A2 Physics P4 Questions ONLY

[Namehere]

Hi, can someone help me allocate (and explain the positions) of each nucleon please? Q8(a)

Thank you very much in advance.

 

[periyasamy]

Hi, can someone help me allocate (and explain the positions) of each nucleon please? Q8(a)

Thank you very much in advance.

The reasoning is fe u must know it is in the max height(peak)
For h the ar is the smallest.
N the zr should be to the right because of higher ar than fe

 

[Champ101]

Fe-56 has the greatest value for binding energy! (ie, it is most stable)
H-2 is the lowest point
Zr-97 is around midway of the graph, probably more towards left.

http://qph.ec.quoracdn.net/main-qimg-f0e7b4a4ff7811ed8e0855900b2c4dd7 (check out the graph)

 

[A*(a*)]

anyone who can explain this? :'(

 

[boxfire1995]

anyone who can explain this? :'(
View attachment 42805

Since ideal gas's internal energy is directly proportional to the temperature, del u will be 0. Wince it is being compressed, the volume decreases and W=+. To make the U=O, Q=-
Heating of a solid without expansion. Heating=+, without expansion means del W is 0 thus overall u=+
As for ice, im a little unsure about it, but still, ill give it a shot. Q=+ as energy is required to weaken the bongs to convert it into water. Since ice is less dense than water, when it will melt, it the density will increase which means volume will decrease, thus W=+ and U is positive.
Just remember, when volume increases, W=-, when volume decreases, W=+.
When energy is supplied, Q=+, when Q=-, energy is given out. And U=del W + del Q.

 

[Namehere]

The reasoning is fe u must know it is in the max height(peak)
For h the ar is the smallest.
N the zr should be to the right because of higher ar than fe

Thank you!

 

[Namehere]

Fe-56 has the greatest value for binding energy! (ie, it is most stable)
H-2 is the lowest point
Zr-97 is around midway of the graph, probably more towards left.

http://qph.ec.quoracdn.net/main-qimg-f0e7b4a4ff7811ed8e0855900b2c4dd7 (check out the graph)

Thank you!

 

[Namehere]

anyone who can explain this? :'(
View attachment 42805

Would you mind (everyone) writing down the question paper and question number, just so anyone can look at it.

 

[Namehere]

Would like some explanations for the following questions:

s11_43: Q3(c), Q9(b, all), Q11(a) & (b), Q12(b)

 

[Champ101]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_4.pdf
Q5(b) 2 PART.Is there an easy method to find the area in this case?????

Damn! we have to consider each boxes? ? ? *_* why god??

 

[periyasamy]

Would like some explanations for the following questions:

s11_43: Q3(c), Q9(b, all), Q11(a) & (b), Q12(b)

Okay for the first question

 

[periyasamy]

for 9b,,,1)gain(memorising stuff)
(11)the attachment
(111)vout=12*(1000/1128)*4.5

 

[Namehere]

Okay for the first question

Thank you! What I don´t get is that you have calcualted the total capacitance to find the total charge? But on the question they ask for the charge one one plate of capacitor X. I simply don´t understand why the total capacitance and times by the p.d. would give the magnitude of the charge ON ONE PLATE OF CAPACITOR X.

 

[Namehere]

for 9b,,,1)gain(memorising stuff)
(11)the attachment
(111)vout=12*(1000/1128)*4.5

For (ii), there is no answer for V1? and for (iii) the ans of Vout you get is 47.9V instead of 0.128V

 

[periyasamy]

Thank you! What I don´t get is that you have calcualted the total capacitance to find the total charge? But on the question they ask for the charge one one plate of capacitor X. I simply don´t understand why the total capacitance and times by the p.d. would give the magnitude of the charge ON ONE PLATE OF CAPACITOR X.

I guess that how it works....The charge on one capacitor represent the charge of the whole thing....Conservation of charge i guess.Not really sure abt it...

 

[Champ101]

Would like some explanations for the following questions:

s11_43: Q3(c), Q9(b, all), Q11(a) & (b), Q12(b)

Hey for the amplifier question, Its an inverting amplifer.
The main formula to be used here is : gain = Vout/Vin = - Rf/Rin.
So, (i) is gain.
(ii), Vout= 0,
so, gain= 0/Vin (Vin = V2-V1)
from here you can say, V2=V1 (using maths, Vin = 0 or V2 - V1=0 or V1=V2)
Thus, we can use the line from strain guage as V1=V2.
therfore, we use potential divider formula to find the potential : i.e, V1 = 1000/(1000+125) * 4.5

(iii) again we use potential divider formula to find V2 as V1 remains constant, but the resistance is 128. Finally we use the gain formula!

I hope this kind of question doesn;t come :|

 

[Namehere]

Q12(b)
UV radiation has shorter wavelength and therefore it means that it has a greater frequency, resulting in higher attenuation.The maximum uninterrupted length when using UV would be less and therefore IR radiation is used ,which has a lower frequency and result in much less atenuation!

Thank you, and sorry for asking so many stupid questions but, why does a shorther wavelength (and therefore a higher frequency) result in higher attenuation?

 

[periyasamy]

For (ii), there is no answer for V1? and for (iii) the ans of Vout you get is 47.9V instead of 0.128V

Sorry for that...
The last part,,,iam sorry.Here's the way....First u get v2--->(1000/1128)*(4.5-0)-----u use the pd concept here.
Then to get v output use the formula given---->12*(4-3.99)
Sorry for the mIstakes. HAHA!!

 

[Namehere]

Hey for the amplifier question, Its an inverting amplifer.
The main formula to be used here is : gain = Vout/Vin = - Rf/Rin.
So, (i) is gain.
(ii), Vout= 0,
so, gain= 0/Vin (Vin = V2-V1)
from here you can say, V2=V1 (using maths, Vin = 0 or V2 - V1=0 or V1=V2)
Thus, we can use the line from strain guage as V1=V2.
therfore, we use potential divider formula to find the potential : i.e, V1 = 1000/(1000+125) * 4.5

(iii) again we use potential divider formula to find V2 as V1 remains constant, but the resistance is 128. Finally we use the gain formula!

I hope this kind of question doesn;t come :|

Thank you! Though I didn´t understand "so, gain= 0/Vin (Vin = V2-V1)" and the jump to using potential divider formula.

 

[Champ101]

I guess that how it works....The charge on one capacitor represent the charge of the whole thing....Conservation of charge i guess.Not really sure abt it...

actually when get the circuit in series, the capacitor store the same charge. So 10miifarad and 20milifarad (after solving) would have the same charge, but they have different Pd,