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# iGCSE EDEXCEL PHYSICS DISCUSSION 2012

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1. (Original post by rtrtrt)
That's still completely correct. So you got 33.33kW?
Oh good! Yeah I got 33.33kW
2. What was the question about calculating energy from a wind turbine? I can't remember it!!!!!!
3. Did anyone else find that this paper didn't test our knowledge that well?
4. (Original post by rtrtrt)
What was the question about calculating energy from a wind turbine? I can't remember it!!!!!!
It said on a windy day 78W of power is generated by the turbine.

Part a) asked you for the equation linking power, energy and time

and then part b) asked how much energy would be generated by the turbine in 10s

(or something to that effect)
5. (Original post by Coco406)
It said on a windy day 78W of power is generated by the turbine.

Part a) asked you for the equation linking power, energy and time

and then part b) asked how much energy would be generated by the turbine in 10s

(or something to that effect)
Yeah it was basically that. Were you meant to convert 24 hours into seconds and then work out the power from there, because that is what I did
6. (Original post by l-g)
Yeah it was basically that. Were you meant to convert 24 hours into seconds and then work out the power from there, because that is what I did
I did that- 0.009J right?

Everyone in school laughed at me for that! But people are saying it's right
7. (Original post by AsUsualImStuck)
Yes, obviously I know that ! I do dual award (but sit the three papers, so 3 x 120 marks per paper is 360!!)

So my total that I have to achieve is ???/360 to get BB ...
So I am asking, what do I have to get out of 360 to get BB?
195 lol
8. (Original post by Coco406)
I don't think it was very clear- it just said how would it compare?

What did you write for it?
Well
Terminal velocity during free fall occurs when the weight of the object is being balanced by air resistance.
The weight in both is the same.
Therefore the speed it travels depends on the time taken for it to reach terminal velocity. The more the time taken the more time the unbalanced force acts upon it
and therefore it keeps on accelerating for longer and therefore the more the speed.
Air resistance depends on the speed (which is what is being questioned), the density of the medium (which remains the same) and the surface area exposed to collisions with the air particles.
When 3 parachutes are working the more the surface area, therefore the more the air resistance, and the less the time it takes for air resistance to balance its weight. Therefore it accelerates for less time and therefore as V=at the speed during terminal velocity is less.
I just wrote "2 parachutes mean less SA, therefore less air resistance, therefore for air resistance to balance weight speed must be greater, as air resistance increases with speed"
9. Energy transferred = Power x time

So i did 78 x 10 = 780J
10. (Original post by StUdEnTIGCSE)
Well
Terminal velocity during free fall occurs when the weight of the object is being balanced by air resistance.
The weight in both is the same.
Therefore the speed it travels depends on the time taken for it to reach terminal velocity. The more the time taken the more time the unbalanced force acts upon it
and therefore it keeps on accelerating for longer and therefore the more the speed.
Air resistance depends on the speed (which is what is being questioned), the density of the medium (which remains the same) and the surface area exposed to collisions with the air particles.
When 3 parachutes are working the more the surface area, therefore the more the air resistance, and the less the time it takes for air resistance to balance its weight. Therefore it accelerates for less time and therefore as V=at the speed during terminal velocity is less.
I just wrote "2 parachutes mean less SA, therefore less air resistance, therefore for air resistance to balance weight speed must be greater, as air resistance increases with speed"
That makes perfect sense! I wrote that because drag force has decrease, it takes longer to reach terminal velocity (but then I think I negated any marks I may have got by saying that the terminal velocity is the same in both cases silly silly silly me)
11. Btw, you know the coin question did the area of the coin stay the same and you had to x the weight by 8 so the equation for pressure of the coin was 0.288 divided by 0.0013 ?
12. (Original post by AM.95)
Energy transferred = Power x time

So i did 78 x 10 = 780J
I guess it depends on your take on the question
Because they had asked you for the equation in the question above, and because the question was 3 marks on its own, I thought that was too simple

78W in one day right? So if you just multiply it by 10, then that would be the power in 10 days...
13. (Original post by AM.95)
Btw, you know the coin question did the area of the coin stay the same and yo had to x the weight by 8 so the equation for pressure of the coin was 0.288 divided by 0.0013 ?
The area has to be either 0.0013 or 0.0013/2.
14. (Original post by Coco406)
What did everyone write for the terminal velocity question and the stupid smoke detector/radon question?
for the alpha particle gas Q about why it was had new health risks in the gas was because one the smoke alarm was a solid so alpha particles didn't have as much kinetic energy as the particles in the Radon gas? and can therefore travel in the human air that is breathed in? I have no idea of that's right or not but I just made a logical guess.. hope I got the marks haha

This was posted from The Student Room's iPhone/iPad App
15. Can remember part a) Energy transferred = Power x time, was b just 780J?
16. (Original post by rtrtrt)
Can remember part a) Energy transferred = Power x time, was b just 780J?
I think it was 0.009J
17. For the coin pressure calculation what were your answers? 200 sumthing? And what was the area? 0.0013 or half that?
18. (Original post by Coco406)
I think it was 0.009J
Its actually 0.009W for 10s.

Energy must be 0.009*10=0.09J?

I just put 780J anyways lol, but that is definetly wrong, it will most certainly be 0.009 or 0.09.
19. (Original post by Relaxedexams)
For the coin pressure calculation what were your answers? 200 sumthing? And what was the area? 0.0013 or half that?
Was the question not just working out the pressure??
20. (Original post by Coco406)
I did that- 0.009J right?

Everyone in school laughed at me for that! But people are saying it's right
I think that was my answer too :P

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