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

Original post by JJBinn

Thanks again, these are perfect

No worries!

Can you think of any other uses of these radiation types?

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Reply 781

Original post by frankiejayx

Anyone know any other disadvantages of a CCD other than expensive and need to be kept cool?

It's difficult to think of other disadvantages I must say - advantages are a lot easier.

Perhaps the sensitivity is the only main disadvantage. Has it come up as a past paper question before?

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Reply 782

Original post by CD223

No worries!

Can you think of any other uses of these radiation types?

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Radiation is used to power pacemakers I think. Can't remember the detail haha, not much use

Reply 783

Original post by JJBinn

Radiation is used to power pacemakers I think. Can't remember the detail haha, not much use

Oh I see! I'm sure it'll be an application of radiation they give us rather than us mentioning it specifically.

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Reply 784

Original post by CD223

Oh I see! I'm sure it'll be an application of radiation they give us rather than us mentioning it specifically.

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Probably, although it could come at the end of a question on radiation after a few calculations involving a certain type of radiation: State and explain a use of *beta/gamma/alpha* radiation for 2 or 3 marks.

Reply 785

Original post by JJBinn

I suppose. Those three would be my safe uses. Not too sure on others! I couldn't name 3 for each type.

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Reply 786

Astrophysics: June 2011, Q1b (ii)
I understand how the angular resolution and angular separation was calculated using:

\theta = \frac{\lambda}{D}

and

\theta = \frac{4.8 \times 10^{3}}{1.4 \times 10^{9}}

But...
Why is the "minimum" angular resolution the smallest angle in this instance?

I've always seen the maximum angular resolution being quoted as the smallest numerical value. So in this case, I thought that when commenting on the discrepancy between the two angles, you could say that:

"The angle calculated in part ii was the minimum angular resolution, so it may resolve two objects at a smaller angle, making the claim valid".

Is there a definitive time when the terms "minimum" and "maximum" resolution are used? Because I've been caught out too many times!

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Reply 787

Jed-Singh

How has everyone been doing on past papers? Ive given up on 2012 and 2013 because they were sooo difficult, like stuff I didn't even know we had to know (if that makes sense, lol). On the first 2 I got an A however so mixed feelings

Reply 788

Original post by Jed-Singh

Have you done June 2014?

I've done June 2010, 2011 and 2014 for PHYA5. Got 2012 and 2013 to go. What made them hard?

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Reply 789

For those doing Astro:

Is

d=\dfrac{1}{p}

for parallax distances to stars an equation we have to remember off by heart?

It came up in June 2014 in my mock today so I was just unsure as I couldn't find it on the formula sheet.

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(edited 8 years ago)

Reply 790

Original post by CD223

For those doing Astro:

Is

d=\dfrac{1}{p}

I wasn't aware of this equation, is p parallex distance? If so what does this even mean?

I thought you work out distances to an object by right angled triangles between the earth, sun, and object in question.

Reply 791

Original post by JJBinn

That's essentially is it. The equation drops out of the right angled triangle geometry.

d

is the distance to the object in parsecs,

p

is the parallax angle in arcseconds

Where:

1 arcsecond = {(\dfrac{1 degree}{3600})}

(edited 8 years ago)

Reply 792

Original post by CD223

That's essentially is it. The equation drops out of the right angled triangle geometry.

d

is the distance to the object in parsecs,

p

is the parallax angle in arcseconds

Where:

1 arcsecond = {(\dfrac{1 degree}{3600})}

That's a relief, thought there was something I'd missed :biggrin:

my teacher today asked us to show that 1/H is the age of the universe. I didn't know we needed to know that until today.

Reply 793

Jed-Singh

Original post by CD223

Have you done June 2014?

I've done June 2010, 2011 and 2014 for PHYA5. Got 2012 and 2013 to go. What made them hard?

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Hi, yes, I got a B on June 2014. But 12/13 had like really difficultly worded questions and I found it difficult to give myself marks. At least I know now where to improve

Reply 794

Original post by Jed-Singh

Hi, yes, I got a B on June 2014. But 12/13 had like really difficultly worded questions and I found it difficult to give myself marks. At least I know now where to improve

That's true! I get my mark back tomorrow 😁

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Reply 795

Original post by JJBinn

That's a relief, thought there was something I'd missed :biggrin:

my teacher today asked us to show that 1/H is the age of the universe. I didn't know we needed to know that until today.

Oh right! Is that involving converting H from

kms^{-1}Mpc^{-1}

into

s^{-1}

Reply 796

Original post by CD223

Oh right! Is that involving converting H from

kms^{-1}Mpc^{-1}

into

s^{-1}

Yeah, you rearrange hubbles law to give 1/H = distance/velocity, which is time. Then you convert 1/h into s^-1 yeah.

Reply 797

Jed-Singh

Original post by CD223

That's true! I get my mark back tomorrow

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Best of luck, I find the easy questions really really easy, but the hard questions are just rock solid impossible. For me theres no inbetween lol :smile:

Reply 798

Original post by JJBinn

Yeah, you rearrange hubbles law to give 1/H = distance/velocity, which is time. Then you convert 1/h into s^-1 yeah.

That makes sense! But then again I guess we could say that discrepancy between estimates of the age of the universe is due to uncertainty in

{H_0}

?

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Reply 799