MedVision ad

General Thoughts: Physics (1 Viewer)

IR

Active Member
Joined
Oct 15, 2014
Messages
255
Gender
Male
HSC
2014
Also qv is the work done on moving across the electric field. So if it was not a stopping voltage it has nothing to no with kinetic energy besides providing it more after liberation.
 

inkdririe

New Member
Joined
Aug 11, 2014
Messages
2
Gender
Female
HSC
2014
Some areas were difficult and others were pretty easy. Astrophysics option was awesome ;)

The KE graph question stumped me too though haha.
 

orcevalm

Member
Joined
Feb 25, 2013
Messages
54
Gender
Female
HSC
2014
how did you guys draw the gravitational potential energy graph? i just drew an increasing PE till 10 and straight line till 40mins dunno if its right
 

Prawnchip

Member
Joined
May 16, 2012
Messages
60
Gender
Female
HSC
2014
how did you guys draw the gravitational potential energy graph? i just drew an increasing PE till 10 and straight line till 40mins dunno if its right
Pretty much but when you drew the increasing GPE did you just draw a straight line? Or was it curved? :/
 

Chris100

Member
Joined
Apr 9, 2013
Messages
108
Gender
Undisclosed
HSC
2014
One must remember that the gravitational potential energy equation is -Gmm/d
So the curve should be negative, that's why the graph given had 0 in the centre to confuse students whether it should be above or below the x-axis
 

Kobweb

New Member
Joined
Jan 21, 2013
Messages
27
Gender
Male
HSC
N/A
Can someone post the paper, or know where to find one? Thanks
 

cashmoe

New Member
Joined
May 21, 2013
Messages
14
Gender
Male
HSC
2014
Best decision of mine: Quanta to Quarks to Age of silicon
 

rated

Member
Joined
Oct 12, 2013
Messages
41
Location
Sydney
Gender
Undisclosed
HSC
2014
Funny how there was no 'impacts on environment' question, most of it was related to society
 

iStudent

Well-Known Member
Joined
Mar 9, 2013
Messages
1,158
Gender
Male
HSC
2014
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.
Yea you have to overcome work function to liberate electrons. It's the voltage that eliminated it (by giving the electrons more energy, if that makes sense). Hence you have to equate work function somehow with voltage. W=qV works perfectly. (they are also both in joules)

The gradient is planck's constant I'm not sure if you can find it that way.

Unless someone else found a better method, I'm sure this is the right one. I'm pretty sure that with a high voltage you can liberate elections without photoelectric effect. Again, shows that voltage does have something to do with electrons overcoming work function.

The circuit is also connected in such a way that the voltage gives the electrons a boost in energy. 4.1 volts is the amount of volts that photoelectric effect occurs at 0Hz. Shows that work function has just been overcome at this value.

I'm rambling but does that makes sense? :s
 
Last edited:

IR

Active Member
Joined
Oct 15, 2014
Messages
255
Gender
Male
HSC
2014
Yea you have to overcome work function to liberate electrons. It's the voltage that eliminated it (by giving the electrons more energy, if that makes sense). Hence you have to equate work function somehow with voltage. W=qV works perfectly. (they are also both in joules)

The gradient is planck's constant I'm not sure if you can find it that way.

Unless someone else found a better method, I'm sure this is the right one. I'm pretty sure that with a high voltage you can liberate elections without photoelectric effect. Again, shows that voltage does have something to do with electrons overcoming work function.
If voltage eliminated it, whats the point of light? It said photoelectric, not thermionic.
 

Brightergull

New Member
Joined
Jul 28, 2013
Messages
20
Gender
Male
HSC
2014
wtfffff!! we were meant to properly plot the graph in the projectile question by shifting a certain amount and shit?? I though it was just meant to represent the shape of the parabola. plus the only thing i did was make sure both projectiles were on the same vertical level -.-
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top