# OCR Physics A G485 - Frontiers of Physics - 18th June 2015Watch

3 years ago
#441
How much and what sort of revision are people doing for this exam?

i.e. how many PP or how many notes, how much time a day?
3 years ago
#442
Can someone please help me with question 1dii in paper June 2010 G485. In the question it mentions that they are connected in parallel so that means both of the capacitors have different charges but why is the voltage calculated by dividing the charge of the capacitor we calculated in the above question by the total capacitance?
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3 years ago
#443
(Original post by sagar448)
Can someone please help me with question 1dii in paper June 2010 G485. In the question it mentions that they are connected in parallel so that means both of the capacitors have different charges but why is the voltage calculated by dividing the charge of the capacitor we calculated in the above question by the total capacitance?
You are right, but total charge is conserved. So the charge you calculated in b)i) is the total charge, but each capacitor gets different amounts of it, when S2 is closed. We are asked to calculate the final p.d across the capacitors so we need total charge/sum of capacitors.
0
3 years ago
#444
(Original post by sagar448)
Can someone please help me with question 1dii in paper June 2010 G485. In the question it mentions that they are connected in parallel so that means both of the capacitors have different charges but why is the voltage calculated by dividing the charge of the capacitor we calculated in the above question by the total capacitance?
I think it's because as soon as switch 2 is closed the pd across the 1.5 microF capacitor is zero. Therefore it's charge is equal to zero using Q=CV. The question states to find the pd across both capacitors, so you do total charge, which is 28.35+0, divided by the total capacitance, which is 1.5+4.5 microF.
0
3 years ago
#445
(Original post by randlemcmurphy)
You are right, but total charge is conserved. So the charge you calculated in b)i) is the total charge, but each capacitor gets different amounts of it, when S2 is closed. We are asked to calculate the final p.d across the capacitors so we need total charge/sum of capacitors.
Oh that makes sense.
0
3 years ago
#446
(Original post by randlemcmurphy)
You are right, but total charge is conserved. So the charge you calculated in b)i) is the total charge, but each capacitor gets different amounts of it, when S2 is closed. We are asked to calculate the final p.d across the capacitors so we need total charge/sum of capacitors.
OHHH how did I miss that? yes!! The charge that I calculated gets distributed to the capacitor so it is the total charge I see, yes. And to think I've done this paper before XD. Thanks bruh. :3
0
3 years ago
#447
(Original post by sagar448)
If you don't mind I would like to know what question that is and from what past paper? Thanks
Q1 June 2011
http://www.ocr.org.uk/Images/62267-q...of-physics.pdf
1
3 years ago
#448

How much and what sort of revision are people doing for this exam?

i.e. how many PP or how many notes, how much time a day?
For this exam only? Maybe an average of 1.5 hours a day. Sometimes 0, sometimes about 3.
0
3 years ago
#449
(Original post by randlemcmurphy)
You are right, but total charge is conserved. So the charge you calculated in b)i) is the total charge, but each capacitor gets different amounts of it, when S2 is closed. We are asked to calculate the final p.d across the capacitors so we need total charge/sum of capacitors.
(Original post by a123a)
I think it's because as soon as switch 2 is closed the pd across the 1.5 microF capacitor is zero. Therefore it's charge is equal to zero using Q=CV. The question states to find the pd across both capacitors, so you do total charge, which is 28.35+0, divided by the total capacitance, which is 1.5+4.5 microF.
For d.i) on that same question, why is the total capacitance just the sum of the capacitors? I'd have thought that because the p.d.s across them are different at t=0 that means they're in series and therefore you have to use the inverse addition formula.
0
3 years ago
#450
(Original post by Elcor)
For d.i) on that same question, why is the total capacitance just the sum of the capacitors? I'd have thought that because the p.d.s across them are different at t=0 that means they're in series and therefore you have to use the inverse addition formula.
Because when both switches are closed the capacitors are in parallel and so the capacitance just add normally. It tells you in the question, calculate the total capacitance is switch S2 is closed.

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3 years ago
#451
(Original post by sagar448)
Because when both switches are closed the capacitors are in parallel and so the capacitance just add normally. It tells you in the question, calculate the total capacitance is switch S2 is closed.

I agree that they're in parallel when both switches are open as they both have the same p.d. (0V). When S2 is closed though one has 6.3V and one has 0V. Different p.d.s so they're in series so use inverse addition formula. Where am I wrong?
0
3 years ago
#452
(Original post by Elcor)
I agree that they're in parallel when both switches are open as they both have the same p.d. (0V). When S2 is closed though one has 6.3V and one has 0V. Different p.d.s so they're in series so use inverse addition formula. Where am I wrong?
Ok so before anything is closed both are uncharged. Then switch S1 closes and so the capacitor 4.5 has a 6.3 pd across it. It specifically states that the capacitors are in parallel and in parallel the capacitance just adds so we get a total of 6uF. The Switch S2 is still open. The charge on the 4.5uF capacitor is 28.35. Now the S2 is closed and S1 is open. We know that the charge on the 1.5uf capacitor was 0 before the switch was closed that means before the 4.5uf capacitor starts to discharge there is only 28.35 of charge available for BOTH capacitors. Therefore that is the total charge. And so to find the total pd across both the capacitors you add the capacitance (they are still in parallel) and you divide the charge by the capacitance to give you the total PD.

Hope that helps.
Ask again if you still don't understand. :3
0
3 years ago
#453
(Original post by Elcor)
For d.i) on that same question, why is the total capacitance just the sum of the capacitors? I'd have thought that because the p.d.s across them are different at t=0 that means they're in series and therefore you have to use the inverse addition formula.
Hmm. I get what you mean. I think you use the sum of the capacitors because in the question it states that another capacitor is added in parallel.
0
3 years ago
#454
Would someone explain when you use and what each variable means in

Z=Pc

Ir/Io=(z2-z1)^2/(z2+z1)^2

Dlamba/lamva=v/c

All on the very last page of formula book
3 years ago
#455

Would someone explain when you use and what each variable means in

Z=Pc

Ir/Io=(z2-z1)^2/(z2+z1)^2

Dlamba/lamva=v/c

All on the very last page of formula book
Z is the acoustic impedance of a substance, rho is its density and c is the speed of ultrasound in the substance.

Ir/Io is the ratio of the intensity of reflected ultrasound waves to the intensity of incident ultrasound waves.

The last equation is for Doppler shift, and I believe it only works for galaxies (someone please correct me if that's wrong), where v is the recessional speed of the galaxy, delta lambda is the change in wavelength of light coming off it and lambda is the original wavelength of the light.
1
3 years ago
#456

Would someone explain when you use and what each variable means in

Z=Pc

Ir/Io=(z2-z1)^2/(z2+z1)^2

Dlamba/lamva=v/c

All on the very last page of formula book
Someone call me out if I'm wrong!

Z is the acoustic impedance; P is the density of the medium, C is the speed of ultrasound in the medium

Delta Lamda is the Change in wavelength of a source that has had its wavelength changed due to motion (increased due to red shift for most)
Lamda is the original/non shifted wavelength, V is the recessional speed of a galaxy and c is the speed of light

Ir is in intensity reflected, Io is the incident intensity, Z are the acoustic impedances of different medium
1
3 years ago
#457
Can anyone expalin how you can use 3/2kt=1/2mv^2 to do the very last question on Jan 2012
3 years ago
#458

Can anyone expalin how you can use 3/2kt=1/2mv^2 to do the very last question on Jan 2012
From that equation we can see that T is directly proportional to v2, so we can say that T/v2 is a constant, we can therefore say that T1/2/v is a constant, you should be able to figure it out from there, by rearranging to form the ratio in terms of the temperature.
0
3 years ago
#459
(Original post by randlemcmurphy)
From that equation we can see that T is directly proportional to v2, so we can say that T/v2 is a constant, we can therefore say that T1/2/v is a constant, you should be able to figure it out from there, by rearranging to form the ratio in terms of the temperature.
Alright thanks that really helped
3 years ago
#460
(Original post by Elcor)
Z is the acoustic impedance of a substance, rho is its density and c is the speed of ultrasound in the substance.

Ir/Io is the ratio of the intensity of reflected ultrasound waves to the intensity of incident ultrasound waves.

The last equation is for Doppler shift, and I believe it only works for galaxies (someone please correct me if that's wrong), where v is the recessional speed of the galaxy, delta lambda is the change in wavelength of light coming off it and lambda is the original wavelength of the light.
I agree with this, but do you know if the Doppler shift equation can be used to determine the speed of stars in other galaxies as well as just galaxies?
0
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