This discussion is now closed.

Check out other Related discussions

AQA PA10 Physics A Synoptic (June 16th)

mlbbrown

what did people get for the amplitude of the last question ?

I think something like 2.2x10^-4 metres, might've been a different question though.

Yes. That's correct.

The elastic collision question was a real bugger because they gave the speed of the nucleus instead of the neutron. So I guess most people who were stuck on the question transposed the two and got incorrect answers. I did that but I found out where I went wrong.

Reply 41

persia90

Fornicis

Ah crap, you're right lol, I didn't bother looking back, surely both should give the same answer though since your working with formulae which you have all the needed answers too, oh and ya, I got 50000A as well :P

well, if you use P = (1/2QV)/t = (1/2x5x3.2x10^6)/0.1 = 8x10^7 W
I use P = VI = 3.2x10^6 x 50000 = 1.6x10^11 W. I use P=VI because I think it asks u for the power of the lighting. Surely the thunder cloud does not transfer all of it stored energy in the single lighting. I am so confused now.
How about the sun cream question? I don't know how to do that either. I said the claim is not reasonable because the sun cream only stops 80% of the UVB so you are still subject to UVA which is more dangerous :frown:

that reason sounds stupid I know :p:

Reply 42

drumbum99

me, myself and I

I certainly did give it my best shot...oh well if i can't cope with this then maybe i am not oxford material :frown:

always doubted myself when i did get the offer anyway...yeah i used 3/2 kT, I'm positive that's correct :smile:

there's no other way lol.
Also why are people talking about converting into Kelvins, it is temperate change, delta T so there is no need..?!?!! right???

you can also use 2RT/3N not sure if the 2 and 3 are the right way round. Anyway it gives you the same value.
And yer delta T would be the same in Kelvin or degrees o.0

Reply 43

sk88

12

me, myself and I

I certainly did give it my best shot...oh well if i can't cope with this then maybe i am not oxford material :frown:

always doubted myself when i did get the offer anyway...yeah i used 3/2 kT, I'm positive that's correct :smile:

there's no other way lol.
Also why are people talking about converting into Kelvins, it is temperate change, delta T so there is no need..?!?!! right???

temperature had to be converted on the speed of sound question, not on the cloud one.

Reply 44

TheTallOne

18

persia90

well, if you use P = (1/2QV)/t = (1/2x5x3.2x10^6)/0.1 = 8x10^7 W

The time was 0.1ms.

So it should be 8x10^10W

But I changed my mind on that question and used P=IV and got double that.

Reply 45

Lozmaster

6

onlinebacon

For the last one I thought that when the machine vibrated the little 1.6N thing fell at like9.81 m/s, so I put 9.81 x 60ms to get 0.14 or something and multiplied that by 17 and got 2.5 :biggrin:

For that question, I used the fact that V is proportional to the Force on the ...whatever that sensor was called. From the graph, and information you could tell that the force was 1.6N with no vibrations, and it increased by 0.4N at max.
then found the mass of the ..sensor thing, (1.6/9.81), used (change in )F=m (change in) a, giving 0.4/(1.6/9.81) and voila, 2.5m/s^-2 (Rounded anyway :P).

Reply 46

onlinebacon

12

Lozmaster

For that question, I used the fact that V is proportional to the Force on the ...whatever that sensor was called. From the graph, and information you could tell that the force was 1.6N with no vibrations, and it increased by 0.4N at max.
then found the mass of the ..sensor thing, (1.6/9.81), used (change in )F=m (change in) a, giving 0.4/(1.6/9.81) and voila, 2.5m/s^-2 (Rounded anyway :P).

Cool! That was a good way to do it. I thought the paper would be a lot harder than it was... it wasn't easy but it didn't kill me so I'm happy.

Reply 47

drumbum99

Lozmaster

For that question, I used the fact that V is proportional to the Force on the ...whatever that sensor was called. From the graph, and information you could tell that the force was 1.6N with no vibrations, and it increased by 0.4N at max.
then found the mass of the ..sensor thing, (1.6/9.81), used (change in )F=m (change in) a, giving 0.4/(1.6/9.81) and voila, 2.5m/s^-2 (Rounded anyway :P).

yay i could do that one, just ******* up the rest

Reply 48

Fornicis

5

persia90

well, if you use P = (1/2QV)/t = (1/2x5x3.2x10^6)/0.1 = 8x10^7 W
I use P = VI = 3.2x10^6 x 50000 = 1.6x10^11 W. I use P=VI because I think it asks u for the power of the lighting. Surely the thunder cloud does not transfer all of it stored energy in the single lighting. I am so confused now.
How about the sun cream question? I don't know how to do that either. I said the claim is not reasonable because the sun cream only stops 80% of the UVB so you are still subject to UVA which is more dangerous :frown:

that reason sounds stupid I know :p:

I said the opposite, since it was asking whether you wouldn't burn for 5 hours, and since without it you would have 100% transmitted, lets say you burn after an hour without, then surely with it you get 20% an hour, which would result in 100% after 5 hours resulting in burns after 5 hours, which means that you can stay in the sun for 5 times as long 5/1 = 5 :P, thats what I saw it as anyway, also sorry for the wall of text :P

Reply 49

persia90

TheTallOne

The time was 0.1ms.

So it should be 8x10^10W

But I changed my mind on that question and used P=IV and got double that.

Yeah, you are right. I use P = VI and get 1.6x10^11. What is the resistance though? I got 0.1 ohm :rolleyes:

Reply 50

mlbbrown

TheTallOne

Yes. That's correct.

I like this answer.

Reply 51

me, myself and I

1

I agree with the 1/2 QV, I spent ages in the exam deliberating over whether or not it was. I used E=QV but from hearing what people have said about one charged plate, I think you're all correct. well done for figuring it out though guys!! :smile:

I reckon we'll get As lol...68% grade boundary!

Reply 52

astarael

0.1ohms sounds about right.

Reply 53

TheTallOne

18

Fornicis

I said the opposite, since it was asking whether you wouldn't burn for 5 hours, and since without it you would have 100% transmitted, lets say you burn after an hour without, then surely with it you get 20% an hour, which would result in 100% after 5 hours resulting in burns after 5 hours, which means that you can stay in the sun for 5 times as long 5/1 = 5 :P, thats what I saw it as anyway, also sorry for the wall of text :P

Hmm.. I barely mentioned about the one that give you sunburn. I talked about the UVA which doesn't get blocked = cancer and death.

Anyway, I put 96% is absorbed.

Reply 54

me, myself and I

1

persia90

Yeah, you are right. I use P = VI and get 1.6x10^11. What is the resistance though? I got 0.1 ohm :rolleyes:

Aye 0.1 ohm, I hope that's right!
I'm not confident with anything I've written in that exam haha but fingers crossed........!!!

Reply 55

me, myself and I

1

TheTallOne

Hmm.. I barely mentioned about the one that give you sunburn. I talked about the UVA which doesn't get blocked = cancer and death.

Anyway, I put 96% is absorbed.

Me too

Reply 56

Fornicis

5

me, myself and I

I agree with the 1/2 QV, I spent ages in the exam deliberating over whether or not it was. I used E=QV but from hearing what people have said about one charged plate, I think you're all correct. well done for figuring it out though guys!! :smile:

I reckon we'll get As lol...68% grade boundary!

Pfft I was the one going on about charged plates so go ME! And ya, 0.1ohms is what I got, just think though guys, we've already done it, theres nothing we can do about it, nice to discuss it though, I'm the only one in my physics class so no-one to talk to :P

Reply 57

persia90

me, myself and I

Me too

How do you work out that Q, I left that one blank :frown:

Reply 58

laurie344

For the elastic collision one I manage to get momentum before = momentum after but only by calculating the speed of the nitrogen after the collision. How were you supposed to do it?