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# AQA Physics Unit 1 PHYA1 20th May 2013 watch

1. (Original post by BenChard)
thanks a lot.

would really appreciate it if someone has a list of units they can share (mega, milli) and their conversions? thanks
Gev = 1000Mev = 1000 000 000 ev Just read this one in the text book now

1A = 1000mA

1k(ohm) = 1000 ohm
2. (Original post by BenChard)
thanks a lot.

would really appreciate it if someone has a list of units they can share (mega, milli) and their conversions? thanks
(Someone help me out here!) Giga?! = 10^9
Mega=10^6
Kilo=10^3
Milli=10^-3
Micro=10^-6
Nano^-9

1ev=1.6x10^-19
1Mev=1.6 x-13
3. (Original post by x-Sophie-x)
(Someone help me out here!) Giga?! = 10^9
Mega=10x^6
Kilo=10^3
Milli=10^-3
Micro=10^-6
Nano^-9
Correct !
4. (Original post by posthumus)
I don't know if I have a good way of remember....

but I remember them because of computer memory kilobytes is 1000 bytes (or km is 1000m) mega is 106 giga 109 and you must have heard of tera- as well 1012.. but to be fair you probably won't come across that

milli is 10-3 (think of how many millimeters there are in a meter - 10 in 1cm & 100cm in 1m... therefore 1000mm in 1 meter)

micro is 10-6

nano is 10-9

Some calculators can convert units for you such as the fx--991ES, but I don't think you should have problems remember these powers
that's really helpful, I do have that calculator but it can only convert between different units like cm to inches thanks though
5. Could we get a circuit question with diodes in them?

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6. (Original post by posthumus)

Ah yes the 6 markers are such a pain, I prefer it when the 6 marker is applying theory (like in unit 4) otherwise it's pointless
I haven't started preparing for unit 1 yet (I know... lol -_-) but I will probably focus on the potential six marker over the weekend, so I'll send you any bullet points I do make But revision guides usually do have the bullet points of pretty much all experiments you should know about...?
Thank you!
Yes, I've been using the CGP revision guide, it's actually pretty good. For some reason I just seem to struggle with them!

(Original post by Jimmy20002012)
Could we get a circuit question with diodes in them?

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Do the specimen paper on the AQA website, they come up
:/
7. (Original post by Xiomara)
Thank you!
Yes, I've been using the CGP revision guide, it's actually pretty good. For some reason I just seem to struggle with them!

Do the specimen paper on the AQA website, they come up
:/
I'm using the CGP book what do you think of the jokes?

Just a off topic question

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8. hey, does anyone have like a list of possible 6 markers??? :s. Or even any tips on revising them?
9. (Original post by sceezy)
hey, does anyone have like a list of possible 6 markers??? :s. Or even any tips on revising them?
Look at the markschemes and learn them

I'll probably copy and paste them all off the markschemes tonight and i'll post on here when I'm done. Also, they're all pretty similar so if you do a few you'll get the hang of them. There's the ones where you have to plot the I-V graphs, or change the temperature of a thermistor and work out the resistance, or the photoelectric effect, excitation and de-excitation, fluorescent lamps, how to find out if the rms values and frequency of a power supply is accurate using an oscilloscope.

I think they're the ones I've come across so far.
10. (Original post by Son234)
I'm using the CGP book what do you think of the jokes?

Just a off topic question

This was posted from The Student Room's iPhone/iPad App
I'm a sucker for cheesy things like that, I quite like them! ;p You?
11. (Original post by Jimmy20002012)
Could we get a circuit question with diodes in them?

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Hello Jimmy! Do you remember me? Yes you can get a diode question. But it is not hard at all! A single diode (only allows current to flow through in 1 direction) requires 0.6V to let the voltage in the battery pass through to the rest of the components.
12. (Original post by StalkeR47)
Hello Jimmy! Do you remember me? Yes you can get a diode question. But it is not hard at all! A single diode (only allows current to flow through in 1 direction) requires 0.6V to let the voltage in the battery pass through to the rest of the components.
Yeah how could I forget you . What happens if you have a reverse diode, isn't it that no current would flow. Also if you has a reversed diode and a bulb in parallel with another bulb, the bulb would not switch on because no current would flow am I right? but what would be the voltage for that diode and bulb? and would the current for the other bulb be at maximum as there would not be any current flowing through the bulb and diode???

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13. How do you reverse a current to see the result in the third quadrant of a graph?
Is a reversed current just a negative current?

14. (Original post by x-Sophie-x)
For the quark combinations of pions and kaons, my teacher suggested 'don't stand under dark umbrellas'

Not sure if that will be any help but hey xD

I don't know of anything else

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Thanks.
15. (Original post by Didierr)
How do you reverse a current to see the result in the third quadrant of a graph?
Is a reversed current just a negative current?
you swap the orientation of the component or biased direction for a diode or change the battery terminals around
16. (Original post by BenChard)
that's really helpful, I do have that calculator but it can only convert between different units like cm to inches thanks though
The fx 991 ms can convert these units. Shift + 9,7,6
17. (Original post by StalkeR47)
Thanks! But older past papers than 2009 have radiation in them instead of electricity.
What do you mean? The old spec PHA3/W papers are called Current Electricity and Elastic Properties of Solids.
18. (Original post by Jimmy20002012)
Yeah how could I forget you . What happens if you have a reverse diode, isn't it that no current would flow. Also if you has a reversed diode and a bulb in parallel with another bulb, the bulb would not switch on because no current would flow am I right? but what would be the voltage for that diode and bulb? and would the current for the other bulb be at maximum as there would not be any current flowing through the bulb and diode???

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Yes m8 you are right! The diode would be reversed bias so the current in the circuit would be negligible. Let's assume we have a 9V battery and it is connected with a diode in it's 'forward bias' and with a filament bulb. You can easily determine the voltage at the bulb which would be 8.4V. This is because the diode requires 0.6V to let voltage pass through so 9-0.6=8.4V. As I said if the diode is reversed bias, the voltage cannot pass through the diode since it wont allow any current to pass through. And therefore, the bulb will not glow...
19. (Original post by StalkeR47)
Yes m8 you are right! The diode would be reversed bias so the current in the circuit would be negligible. Let's assume we have a 9V battery and it is connected with a diode in it's 'forward bias' and with a filament bulb. You can easily determine the voltage at the bulb which would be 8.4V. This is because the diode requires 0.6V to let voltage pass through so 9-0.6=8.4V. As I said if the diode is reversed bias, the voltage cannot pass through the diode since it wont allow any current to pass through. And therefore, the bulb will not glow...
What about a resistor would voltage go through that if diode is reversed?

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20. (Original post by Jimmy20002012)
Yeah how could I forget you . What happens if you have a reverse diode, isn't it that no current would flow. Also if you has a reversed diode and a bulb in parallel with another bulb, the bulb would not switch on because no current would flow am I right? but what would be the voltage for that diode and bulb? and would the current for the other bulb be at maximum as there would not be any current flowing through the bulb and diode???

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Hey Jimmy! I think you re still a bit confused. Let me clear that...... Consider a forward diode with a bulb and a battery with a conventional current towards the flat end of the diode. (try to sketch this), now the bulb will be on. Draw a same diagram with the conventional current towards the pointed edge of the diode, the bulb will be off because there will be no current. This is the state in which the polarity is reversed compared to you first diagram and therefore, for the voltage, the diode is reversed bias. Got it?

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