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PHYA5 ~ 20th June 2013 ~ A2 Physics

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Original post by Pinkhead
A) p1v1=p2v2

(1.20 x 10^-4 +x)(110000) = (1.20 x 10^-4)(135000)

Solve for x.

B) 110000(1.2 x10^-4 +2.7x10-5 - x) = (1.20 x 10^-4 - x)(141000)
Where x is the difference in volume due to the powder. Solve for x and you should get 2.32 x10^-5.
Then it's a simple case of mass/volume =density.


hey.. i dont understand why.. you multiply 135kp with 1.20*-4
i thought you wuld;
(1.20 x 10^-4 +x)(135000) = (1.20 x 10^-4)(110000)
also i don't get the correct answer of 2.32 i keep gettting the wrong one..
(edited 10 years ago)
Reply 881
Original post by jarasta
Input power is the power you get from the fuel, which is the calorific value * rate of fuel flow
Indicated work per cycle is the work done per cycle, which is just the area of the loop. What the enging should give you ideally, you really wont have your output=the indicated workdone usually though.
Indicated power is the power produced by the whole cycle, the formula for this is on the formula sheet, its something like. Area of the loop*number of cylinders*rotations per second


Thanks, but I know that. What I'm asking is why each two revolution perform 350J of work when the cycle is just for one revolution and the area is 350J?

Here's the question again. Any help would be greatly appreciated.

Screen Shot 2013-06-16 at 15.22.55.png
Screen Shot 2013-06-16 at 15.22.59.png

Okay, so this is quite a simple question really for applied physics. The one thing I don't get is why for the final question, you are supposed to halve it because 'each two revolutions produce 350J'. How does the area signify two revolutions, and why does the indicated power use two revolutions instead of one?
Reply 882
By the way, here's a link to the past specification for people looking for more revision resources. Very useful! (although I can't vouch for their relevance for Astro, Turning points or medical, the spec seems is very similar for applied and nuclear).
Reply 883
why is it that on the moon, all gas molecules released on its surface would eventually escape?

I know that gas molecules move randomly and you will always get some that have sufficient velocity (kinetic energy) to escape the moon, but how exactly do these molecules gain energy? do they just keep bouncing around until they have enough kinetic energy to escape?

but how would that explain the idea that energy kinetic energy must be conversed for ideal gas?

If there is a fixed amount of gas on the surface of the moon, and some happen to already have enough velocity to escaped the moon completely, there would be a loss of kinetic energy from that original cluster of gas, so hows does every gas molecule eventually gain enough energy (seemingly from nowhere) that enables them all to have enough velocity to escape the moon?
(edited 10 years ago)
Reply 884
Original post by mrwho
Thanks, but I know that. What I'm asking is why each two revolution perform 350J of work when the cycle is just for one revolution and the area is 350J?

Here's the question again. Any help would be greatly appreciated.

Screen Shot 2013-06-16 at 15.22.55.png
Screen Shot 2013-06-16 at 15.22.59.png

Okay, so this is quite a simple question really for applied physics. The one thing I don't get is why for the final question, you are supposed to halve it because 'each two revolutions produce 350J'. How does the area signify two revolutions, and why does the indicated power use two revolutions instead of one?


There are 4 cylinders, but only two will be providing power on any single revolution (to put it another way, each cylinder provides power once every two revolutions... or a 'cycle' is two full revolutions)
Am I right in thinking, that for intensity of balmer series (hydrogen absorption lines) is dependent on how hot the star is? (i.e the number of hydrogen atoms at n = 2 level is dependent on temperature)

so is it right that the light photons coming out and hitting the hydrogen atoms, are the same in all stars?
Reply 886
Original post by The H
Can anyone doing turning points explain why there is a resolution limit in TEM microscopes?


There are two reasons, I think..

The 1st is that, due to the sample having a thickness, even though it is relatively small, the electrons in the beam are slowed down due to collisions with the sample molecules, and so there will be more diffraction and less detail.

2nd is due to Lens aberrations - Basically, the magnetic field may focus several electrons from a given point to various positions on the screen, rather than the same position, which would make the image more blurry. This can happen because the electrons may have slightly different speeds (I'd guess because some electrons will have collided with more of the sample's molecules than others, so will have lost more kinetic energy).

Hope this helps!
Reply 887
Original post by fwook
There are two reasons, I think..

The 1st is that, due to the sample having a thickness, even though it is relatively small, the electrons in the beam are slowed down due to collisions with the sample molecules, and so there will be more diffraction and less detail.

2nd is due to Lens aberrations - Basically, the magnetic field may focus several electrons from a given point to various positions on the screen, rather than the same position, which would make the image more blurry. This can happen because the electrons may have slightly different speeds (I'd guess because some electrons will have collided with more of the sample's molecules than others, so will have lost more kinetic energy).

Hope this helps!


Nice, thanks!
Original post by TauMuon
"If AQA are generous with this one"

Bwahahahaha, don't make me laugh...

Every single AQA paper I've done so far has been nasty :'(


Yeah, AQA seem to hate us this summer...
Not got high hopes for Thursdays paper :frown:
Reply 889
How and when do we use dN/dt=Lamda*N?

See given example..

Why is dt replaced with lamda even though we have the value of t (10 minutes from question) and is dN always the initial count rate? dndt.png
For those doing astrophysics:
When do I use the negative sign in the Doppler equation? thanks!
Reply 891
Original post by MSI_10
How and when do we use dN/dt=Lamda*N?

See given example..

Why is dt replaced with lamda even though we have the value of t (10 minutes from question) and is dN always the initial count rate? dndt.png


In this exam I've never had to use any derivatives.
For the second part you use the equation C = C0 e-(lambda)t to determine a value for lambda.
For the third part you use equation A = N(lambda) to work out N, the number of undecayed nuclei, where A is the count rate.

I got 81 = 107e-600(lambda) for the second part, so lambda = 4.64x10-4

I got N = 107 / 4.64x10-4 = 230603 for the third part.
Reply 892
Original post by fuzzybear
Am I right in thinking, that for intensity of balmer series (hydrogen absorption lines) is dependent on how hot the star is? (i.e the number of hydrogen atoms at n = 2 level is dependent on temperature)

so is it right that the light photons coming out and hitting the hydrogen atoms, are the same in all stars?


The balmer series is observed from almost all starts but is most intense in the B and A spectral classes.

The link below shows a graph of the relative strengths of the absorption lines - a usual graph to have visualised I think!

http://www.star.ucl.ac.uk/~pac/stellar_lines.gif
Reply 893
Original post by Calm1
In this exam I've never had to use any derivatives.
For the second part you use the equation C = C0 e-(lambda)t to determine a value for lambda.
For the third part you use equation A = N(lambda) to work out N, the number of undecayed nuclei, where A is the count rate.

I got 81 = 107e-600(lambda) for the second part, so lambda = 4.64x10-4

I got N = 107 / 4.64x10-4 = 230603 for the third part.


yeah those are the correct answers.

Okay think I got it now; the derivative is equal to Activity right?
Reply 894
Original post by TeddyBasherz
For those doing astrophysics:
When do I use the negative sign in the Doppler equation? thanks!


For the Doppler equations just remember that it is the emitted frequency/wavelength subtract the laboratory frequency/wavelength all divided by the emitted frequency/wavelength

(Emitted - Laboratory observed) / Emitted

For the use of the negative sign just look at the formula sheet - the negative is used when dealing with the wavelengths. I'd assume the use of the negative has something to do with the relationship between frequency and wavelength? - as [delta frequency divided by frequency] is equal to [minus delta wavelength divided by wavelength].

A long winded reply that probably doesn't answer your question - sorry!
Reply 895
Original post by MSI_10
yeah those are the correct answers.

Okay think I got it now; the derivative is equal to Activity right?


I assume the count rate is equal to the activity (that's what I used in your question).

I've been doing both maths and physics and have only had to use derivatives and differential equations in maths? Unless I misunderstand what you're saying eheh
Reply 896
Original post by Calm1
I assume the count rate is equal to the activity (that's what I used in your question).

I've been doing both maths and physics and have only had to use derivatives and differential equations in maths? Unless I misunderstand what you're saying eheh


Count rate is proportional to activity, but not equal.
Reply 897
Anyone predict anything specific to come up that hasn't come up previously for a while?
Reply 898
so theres two long answer questions in unit 5......

and I was starting to think that this was easier than unit 4

unit 4 is definitely better than this, only one 6 markers, and the grade boundaries nicer as well
Original post by TauMuon
"If AQA are generous with this one"

Bwahahahaha, don't make me laugh...

Every single AQA paper I've done so far has been nasty :'(


Core 3 and Biology unit 5 were both pretty good, there's hope! :biggrin:

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