AQA Physics PHYA5 - 28th June 2016 [Exam Discussion Thread] Watch

aditisb1
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#421
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#421
(Original post by cowie)
Does Hubble's law assume universe is expanding at a constant acceleration?
yes it assumes that the rate of expansion of the universe is constant
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Ainsleyy
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#422
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Astrophysics question

Hey guys could I get some help with question 4b Jun14? Never really seen anything like it. Question is in the spoiler
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Vikingninja
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#423
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(Original post by Ainsleyy)
Astrophysics question

Hey guys could I get some help with question 4b Jun14? Never really seen anything like it. Question is in the spoiler
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Theta in radians = 1AU/distance.

Rearrange to find distance. Also 1 degree (need to put into radians) = 60 arc minutes. So need to find 0.002 arc seconds in degrees then radians.
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Ainsleyy
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(Original post by Vikingninja)
Theta in radians = 1AU/distance.

Rearrange to find distance. Also 1 degree (need to put into radians) = 60 arc minutes. So need to find 0.002 arc seconds in degrees then radians.
Are you using arc length = r x theta to get to theta = 1AU/distance?
If so how do we know that the arc length is 1AU?
Is the 100 000 stars figure just a distraction?
And do we have to turn the arc seconds into radians because you need to use radians for arc length = r x theta?
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marcusman97
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#425
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(Original post by Ainsleyy)
Are you using arc length = r x theta to get to theta = 1AU/distance?
If so how do we know that the arc length is 1AU?
Is the 100 000 stars figure just a distraction?
And do we have to turn the arc seconds into radians because you need to use radians for arc length = r x theta?
I too am really confused by this question
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Vikingninja
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#426
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(Original post by Ainsleyy)
Are you using arc length = r x theta to get to theta = 1AU/distance?
If so how do we know that the arc length is 1AU?
Is the 100 000 stars figure just a distraction?
And do we have to turn the arc seconds into radians because you need to use radians for arc length = r x theta?
Tan theta = opposite/adjacent, though because of small angle approximations you can just use theta but it needs to be in radians so yeah you need to turn arc seconds into it. Yeah the 100000 stars figure is just a distraction.

With parallax it forms a right angled triangle where distance (adjacent) is the star to the sun and the opposite is the distance between the Sun and Earth which is 1AU. Also make sure that these are all in m or the same unit.
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marcusman97
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(Original post by Vikingninja)
Tan theta = opposite/adjacent, though because of small angle approximations you can just use theta but it needs to be in radians so yeah you need to turn arc seconds into it. Yeah the 100000 stars figure is just a distraction.

With parallax it forms a right angled triangle where distance (adjacent) is the star to the sun and the opposite is the distance between the Sun and Earth which is 1AU. Also make sure that these are all in m or the same unit.
But what does with a precision of 0.002 arc seconds mean? Is that the uncertainty in the measurement?
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Vikingninja
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(Original post by marcusman97)
But what does with a precision of 0.002 arc seconds mean? Is that the uncertainty in the measurement?
No that's the smallest angle subtended for which the furthest distance it can observe a star.
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Ac19951
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Can someone explain why using 5 um as the wavelength is wrong in this question? 1bii June15
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Ainsleyy
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(Original post by Vikingninja)
Tan theta = opposite/adjacent, though because of small angle approximations you can just use theta but it needs to be in radians so yeah you need to turn arc seconds into it. Yeah the 100000 stars figure is just a distraction.

With parallax it forms a right angled triangle where distance (adjacent) is the star to the sun and the opposite is the distance between the Sun and Earth which is 1AU. Also make sure that these are all in m or the same unit.
Ahh okay I just turned 0.002 arc seconds into radians and then did 1AU/theta and I got 1.56x10^19 and the mark scheme says 2x10^19 so I guess I would have got the mark but just wondering they say to use the method in the spoiler do you know what they are doing?
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-jordan-
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(Original post by Cheesecake Ali)
Is it just me that finds electron microscopes in turning point extremely hard to learn cuz it's so boring 😪


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You aren't alone, came up in 2014 though and it seems rare so not expecting it.
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twallien
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#432
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(Original post by Charlie1523)
Can anyone explain question 1c to me please? thanks http://filestore.aqa.org.uk/subjects...C-QP-JUN12.PDF
When the process happens quickly one can assume it's adiabatic because it happens too quickly for energy to transfer out of the system (ie the air gun warming up the surrounding air). However if the compression happens slowly then energy can transfer so one can assume the temperature remains constant (because as the air heats up it has time to transfer the heat out of the system, thus the temp remains near constant). Now if one looks at the graphs of adiabatic and isothermal for a given pressure the volume must be smaller for an isothermal.
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Vikingninja
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#433
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(Original post by Ainsleyy)
Ahh okay I just turned 0.002 arc seconds into radians and then did 1AU/theta and I got 1.56x10^19 and the mark scheme says 2x10^19 so I guess I would have got the mark but just wondering they say to use the method in the spoiler do you know what they are doing?
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"Give to appropriate number sf"

It's correct technically, the equation in the markscheme gives a similar answer to you but it's then changed to lower sf.

You don't need to use the exact same equation. Not sure what that equation is but 1 parsec is the distance if theta is 1 arc second so I think that it's something about them being proportional.
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Ainsleyy
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#434
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(Original post by Vikingninja)
"Give to appropriate number sf"

It's correct technically, the equation in the markscheme gives a similar answer to you but it's then changed to lower sf.

You don't need to use the exact same equation. Not sure what that equation is but 1 parsec is the distance if theta is 1 arc second so I think that it's something about them being proportional.
Okay thanks man you helped alot!
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thestarter
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#435
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6 mark predictions?
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marioman
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#436
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(Original post by thestarter)
6 mark predictions?
Experiment to detemine the specific heat capacity of a substance?

Just a random guess.
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Vikingninja
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#437
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(Original post by Ainsleyy)
Okay thanks man you helped alot!
When you asked about 1AU being arc length I think that is how the small angle approximation occurs since it's an approximation and because the angle is so small then the distance between Earth and the sun is very close to being an arc length if a circle is formed from many of the angle sub tended.
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Lamassu
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#438
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#438
Do you get the method marks if you use a different method to what's on the mark scheme but get the right answer?
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arrow_h
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(Original post by Ac19951)
Can someone explain why using 5 um as the wavelength is wrong in this question? 1bii June15
It says the smallest angle it can resolve is 3.3x10^7.

You know theta=lambda/D therefore... if lambda is greater theta is greater and if lamda is smaller theta is smaller.

As it gives the smallest angle, you have to use the smaller value of lambda
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Seclusion
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#440
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#440
Can someone explain how electron diffraction calculates the nuclear radius?
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