# PHYA5 ~ 20th June 2013 ~ A2 Physics Watch

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#581

(Original post by

Just curious, how we're you guys taught the applied topic? Did you have "proper" lessons? Because we just had half an hour in a lunchtime every week... XD

**iCiaran**)Just curious, how we're you guys taught the applied topic? Did you have "proper" lessons? Because we just had half an hour in a lunchtime every week... XD

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#582

(Original post by

I think it is because the ice initially -14 degrees in an insulated room and after 30 s the ice is 0 degrees.But now we have an uninsulated room at 25 degrees more ice will turn into water as heat energy from the surrounding is absorbed by the ice.So the final temp will be higher as thermal energy is transferred from a warmer body to a cooler one.

I hope that makes sense if any

Smith

**smith50**)I think it is because the ice initially -14 degrees in an insulated room and after 30 s the ice is 0 degrees.But now we have an uninsulated room at 25 degrees more ice will turn into water as heat energy from the surrounding is absorbed by the ice.So the final temp will be higher as thermal energy is transferred from a warmer body to a cooler one.

I hope that makes sense if any

Smith

I really hope this was like a 3 marker

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#583

(Original post by

Thanks for the response I think I get it now I really need to think more aha... & I assumed it was a 1 marker, but I guess there's more lines for that question! All the ice turned into water anyway in the previous procedure, but it will just take less than 30 seconds this time right ? A bit more energy is being transferred each second because on addition to the heater the room is also providing heat energy.

I really hope this was like a 3 marker

**posthumus**)Thanks for the response I think I get it now I really need to think more aha... & I assumed it was a 1 marker, but I guess there's more lines for that question! All the ice turned into water anyway in the previous procedure, but it will just take less than 30 seconds this time right ? A bit more energy is being transferred each second because on addition to the heater the room is also providing heat energy.

I really hope this was like a 3 marker

Could you help me out on this question please

Thanks,

Smith

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#584

(Original post by

Oh really? That sounds bad, we had proper lessons for it.

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**D4rth**)Oh really? That sounds bad, we had proper lessons for it.

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#585

(Original post by

If that sounds bad how does having 4 changes of teacher over the 2 years sound? We're on our 5th teacher now

**iCiaran**)If that sounds bad how does having 4 changes of teacher over the 2 years sound? We're on our 5th teacher now

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#586

(Original post by

OMG, never thought of that Yh I think you are very right

Thanks for bringing that up, I didn't even consider the initial conditions of the procedure!

**posthumus**)OMG, never thought of that Yh I think you are very right

Thanks for bringing that up, I didn't even consider the initial conditions of the procedure!

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#587

I could've potentially had 4 teachers over the two years (as it turns out I had 3)

But 2 for AS, one teaching all the mechanics for unit 2 (who I love) and one teaching all the waves and all that (who I hated)

I also had the one that I hated for all the particle physics and the one I loved for the other bit of unit 1.

Then for A2 fortunately the teacher I hated doesn't teach A2 and I kept the one I loved, as well as gaining one I had at GCSE who I loved too

1 taught all the force and gravitational fields part for unit 4 and the other all the electric fields and capacitors and all that

Then for unit 5 one teaches all the core bit and one the applied (turning points)

I spend roughly 2 and a half hours of lessons with them a week (before I finished obviously aha) and we could ask any teacher about anything

Could've easily got two new ones this year instead of keeping one of them

(This happened to me in chemistry, one of my teachers retired and I got moved groups to even out numbers and because of the subject I dropped, one of them is a pretty much new qualified teacher and I've hated her all year aha, safe to say I now suck at chemistry )

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But 2 for AS, one teaching all the mechanics for unit 2 (who I love) and one teaching all the waves and all that (who I hated)

I also had the one that I hated for all the particle physics and the one I loved for the other bit of unit 1.

Then for A2 fortunately the teacher I hated doesn't teach A2 and I kept the one I loved, as well as gaining one I had at GCSE who I loved too

1 taught all the force and gravitational fields part for unit 4 and the other all the electric fields and capacitors and all that

Then for unit 5 one teaches all the core bit and one the applied (turning points)

I spend roughly 2 and a half hours of lessons with them a week (before I finished obviously aha) and we could ask any teacher about anything

Could've easily got two new ones this year instead of keeping one of them

(This happened to me in chemistry, one of my teachers retired and I got moved groups to even out numbers and because of the subject I dropped, one of them is a pretty much new qualified teacher and I've hated her all year aha, safe to say I now suck at chemistry )

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#588

(Original post by

It was a 2 marker .Anyways I'm glad you get it now

Could you help me out on this question please

Thanks,

Smith

**smith50**)It was a 2 marker .Anyways I'm glad you get it now

Could you help me out on this question please

Thanks,

Smith

so 390.406 + 1.675 - (149.357 + 239.056 + 2x1.675)

and you get a mass difference of 3.18 x 10^-27kg.

Now, for a uranium atom, there are 235g per mole. So if you want to find the mass change for 0.5kg, you do 500/235 x (6.0 x 10^23) and this will give you the number of uranium atoms in 0.5kg. You'll get an answer of 1.276 x 10^24 atoms.

Then you just multiply the mass difference if one uranium atom disintegrates by the number of uranium atoms you have, so (3.18 x 10^-27) x (1.276 x 10^24) and you'll get a final answer of 4.06 x 10^-4kg, or 406g.

Hope that helps!

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#589

**smith50**)

It was a 2 marker .Anyways I'm glad you get it now

Could you help me out on this question please

Thanks,

Smith

Energy before - energy after

energy = mc^2

mc^2 - mc^2 = [0.5 x (3x10

^{8})

^{2}+ uc

^{2}] - 236u x (3x10

^{8})

^{2}

Note that u = 1.661 x 10^-27

& 236 is the number of nucleons on the other side

I ended up with 4.5 x 10^16 J

m=E/c^2

Brilliant I got a god damn mass of 0.5kg

Previously I didn't include uc^2 on the left but still got the exact same energy so maybe the calculator just rounds it up since it can't show that many figures

I'm going to use u=931.3 MeV now...

2.8125 x 10^29 MeV + 931.3 MeV - 236 x 931.3 = 2.815 x 10^29 MeV

m=2.815 x 10^35 x 1.6 x 10^-19 / (3 x 10^8)^2 = 0.5 kg

Okay I'm a moron... not getting this one right lol, hopefully someone else will be able to help. Keep getting mass difference of 0.5 kg which I'd be surprised if it was right... might have to use moles or something

EDIT: Ah I see Amish has done it already above

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#590

How are you feeling about it? I hope they don't ask too much about the ear, that's my worst topic, and some of the lens calculations are a pretty confusing, but all in all I don't find it too hard compared to unit 5 section A and 4. Helps that the grade boundaries have been ridiculously low the last few years too!

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#591

(Original post by

I initially thought about working with moles, but I've gone for this method...

Energy before - energy after

energy = mc^2

mc^2 - mc^2 = [0.5 x (3x10

Note that u = 1.661 x 10^-27

& 236 is the number of nucleons on the other side

I ended up with 4.5 x 10^16 J

m=E/c^2

Brilliant I got a god damn mass of 0.5kg

Previously I didn't include uc^2 on the left but still got the exact same energy so maybe the calculator just rounds it up since it can't show that many figures

I'm going to use u=931.3 MeV now...

2.8125 x 10^29 MeV + 931.3 MeV - 236 x 931.3 = 2.815 x 10^29 MeV

m=2.815 x 10^35 x 1.6 x 10^-19 / (3 x 10^8)^2 = 0.5 kg

Okay I'm a moron... not getting this one right lol, hopefully someone else will be able to help. Keep getting mass difference of 0.5 kg which I'd be surprised if it was right... might have to use moles or something

EDIT: Ah I see Amish has done it already above

**posthumus**)I initially thought about working with moles, but I've gone for this method...

Energy before - energy after

energy = mc^2

mc^2 - mc^2 = [0.5 x (3x10

^{8})^{2}+ uc^{2}] - 236u x (3x10^{8})^{2}Note that u = 1.661 x 10^-27

& 236 is the number of nucleons on the other side

I ended up with 4.5 x 10^16 J

m=E/c^2

Brilliant I got a god damn mass of 0.5kg

Previously I didn't include uc^2 on the left but still got the exact same energy so maybe the calculator just rounds it up since it can't show that many figures

I'm going to use u=931.3 MeV now...

2.8125 x 10^29 MeV + 931.3 MeV - 236 x 931.3 = 2.815 x 10^29 MeV

m=2.815 x 10^35 x 1.6 x 10^-19 / (3 x 10^8)^2 = 0.5 kg

Okay I'm a moron... not getting this one right lol, hopefully someone else will be able to help. Keep getting mass difference of 0.5 kg which I'd be surprised if it was right... might have to use moles or something

EDIT: Ah I see Amish has done it already above

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#592

(Original post by

I keep trying what you did, and the calculator kept giving 0.5kg! I was like eh?! Then I tried using moles and it worked! Nifty question that!

**amish123**)I keep trying what you did, and the calculator kept giving 0.5kg! I was like eh?! Then I tried using moles and it worked! Nifty question that!

EDIT: Also may I ask where you got the more accurate values for atomic mass from? Are they given to us in the exam in some form of table or something.. ? Thanks

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#593

(Original post by

Not so great, but I think the paper in general was okay... I just didn't really prepare for it, and wasn't expecting to improve anyway... I've had way too many exams

Was busy preparing for C3 as well, but that turned out to be a frickin' disaster lol

How'd you find PHYA4?

**posthumus**)Not so great, but I think the paper in general was okay... I just didn't really prepare for it, and wasn't expecting to improve anyway... I've had way too many exams

Was busy preparing for C3 as well, but that turned out to be a frickin' disaster lol

How'd you find PHYA4?

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#594

**amish123**)

I keep trying what you did, and the calculator kept giving 0.5kg! I was like eh?! Then I tried using moles and it worked! Nifty question that!

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#595

**posthumus**)

I initially thought about working with moles, but I've gone for this method...

Energy before - energy after

energy = mc^2

mc^2 - mc^2 = [0.5 x (3x10

^{8})

^{2}+ uc

^{2}] - 236u x (3x10

^{8})

^{2}

Note that u = 1.661 x 10^-27

& 236 is the number of nucleons on the other side

I ended up with 4.5 x 10^16 J

m=E/c^2

Brilliant I got a god damn mass of 0.5kg

Previously I didn't include uc^2 on the left but still got the exact same energy so maybe the calculator just rounds it up since it can't show that many figures

I'm going to use u=931.3 MeV now...

2.8125 x 10^29 MeV + 931.3 MeV - 236 x 931.3 = 2.815 x 10^29 MeV

m=2.815 x 10^35 x 1.6 x 10^-19 / (3 x 10^8)^2 = 0.5 kg

Okay I'm a moron... not getting this one right lol, hopefully someone else will be able to help. Keep getting mass difference of 0.5 kg which I'd be surprised if it was right... might have to use moles or something

EDIT: Ah I see Amish has done it already above

(Original post by

Right first work out the total mass change if one uranium atom disintegrates,

so 390.406 + 1.675 - (149.357 + 239.056 + 2x1.675)

and you get a mass difference of 3.18 x 10^-27kg.

Now, for a uranium atom, there are 235g per mole. So if you want to find the mass change for 0.5kg, you do 500/235 x (6.0 x 10^23) and this will give you the number of uranium atoms in 0.5kg. You'll get an answer of 1.276 x 10^24 atoms.

Then you just multiply the mass difference if one uranium atom disintegrates by the number of uranium atoms you have, so (3.18 x 10^-27) x (1.276 x 10^24) and you'll get a final answer of 4.06 x 10^-4kg, or 406g.

Hope that helps!

**amish123**)Right first work out the total mass change if one uranium atom disintegrates,

so 390.406 + 1.675 - (149.357 + 239.056 + 2x1.675)

and you get a mass difference of 3.18 x 10^-27kg.

Now, for a uranium atom, there are 235g per mole. So if you want to find the mass change for 0.5kg, you do 500/235 x (6.0 x 10^23) and this will give you the number of uranium atoms in 0.5kg. You'll get an answer of 1.276 x 10^24 atoms.

Then you just multiply the mass difference if one uranium atom disintegrates by the number of uranium atoms you have, so (3.18 x 10^-27) x (1.276 x 10^24) and you'll get a final answer of 4.06 x 10^-4kg, or 406g.

Hope that helps!

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#596

Hi, is anyone here who does the applied option and has done the June 10 paper? On question 4 a) the applied section, it asks you to estimate the work done by the air, In the AQA booklet it says that you need to find out the area under the graph between the 2 volumes, so in this case it will be between the volume of 1.2 and 3.0, and pressure between 0 and 3.6, but if you do that according to the marks scheme you will only get 2 marks, you get 3 marks if you do it between the pressure if 1.0 to 3.6!! I asked my teacher he has no idea why the mark scheme is inconsistent to book so can anyone here help me out? Thanks

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#597

For those doing Astrophysics,

I'm struggling with magnitudes! Please can someone give me a simplified definition for both apparent and absolute? And applications on how it can be tested on you in the exam?

It's very difficult, AQA's notes are appalling! (too much unnecessary information!)

I'm struggling with magnitudes! Please can someone give me a simplified definition for both apparent and absolute? And applications on how it can be tested on you in the exam?

It's very difficult, AQA's notes are appalling! (too much unnecessary information!)

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#598

(Original post by

Me

How are you feeling about it? I hope they don't ask too much about the ear, that's my worst topic, and some of the lens calculations are a pretty confusing, but all in all I don't find it too hard compared to unit 5 section A and 4. Helps that the grade boundaries have been ridiculously low the last few years too!

**jonnyb123**)Me

How are you feeling about it? I hope they don't ask too much about the ear, that's my worst topic, and some of the lens calculations are a pretty confusing, but all in all I don't find it too hard compared to unit 5 section A and 4. Helps that the grade boundaries have been ridiculously low the last few years too!

For the eye calculations just say that for example a diverging lens for a myopic eye will make a point at infinity appear to be at the near point of that eye (eg 10m) so the eye can see it, so make u=infinity and v=-10 ? Does that help?

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#599

(Original post by

For those doing Astrophysics,

I'm struggling with magnitudes! Please can someone give me a simplified definition for both apparent and absolute? And applications on how it can be tested on you in the exam?

It's very difficult,

**franko06**)For those doing Astrophysics,

I'm struggling with magnitudes! Please can someone give me a simplified definition for both apparent and absolute? And applications on how it can be tested on you in the exam?

It's very difficult,

**AQA's notes are appalling! (too much unnecessary information!)**apparent magnitude is how bright the stars appear to be

absolute magnitude is an objective measurement of brightness, it is the how bright the stars would appear if they were at 10 parsecs away from the earth

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#600

(Original post by

I think it'd be much lower than 40 degrees Celsius because since it's not insulated... it will reach thermal equilibrium with the room... so the water will reach room temperature (or a little bit above it), but definitely lower than 40 degrees I would have thought.

**posthumus**)I think it'd be much lower than 40 degrees Celsius because since it's not insulated... it will reach thermal equilibrium with the room... so the water will reach room temperature (or a little bit above it), but definitely lower than 40 degrees I would have thought.

(Original post by

It would be lower as your calculations take no account for heat loss to the environment

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**laser174572**)It would be lower as your calculations take no account for heat loss to the environment

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(Original post by

Wouldn't it be higher, because the average temperature of the ice/water throughout the experiment is below the room temperature, and so there will be a net flow of heat into the water/ice?

**SpiggyTopes**)Wouldn't it be higher, because the average temperature of the ice/water throughout the experiment is below the room temperature, and so there will be a net flow of heat into the water/ice?

Although I kind of understand that since the temperature of the water is mostly under 25 degrees, heat will be transferred from surroundings to water as well as from the heater, but it doesn't explain why thermal equilibrium wouldn't be at around 25 degrees. Once the water exceeds 25 degrees, wouldn't it just start losing heat instead of gaining temperature?

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