# Need Urgent Physics Help PLEASE!

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Thread starter 7 months ago
#1
Many Thanks to anyone in advance!!

Question Paper:
https://filestore.aqa.org.uk/sample-...2-QP-JUN17.PDF
Mark Scheme:
https://filestore.aqa.org.uk/sample-...W-MS-JUN17.PDF

Also I don't know how to do any of these, which is getting on my mind...
Q: 10,11,12,13,19, 23, 31

For question 6.3, how do you know the Gravitational Potential is lower??
Last edited by 234UncleBob; 7 months ago
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Thread starter 7 months ago
#2
(Original post by 234UncleBob)
Many Thanks to anyone in advance!!

Question Paper:
https://filestore.aqa.org.uk/sample-...2-QP-JUN17.PDF
Mark Scheme:
https://filestore.aqa.org.uk/sample-...W-MS-JUN17.PDF

Also I don't know how to do any of these, which is getting on my mind...
Q: 10,11,12,13,19, 23, 31

For question 6.3, how do you know the Gravitational Potential is lower??
..
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reply
7 months ago
#3
Hi, I will give a few clues to help out!
10: moles = mass * molar mass ... Not sure what else to say really...
11: An ideal gas has no intermolecular forces, so it internal energy is purely kinetic, this can be calculated with formulas!
12. M/r^2 must be the same for both... and define the distance from Pluto to d as X...
13. F = - GMm / R^2 you know F and R at the closest point, so find -GMm then divide by R^2 where R is furthest away.
19. I don't know... Haven't learnt about capacitors yet sorry!
23. Again, not looked at that yet...
31. Use Nuclear radius equation in formula book I think!

with regards to 6.3, I also find potential a tricky subject. Here is a great link I find helpful! http://www.physbot.co.uk/gravity-fie...otentials.html
I think this quote comes close to answering your question: potential well'; you need energy input into the system to escape this well. (Hence why as we get into a stronger field, potential decreases) hope that makes some sense. I can always give worked solutions if not!)
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7 months ago
#4
(Original post by tande33)
...
10: moles = mass * molar mass ... Not sure what else to say really...
.....
I think you mean moles = mass/molar mass instead of moles = mass * molar mass .
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7 months ago
#5
(Original post by 234UncleBob)
Many Thanks to anyone in advance!!

Question Paper:
https://filestore.aqa.org.uk/sample-...2-QP-JUN17.PDF
Mark Scheme:
https://filestore.aqa.org.uk/sample-...W-MS-JUN17.PDF

Also I don't know how to do any of these, which is getting on my mind...
Q: 10,11,12,13,19, 23, 31

For question 6.3, how do you know the Gravitational Potential is lower??
I understand you need help urgently due to exam but it does not give you the excuse of not following guidelines.

Q11
Find the mass of one CO2 molecule.
The number of molecules is the mass of the given gas divided by the mass of one CO2 molecule.
Some simple logical thinking can help to solve such question.

Q12
In thermal physics, you should remember that in an ideal gas, the average KE of one particle is directly proportional to the temperature and this formula is found in the data booklet or datasheet.
In 1 mole of an ideal gas, there are 6.02 × 1023 of particles.

Q19
How do you compute the new energy stored when the p.d. is doubled?
The formula for the energy stored in a capacitor is given in the datasheet.
Recommend that you re-study your capacitor notes.

Q23
First find the rms current from the mean power and rms voltage, then compute the peak current from the simple equation that relate the rms current Irms and peak current I0. You can find the simple equation from the datasheet.
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Thread starter 7 months ago
#6
(Original post by tande33)
Hi, I will give a few clues to help out!
10: moles = mass * molar mass ... Not sure what else to say really...
11: An ideal gas has no intermolecular forces, so it internal energy is purely kinetic, this can be calculated with formulas!
12. M/r^2 must be the same for both... and define the distance from Pluto to d as X...
13. F = - GMm / R^2 you know F and R at the closest point, so find -GMm then divide by R^2 where R is furthest away.
19. I don't know... Haven't learnt about capacitors yet sorry!
23. Again, not looked at that yet...
31. Use Nuclear radius equation in formula book I think!

with regards to 6.3, I also find potential a tricky subject. Here is a great link I find helpful! http://www.physbot.co.uk/gravity-fie...otentials.html
I think this quote comes close to answering your question: potential well'; you need energy input into the system to escape this well. (Hence why as we get into a stronger field, potential decreases) hope that makes some sense. I can always give worked solutions if not!)
(Original post by Eimmanuel)
Q11
Find the mass of one CO2 molecule.
The number of molecules is the mass of the given gas divided by the mass of one CO2 molecule.
Some simple logical thinking can help to solve such question.

Q12
In thermal physics, you should remember that in an ideal gas, the average KE of one particle is directly proportional to the temperature and this formula is found in the data booklet or datasheet.
In 1 mole of an ideal gas, there are 6.02 × 1023 of particles.

Q19
How do you compute the new energy stored when the p.d. is doubled?
The formula for the energy stored in a capacitor is given in the datasheet.
Recommend that you re-study your capacitor notes.

Q23
First find the rms current from the mean power and rms voltage, then compute the peak current from the simple equation that relate the rms current Irms and peak current I0. You can find the simple equation from the datasheet.
Many thanks for the Help from both of you and the guidelines, I now know how to properly start my next thread.

I always find it better to learn the method, rather than just to copy an answer, because then you can apply it to any context, so these clues/ worked answers are really helpful, so Many Thanks!!

The was helpful thanks!! With 6.3 I think I get it, the ball accelerates, meaning more Ek, so more energy to input into the system to escape it, but the ball is entering it, so surely it would lose energy; a worked example would help me understand it more, if you have time.

My attempts:
10) I get it know, 2.2 x 1000 to get into grams, 12+16x2 = 44 which is the Mr. 2200/44=50 x Avogadro = C
11) KE = 3/2KT, Plug in numbers to get D,
GMm / r^2 = F
13) F=GMm/r^2 = 1/(2.1x10^11)^2 = 2.26x10^-23 / (2.5x10^11)^2 = 3.628x10^-46 , which is waay off
where am I going wrong?

19) I used the equation Q=CV first, Q is proportional to V, so 2v = 2Q, So Ek = 0.5 Q^2/C turns into 0.5 = 2Q^2 /C. I've tried using the other Ek equations like Ek = 0.5 CV^2 but then i don't get Q and C in the question, i don't know how to get the 3/2 fraction.

23) Ah, okay, I did it by converting everything into Peak straight away and then use it to find the Peak current, must have made a mistake when converting. Just double checking Mean Power is like saying RMS power?
Last edited by 234UncleBob; 7 months ago
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7 months ago
#7
(Original post by 234UncleBob)

My attempts:

19) I used the equation Q=CV first, Q is proportional to V, so 2v = 2Q, So Ek = 0.5 Q^2/C turns into 0.5 = 2Q^2 /C. I've tried using the other Ek equations like Ek = 0.5 CV^2 but then i don't get Q and C in the question, i don't know how to get the 3/2 fraction.
Not sure how you get 0.5 = 2Q^2 /C.
The amount of charge is doubled, so the new energy stored is Enew =(0.5(2Q)^2)/C. Pay attention to the brackets.

(Original post by 234UncleBob)
23) Ah, okay, I did it by converting everything into Peak straight away and then use it to find the Peak current, must have made a mistake when converting. Just double checking Mean Power is like saying RMS power?
Mean power is computed from RMS current and voltage, so you can say is RMS power.
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Thread starter 7 months ago
#8
(Original post by Eimmanuel)
Not sure how you get 0.5 = 2Q^2 /C.
The amount of charge is doubled, so the new energy stored is Enew =(0.5(2Q)^2)/C. Pay attention to the brackets.
That's what I meant: Ek = 0.5 2Q^2 / C.
This Equation doesn't get me 3/2 though.

Is the first step what i did to use CV=Q correct? And should i be think about for the next step?
Last edited by 234UncleBob; 7 months ago
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7 months ago
#9
(Original post by 234UncleBob)
That's what I meant: Ek = 0.5 2Q^2 / C.
This Equation doesn't get me 3/2 though.

Is the first step what i did to use CV=Q correct? And should i be think about for the next step?
I am not sure why you keep writing Ek.

(2Q)^2 IS NOT EQUAL 2Q^2

Read the question a few times to know what to compute.

Spoiler:
Show

What is the change in the energy stored by the capacitor?
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Thread starter 7 months ago
#10
(Original post by Eimmanuel)
I am not sure why you keep writing Ek.

(2Q)^2 IS NOT EQUAL 2Q^2

Read the question a few times to know what to compute.

Spoiler:
Show

What is the change in the energy stored by the capacitor?
Silly me, I meant energy of the Capacitor, not Kinteic energy, silly me.

I need to work out the CHANGE in ENERGY.

E = 0.5 Q^2/c at first
Double p.d, so E = 0.5CV^2 = 2 squared is 4 x 0.5 = 2
So 2 - 0.5 = 1.5 which is 3/2 which is 3/2Q^2/C.
Thank You, thank you for persisting and putting up with me. I don't know why i was making so many mistakes and why it took me longer than usual to understand that, maybe i'm tired or i am actually stressed with out realising it, but nevertheless many many THANKS!!!
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7 months ago
#11
(Original post by 234UncleBob)

My attempts:

13) F=GMm/r^2 = 1/(2.1x10^11)^2 = 2.26x10^-23 / (2.5x10^11)^2 = 3.628x10^-46 , which is waay off
where am I going wrong?

GMm is NOT EQUAL TO 1.
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Thread starter 7 months ago
#12
(Original post by Eimmanuel)
GMm is NOT EQUAL TO 1.
I meant to put

13) GMm = F/r^2 = 1/(2.1x10^11)^2 = 2.26x10^-23 / (2.5x10^11)^2 = 3.628x10^-46 , which is waay off
where am I going wrong?
0
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7 months ago
#13
(Original post by 234UncleBob)
I meant to put

13) GMm = F/r^2 = 1/(2.1x10^11)^2 = 2.26x10^-23 / (2.5x10^11)^2 = 3.628x10^-46 , which is waay off
where am I going wrong?
Again,
F= GMm/r^2 => GMm = F*r^2 NOT GMm = F/r^2
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Thread starter 7 months ago
#14
(Original post by Eimmanuel)
Again,
F= GMm/r^2 => GMm = F*r^2 NOT GMm = F/r^2
Thanks
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