mark.
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    (Original post by TheRAG)
    1) (i) Closest distance was sqrt 2.57^2 +1^2 x 1.5x10 11 = 4x10 11(ii) Show that theta = 3.12x10 -6 radians(iii)Draw a Casegrain telescope (the correct diagram is shown in the June2011Mark scheme)(iv)The crater can be resolved in detail because the angle subtended is larger thanthe theoretical min. resolving limit2) (i) ~Apparent magnitude graph shows the variation in apparent magnitude of theeclipsing binaries.~The deeper troughs show that the less luminous star is eclipsing the moreluminous star and visa versa for the shallower troughs.~The maximum height of the graph occurs when the two star are equidistant fromthe earth ergo one quarter further in there orbit from an eclipse.~Period can be calculated as the time between two consecutive troughs of equaldepth. Therefore 5-1=4 days.~Wavelength analysis graph shows the variation in wavelength of the Balmer linecentered on ~628 nm. ~Change in wavelength is a consequence of the Doppler effect when the star ismoving towards or away from the earth in its orbit.~Red shift occurs when the star is moving away us therefore an increase in thewavelength of the light and visa versa for blue shift.~Orbital speed may be calculated by z=delta.lamda/lamda and then feeding thatanswer back into z=v/c.(ii) m - M = 5log25/10 = -0.00973)(i) T = 16111K(ii) The star was a white dwarf (iii) H balmer lines because the star's atmosphere was hot enough to havehydrogen in it's n=2 state andthe deep prominent absorption lines in the Balmerseries in the absorption spectra of stars with this surface temperature.4)(i) A black hole is an object whose escape velocity is greater than c(ii) Schwarzschild Radius, can't remember.(iii) 4.96x10 17 s
    The mass given on the Schwartzchild radius question was 1x10^10, therefore a radius of ~3x10^13 m is to be calculated.
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    (Original post by Jamesbe1993)
    The mass given on the Schwartzchild radius question was 1x10^10, therefore a radius of ~3x10^13 m is to be calculated.
    Thanks I remember this

    Were we supposed to calculate v fir the six marker?
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    Once again, thanks a lot for your help increating this
    1) (i) Closest distance was sqrt 2.57^2 +1^2 x 1.5x10 11 = 4x10 11(ii) Show that theta = 3.12x10 6 radians(iii)Draw a Cassegrain telescope (the correct diagram is shown in the June2011Mark scheme)(iv)The crater can be resolved in detail because the angle subtended is larger thanthe theoretical min. resolving limit2) (i) ~Apparent magnitude graph shows the variation in apparent magnitude of theeclipsing binaries.~The deeper troughs show that the less luminous star is eclipsing the moreluminous star and visa versa for the shallower troughs.~The maximum height of the graph occurs when the two star are equidistant fromthe earth ergo one quarter further in there orbit from an eclipse.~Period can be calculated as the time between two consecutive troughs of equaldepth. Therefore 5-1=4 days.~Wavelength analysis graph shows the variation in wavelength of the Balmer linecentered on ~628 nm. ~Change in wavelength is a consequence of the Doppler effect when the star ismoving towards or away from the earth in its orbit.~Red shift occurs when the star is moving away us therefore an increase in thewavelength of the light and visa versa for blue shift.~Orbital speed may be calculated by z=delta.lamda/lamda and then feeding thatanswer back into z=v/c.(ii) m - M = 5log25/10 = -0.00973)(i) T = 16111K(ii) The star was a white dwarf (iii) H balmer lines because the star's atmosphere was hot enough to havehydrogen in it's n=2 state andthe deep prominent absorption lines in the Balmerseries in the absorption spectra of stars with this surface temperature.4)(i) A black hole is an object whose escape velocity is greater than c(ii) Schwarzschild Radius was 3x10^13 m(iii) 4.96x10 17 s
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    (Original post by TheRAG)
    Thanks I remember this

    Were we supposed to calculate v fir the six marker?
    It did state appropriate calculations are necessary in the question. I did calculate it to ~113 Kms-1. Which as orbital speeds go would be correct in real world application.
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    (Original post by TheRAG)
    Once again, thanks a lot for your help increating this
    1) (i) Closest distance was sqrt 2.57^2 +1^2 x 1.5x10 11 = 4x10 11(ii) Show that theta = 3.12x10 6 radians(iii)Draw a Cassegrain telescope (the correct diagram is shown in the June2011Mark scheme)(iv)The crater can be resolved in detail because the angle subtended is larger thanthe theoretical min. resolving limit2) (i) ~Apparent magnitude graph shows the variation in apparent magnitude of theeclipsing binaries.~The deeper troughs show that the less luminous star is eclipsing the moreluminous star and visa versa for the shallower troughs.~The maximum height of the graph occurs when the two star are equidistant fromthe earth ergo one quarter further in there orbit from an eclipse.~Period can be calculated as the time between two consecutive troughs of equaldepth. Therefore 5-1=4 days.~Wavelength analysis graph shows the variation in wavelength of the Balmer linecentered on ~628 nm. ~Change in wavelength is a consequence of the Doppler effect when the star ismoving towards or away from the earth in its orbit.~Red shift occurs when the star is moving away us therefore an increase in thewavelength of the light and visa versa for blue shift.~Orbital speed may be calculated by z=delta.lamda/lamda and then feeding thatanswer back into z=v/c.(ii) m - M = 5log25/10 = -0.00973)(i) T = 16111K(ii) The star was a white dwarf (iii) H balmer lines because the star's atmosphere was hot enough to havehydrogen in it's n=2 state andthe deep prominent absorption lines in the Balmerseries in the absorption spectra of stars with this surface temperature.4)(i) A black hole is an object whose escape velocity is greater than c(ii) Schwarzschild Radius was 3x10^13 m(iii) 4.96x10 17 s
    I did the same thing but that's wrong. How many marks out of 3 do you guys reckon we would get for it?
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    (Original post by TheRAG)
    Once again, thanks a lot for your help increating this
    1) (i) Closest distance was sqrt 2.57^2 +1^2 x 1.5x10 11 = 4x10 11(ii) Show that theta = 3.12x10 6 radians(iii)Draw a Cassegrain telescope (the correct diagram is shown in the June2011Mark scheme)(iv)The crater can be resolved in detail because the angle subtended is larger thanthe theoretical min. resolving limit2) (i) ~Apparent magnitude graph shows the variation in apparent magnitude of theeclipsing binaries.~The deeper troughs show that the less luminous star is eclipsing the moreluminous star and visa versa for the shallower troughs.~The maximum height of the graph occurs when the two star are equidistant fromthe earth ergo one quarter further in there orbit from an eclipse.~Period can be calculated as the time between two consecutive troughs of equaldepth. Therefore 5-1=4 days.~Wavelength analysis graph shows the variation in wavelength of the Balmer linecentered on ~628 nm. ~Change in wavelength is a consequence of the Doppler effect when the star ismoving towards or away from the earth in its orbit.~Red shift occurs when the star is moving away us therefore an increase in thewavelength of the light and visa versa for blue shift.~Orbital speed may be calculated by z=delta.lamda/lamda and then feeding thatanswer back into z=v/c.(ii) m - M = 5log25/10 = -0.00973)(i) T = 16111K(ii) The star was a white dwarf (iii) H balmer lines because the star's atmosphere was hot enough to havehydrogen in it's n=2 state andthe deep prominent absorption lines in the Balmerseries in the absorption spectra of stars with this surface temperature.4)(i) A black hole is an object whose escape velocity is greater than c(ii) Schwarzschild Radius was 3x10^13 m(iii) 4.96x10 17 s
    Sorry for correction but 1a is incorrect. The maximum distance occurs when the earth is on the directly opposite side of the sun therefore (2.57+1)(1.5x10^11)=5.36x10^11
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    (Original post by Reverse Swing)
    I did the same thing but that's wrong. How many marks out of 3 do you guys reckon we would get for it?
    1 at most. Use of the value of an AU in meters.
 
 
 
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