General Thoughts: Physics (1 Viewer)

[screwuhsc]

My teacher had a look at it and he said that the diagram was misleading!!! most of the state won't get it

U srs ahhaahahhaha lolb

 

[iStudent]

Got caught stupidly on the power loss question and then the stopping voltage. Ugh.... other than that pretty fair paper, although the wording of some of the multiple choice questions was really iffy. (eg. there are fewer high energy photons at high frequencies, no commas.)

For the power loss q you used VI=VI to find current in secondary coil then you used P=I^2R to find power loss.

 

[Beterban]

The projectile q was 27 spaces down. around 3 spaces up

yep sot the same
for the horizontal shift it 3 units right each second right?

 

[iStudent]

yep sot the same
for the horizontal shift it 3 units right each second right?

Yep

 

[Beterban]

Yep

or the votage equal 0 one, do you just shift the graph 4.1 down?

 

[iStudent]

lol idk if i got the projectile one right. option was alright except for the binding energy one :/ i also didnt know how to do the stopping voltage one like i drew another parallel line next to it and probs calculated the whole frequency wrong

Fission fusion for binding energy q.
I didn't get the debroglie q :/ so random lol

 

[iStudent]

No. I posted solution earlier. 9.9 threshold frequency.

 

[IR]

The threshold freq was around 9.9
radiation frequency was 12.8 ish

I'll admit I didn't get the q till the last 10 minutes.
You find stopping voltage as 4.1
so work function was qV
Then you find threshold where KE=0
so hf=W. Find f your threshold frequency. Draw a line parallel then rest is easy

If u looked carefully it wasn't stopping voltage. But it rather was similar to a photocell voltage. The positive plate was on the other end of photoelectric plate

 

[iStudent]

yea I know. But at 4.1V you got the graph with the frequency starting at 0 so this follows that 4.1V is the value for stopping voltage (it's not called "stopping voltage" but that's what its value is).

I'll try make it more clear. The graph starts from 0Hz. This means that the voltage has exactly eliminated the work function. So your work function is 4.1q. Sort of like stopping voltage but not really. Yea nvm it's not called stopping voltage
All good I didn't write that there!

Rest of the working out is fine. I'll just edit...

 

[Lethal Toxin]

LOL for the 0v question, i drew a random line, had no clue what it was asking

 

[Prawnchip]

Please tell me I'm not the only one who found the exam difficult - got so screwed over by the projectile crap and the weird KE stuff
:alone:

 

[sisabella]

someone post the exam?

 

[rated]

Pretty okay exam if you ask me, though I'm pretty shit at physics... Got thrown off by the 0V question and a few graphing ones

 

[rimatiger]

q2q incident photon question rekt me, the rest of my elective was ok

exam was mostly fair though, harder than 2013 and 1 or 2 bullshit questions but mostly ok

 

[armageddonmuscle55]

what did u write about the space one

 

[orcevalm]

studied so hard for relativity yet only one q in it. and btw for the projectile, if you wrote the same horizontal velocity in ur working but didnt draw it like that do u get a mark or no? it was 4 marks

 

[IR]

For the power loss q you used VI=VI to find current in secondary coil then you used P=I^2R to find power loss.

Yep thats right. You could also have used Vp/Vs = Is/Ip to find the current and then sub into P=I^2R[/QUOTE]

Precisely

 

[orcevalm]

Fission fusion for binding energy q.
I didn't get the debroglie q :/ so random lol

yea so i just talked about standing waves and matter waves hahaahah

 

[schmickrx3]

Yep thats right. You could also have used Vp/Vs = Is/Ip to find the current and then sub into P=I^2R

Precisely[/QUOTE]

Hahaha so weird, that is not my original post, but that is my comment and it somehow has disappeared.

 

[IR]

yea I know. But at 4.1V you got the graph with the frequency starting at 0 so this follows that 4.1V is the value for stopping voltage (it's not called "stopping voltage" but that's what its value is).

I'll try make it more clear. The graph starts from 0Hz. This means that the voltage has exactly eliminated the work function. So your work function is 4.1q. Sort of like stopping voltage but not really. Yea nvm it's not called stopping voltage
All good I didn't write that there!

Rest of the working out is fine. I'll just edit...

Work function is a constant and has absolutely nothing to do with external field. It's the energy required to liberate. The only thing 0 sows that it was provided the exact work function on from light. I think it has something to do with. Gradient.