You are Here: Home >< Physics

AQA Physics Unit 1 PHYA1 20th May 2013 Watch

1. (Original post by StalkeR47)
Are you guys still talking about intensity and frequency?
Yeah
2. For the oscilloscope question I wrote that the time base should be adjusted so that the time was 4ms per division instead of 2ms per division. Is that right? If so, how many marks would I get?
3. Anyone know for the conservation of energy if I've proved it but not wrote anything would I at least get 1 mark?
4. (Original post by 123456789012)
For the oscilloscope question I wrote that the time base should be adjusted so that the time was 4ms per division instead of 2ms per division. Is that right? If so, how many marks would I get?
It was 1ms per div not 2. If you counted the squares 10ms was at the 10th square
So it should have been 1ms to 2ms or 2.5ms
5. i really dont know why im still here, im not really finding out if i got things right,im only finding out that i might have got them wrong and reading better answers then mine, this is depressing! hahahhaa
i really dont know why im still here, im not really finding out if i got things right,im only finding out that i might have got them wrong and reading better answers then mine, this is depressing! hahahhaa
What questions were you worried about? Let's see if you got similar to mine
7. (Original post by Qari)
It was 1ms per div not 2. If you counted the squares 10ms was at the 10th square
So it should have been 1ms to 2ms or 2.5ms
it wanted 2 waves in 10 squares so if at 1ms/div there was 1 wave.... *three dots in triangle* 2ms/div will give you 2 waves..think that KO
8. (Original post by Qari)
It was 1ms per div not 2. If you counted the squares 10ms was at the 10th square
So it should have been 1ms to 2ms or 2.5ms
it had to be at least 2ms per division to fit two waveforms on? each wavelength was 10ms and there was only 10 squares.
9. (Original post by Qari)
It was 1ms per div not 2. If you counted the squares 10ms was at the 10th square
So it should have been 1ms to 2ms or 2.5ms
Oh no! Would I get even a single mark though? Say if I got around 60 raw marks, how many UMS do you reckon that would be?
10. (Original post by amar96)
it wanted 2 waves in 10 squares so if at 1ms there was 1 wave.... *three dots in triangle* 2ms/div will give you 2 waves..think that KO
Someone said it was 2ms and they did to 4ms and would they have been able to see the 2 waves completely?
11. (Original post by 123456789012)
Oh no! Would I get even a single mark though? Say if I got around 60 raw marks, how many UMS do you reckon that would be?
here you are check,
12. aaaaammmmamamam yaun yaun yaun I am really tired. damit, I needed to do more revision on sociology.
13. (Original post by StalkeR47)
aaaaammmmamamam yaun yaun yaun I am really tired. damit, I needed to do more revision on sociology.
Just realies my mate has that exam tomorrow
14. (Original post by Qari)
It was 1ms per div not 2. If you counted the squares 10ms was at the 10th square
So it should have been 1ms to 2ms or 2.5ms
it was 2ms per square because it wanted 2 cycles. each cycle was 10ms and therefore 2 cycles would be 20ms. and if we divide this by the number of horizontal squares (10) we get 2ms per square.
15. OMG STOP TALKING ABOUT this oscilloscope. you are making me fall asleep. you got what you got. you did what you could. and you cannot change it, can u? so shhhh and wait for the results day.
16. (Original post by dr.zoidberg)
it was 2ms per square because it wanted 2 cycles. each cycle was 10ms and therefore 2 cycles would be 20ms. and if we divide this by the number of horizontal squares (10) we get 2ms per square.
I'm saying Figure 2 not what we had to change
17. I swear the question said that there were 8 vertical divisions . So if 1 complete cycle was 10ms, 2 complete cycles would be 20ms, 20/8= 2.5ms per division?
18. (Original post by v1k2a3t4)
I swear the question said that there were 8 vertical divisions . So if 1 complete cycle was 10ms, 2 complete cycles would be 20ms, 20/8= 2.5ms per division?
Yes, but you are interested in the horizontal divisions for the time base, as the time base is on the horizontal axes.
19. (Original post by Kayakayakaya)
question 4, about the frequency being reduced, surely this has no effect unless the frequency is below the threshold frequency?
I think that lowering the frequency would mean less photon energy, hence KEmax is less because hf=KEmax+work function (constant). So the electrons have a lower kinetic energy.
20. (Original post by v1k2a3t4)
I swear the question said that there were 8 vertical divisions . So if 1 complete cycle was 10ms, 2 complete cycles would be 20ms, 20/8= 2.5ms per division?
This is what I got

Posted from TSR Mobile

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: May 14, 2014
Today on TSR

What should I do?

Am I doomed because I messed up my mocks?

Discussions on TSR

• Latest
• See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• Poll
Useful resources
Discussions on TSR

• Latest
• See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE