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

Scroll to see replies

Reply 2000

[br]d = \dfrac{1}{p}[br]

This equation relates the maximum distance, d, measurable (the precision) of a telescope to the parallax angle it is capable of resolving, p.

Subbing in the numbers gives:

[br]d = \dfrac{1}{0.002} = 500 pc[br]

As it asks for the distance in metres,

[br]d = 500 \times 3.08 \times 10^{16}m[br]

[br]\therefore d = 2 \times 10^{19}m\ \text{(1 sf)}[br]

Posted from TSR Mobile

this d = 1/p business.. It's not in formula book ? Where can I find notes on it or how they got the equation (derive)

Reply 2001

000alex

d=1/p isn't on the spec anymore?

Reply 2002

johnh545

Original post by johnh545

I need help with this question.
i dont really understand what it means by 1 in 400 detected on the facing surface of the detector.
were told earlier in the question that the detector has area 1.5x10^-3 and the count rate is 0.62

Anyone?

Reply 2003

betbi3etwerrd

Original post by 000alex

d=1/p isn't on the spec anymore?

I agree.

I did it a different way but you can get 1.55 which is approximately 2 (their answer 1 s.f since they say so) or you can get 3 depending on the question I ask in this diagram:

(edited 8 years ago)

Reply 2004

CD223

Original post by frankiejayx

Could anyone explain what we need to know of the parallax angle?

The parallax angle is half the angular shift in the line of sight over six months drawn directly from Earth to the object of interest.

The angular separation to an object is measured over 6 months when the earth has rotated about its axis and measurements are taken from diametrically opposite positions on earth.

Nearby stars appear to move against the backdrop of more distant stars, allowing this angular separation to be measured.

Once you have the angular separation, astronomers halve this number to get the parallax angle.

The formula

[br]d = \dfrac{1}{p}[br]

Can then be used. By subbing in your value of p you can determine the distance to the object in parsecs :smile:

ImageUploadedByStudent Room1434541643.371168.jpg

More info can be found here:
http://www2.jpl.nasa.gov/teachers/attachments/parallax.html

Posted from TSR Mobile

Reply 2005

Protoxylic

Original post by Master Sam

http://filestore.aqa.org.uk/subjects/AQA-PHYA51-QP-JUN14.PDF

hey guys can anyone explain why the markscheme doesnt take the binding energy of neutrons into account in Q 1b (ii) ? confused :frown:

Are the neutrons bounded to anything? They are just neutrons so they have a binding energy of zero

Reply 2006

mrno1324

What's the point of the 97% U-238 in fuel rods if it's not fissionable?

Reply 2007

CD223

Original post by EconFan_73

Thanks CD223. I understand how to get the answer once you know that d=1/p, it's just I had never seen that before anywhere, neither in the textbook nor spec notes online. At least I'll know it now if it comes up tomorrow.

P.s Cheers for your notes in the op, was useful to have a quick read through as there's little bits in the Nelson Thornes book not covered in the CGP one :smile:

Original post by betbi3etwerrd

this d = 1/p business.. It's not in formula book ? Where can I find notes on it or how they got the equation (derive)

Original post by 000alex

d=1/p isn't on the spec anymore?

Nope. It's not in the formulae sheet. You're expected to remember it, as evidenced by last year's question.

Posted from TSR Mobile

Reply 2008

Plasmapause

Does anyone have a way to remember the order of star types/spectral classes or whatever on a Hertzsprung - Russell diagram?

(edited 8 years ago)

Reply 2009

CD223

Original post by mrno1324

What's the point of the 97% U-238 in fuel rods if it's not fissionable?

Natural uranium is not found in the more-fissionable form of U-235 directly. U-238 fuel has to be enriched with U-235 to obtain a fissionable fuel.

It's extremely difficult to get a fuel with a higher % of U-235. I also imagine it would be fairly dangerous as the reaction could spiral out of control if more fissionable material was used.

Posted from TSR Mobile

Reply 2010

CD223

Original post by Plasmapause

Does anyone have a way to remember the order of star types/spectral classes or whatever on a Hertzsprung - Russell diagram?

Oh
Be
A
Fine
Girl/Guy
Kiss
Me

I use that for the order haha!

Posted from TSR Mobile

Reply 2011

frankiejayx

Original post by CD223

The parallax angle is half the angular shift in the line of sight over six months drawn directly from Earth to the object of interest.

Can then be used. By subbing in your value of p you can determine the distance to the object in parsecs :smile:

Posted from TSR Mobile

Thanks

Reply 2012

Plasmapause