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February 21, 2026, 11:51:19 pm

Author Topic: Rohitpi's Physics Questions Thread  (Read 5953 times)  Share 

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pi

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Re: Rohitpi's Physics Questions Thread
« Reply #30 on: February 19, 2011, 09:25:07 pm »
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Damn haha, oh well you can forget about it now XD
(but just confirm it wont be on your sacs, not that it should be)


I'll ask the Physics the co-ordinator just to be sure. But if its not the exam, thats a good thing!

pi

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Re: Rohitpi's Physics Questions Thread
« Reply #31 on: March 20, 2011, 06:41:08 pm »
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New question, its attached. I just need help with parts c) and d).

(read Q27 first btw)

Thanks!


(btw thanks Whatlol for the PM help before)
« Last Edit: March 20, 2011, 07:50:52 pm by Rohitpi »

pi

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Re: Rohitpi's Physics Questions Thread
« Reply #32 on: March 20, 2011, 07:39:15 pm »
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BUMP!

And another question, any help will be greatly appreciated for parts c), d) and e)

I have no idea what to do for part c), and hence can't get parts d) and e)

These formulas might help: and


btw, do we need to know these formulas for the exam?!?
« Last Edit: March 20, 2011, 07:55:13 pm by Rohitpi »

Whatlol

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Re: Rohitpi's Physics Questions Thread
« Reply #33 on: March 20, 2011, 08:46:36 pm »
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P = NE
N= P/E
N= 2.43x10^13 photons.

(If one photon has an energy E then how many photons hitting the surface every second produce power P )

Try the next questions now, if you cant get them id help, also i doubt you will get this in an exam.
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Whatlol

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Re: Rohitpi's Physics Questions Thread
« Reply #34 on: March 20, 2011, 09:10:32 pm »
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For the question 27 , part d

what you needa do is look at the graph to find what current you need for 1.7 volts.

its under 10mA. so lets just say about 8 mA

now current running through resistor and LED is the same so

Vs-1.7 = 8x10^-3 x 100
Vs = 2.5V

and part c.... no idea... because the current is determined by the voltage running through it, but how can we determine that. if you assume it was the same voltage running through it as before you would get 80mA... which would mean its operating brightly but with poor life. hmm sorry
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pi

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Re: Rohitpi's Physics Questions Thread
« Reply #35 on: March 21, 2011, 06:54:27 pm »
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P = NE
N= P/E
N= 2.43x10^13 photons.

(If one photon has an energy E then how many photons hitting the surface every second produce power P )

Try the next questions now, if you cant get them id help, also i doubt you will get this in an exam.

Thanks! I got the other ones too. You can also use the definition of poser (joules/sec) for that one too (found out in school).

Thanks for question 27, I still don't know part c. But from what you said, doesn't look like either could be VCAA questions anyway.

THANKS AGAIN!

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Re: Rohitpi's Physics Questions Thread
« Reply #36 on: March 21, 2011, 07:02:44 pm »
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P = NE
N= P/E
N= 2.43x10^13 photons.

(If one photon has an energy E then how many photons hitting the surface every second produce power P )

Try the next questions now, if you cant get them id help, also i doubt you will get this in an exam.

Thanks! I got the other ones too. You can also use the definition of poser (joules/sec) for that one too (found out in school).

Thanks for question 27, I still don't know part c. But from what you said, doesn't look like either could be VCAA questions anyway.

THANKS AGAIN!

yea .... part c kinda dodgy to be honest, i reakon they have just made some pretty rash assumption in it.
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Re: Rohitpi's Physics Questions Thread
« Reply #37 on: March 26, 2011, 03:32:45 pm »
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Could someone please give me a definition of the 'switch-on voltage' (of a standard diode)?

Thanks (in advance)


EDIT: Don't worry, found one
« Last Edit: March 26, 2011, 04:35:27 pm by Rohitpi »

schnappy

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Re: Rohitpi's Physics Questions Thread
« Reply #38 on: March 27, 2011, 09:06:24 pm »
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It's the voltage across the diode when a current passes through it, I think. 'switch-on voltage' isn't lingo I've seen in terms of diodes before.