Login

Welcome, Guest. Please login or register.

March 29, 2024, 09:48:33 am

Author Topic: VCE Physics Question Thread!  (Read 603390 times)  Share 

0 Members and 2 Guests are viewing this topic.

eddybaha

  • Victorian
  • Trailblazer
  • *
  • Posts: 43
  • Respect: 0
  • School: GWSC
  • School Grad Year: 2013
Re: VCE Physics Question Thread!
« Reply #375 on: November 12, 2013, 09:59:20 pm »
0
flux should be at a maximum inside the loop i believe
Mathematical Methods:44 (2012)
English Language, Specialist Maths, Physics, Chemistry, Extension Maths (monash) (2013)
Hard work beats talent when talent fails to work hard.

~T

  • Victorian
  • Trendsetter
  • **
  • Posts: 197
  • Respect: +4
  • School: St Patrick's College
  • School Grad Year: 2013
Re: VCE Physics Question Thread!
« Reply #376 on: November 12, 2013, 09:59:34 pm »
+2
Hi guys, on the 2011 Exam 2 Electric Power Q.11, I've been talking with my friend from school and we can't figure the graph out. We looked at the VCAA solution and we get what they're saying but we were thinking something else - I put our train of thought in the attachment below.

Pretty much it's that since the flux will reduce once the magnet is actually inside the loop shouldn't that cause some emf generation, spiking in the opposite direction? The attachment will make what I'm saying clearer hopefully (even if it is nonsense lol)

Any advice would be great, thanks :)
I'm a little confused but I think I see where you're coming from. The only thing is that the flux doesn't reduce when the magnet is inside the loop. Flux still "runs through" the inside of a magnet, and the field lines are still there when this magnet is in the centre of the loop.

At every point with the magnet approaching, going through, and leaving the loop, there is a leftwards flux, yes? It never changes direction. Now, given this constant direction, all that matters is whether this flux is increasing or decreasing. The field lines are closest together when near the magnet, so the flux will be greatest when the magnet is inside the loop. It will increase up until this point, and then decrease away as the magnet leaves.

I hope I've clarified, but I'm not sure it was super clear sorry :-\
ATAR: 99.95
Specialist 50 | Methods 50 | Physics 50 | Further 49 | Literature 48 | Music Style/Composition 41

2014 - 2016: Bachelor of Science (Chancellor's Scholars' Program) at The University of Melbourne

I will be tutoring in Melbourne this year. Methods, Specialist, and Physics. PM me if you are interested :)

Lachjames

  • Victorian
  • Fresh Poster
  • *
  • Posts: 4
  • Respect: 0
  • School: Waverley Christian College
Re: VCE Physics Question Thread!
« Reply #377 on: November 12, 2013, 10:05:32 pm »
0
Thanks for that, I think I get it now :)

papertowns

  • Victorian
  • Trailblazer
  • *
  • Posts: 35
  • Respect: 0
Re: VCE Physics Question Thread!
« Reply #378 on: November 13, 2013, 01:29:12 am »
0
PLeaseee:
A beam of red light of frequency 4 × 10^14 Hz is found to deliver 7.54 × 10^19 photons per second.

What would be the power of the beam?

silverpixeli

  • ATAR Notes Lecturer
  • Victorian
  • Forum Leader
  • ****
  • Posts: 855
  • Respect: +110
Re: VCE Physics Question Thread!
« Reply #379 on: November 13, 2013, 06:50:21 am »
0
If anyone is here :)

I'm not sure, sorry, hopefully somebody catches this before the exam :(

I tried rotating it both ways in my head and both times got current from X to Y initially, not sure if it's a bad question or if I just stuffed up the logic somewhere.

PLeaseee:
A beam of red light of frequency 4 × 10^14 Hz is found to deliver 7.54 × 10^19 photons per second.

What would be the power of the beam?

P=E/t

E = number of photons x single photon energy
= 7.54 x 10^19 x hf
= 7.54 x 10^19 x 6.63 x 10^-34 x 4 x 10^14

so P=(that)/1=that

EDIT: I used the Js version of h because we want energy in Joules so that we can have Power in Watts (Joules/second), if we had have used the eVs value of h (the 4.14x10^-15 one) we would have got energy in eV meaning power in eV/s which isn't a standard unit for power, afaik.

Hopefully that makes sense, the key idea is that the energy of a beam is the number of photons times the energy of one photon :)

Good luck today everyone!
« Last Edit: November 13, 2013, 06:55:59 am by silverpixeli »
ATAR 99.80 :: Methods [50] | Physics [50+Premier's] | Specialist [47] | Software [48] | English [42] | Legal [39 '12]
+ Australian Student Prize

ATAR Notes Specialist/Methods/Physics Lecturer
ATAR Notes Specialist Maths Webinar Presenter

Robert123

  • Victorian
  • Forum Obsessive
  • ***
  • Posts: 201
  • Respect: +5
  • School: Kyabram P-12 College
Re: VCE Physics Question Thread!
« Reply #380 on: November 13, 2013, 07:31:23 am »
0
If anyone is here :)

Ok, to solve this we use lenz's law and the right hand slap rule. Let start off by analysining what would happen if we rotate it clockwise.
Firstly side XY would go up and due to lenzs law we know it would create a force in the opposite direction (that is down).
So using our right hand slap rule, palm  down and fingers point to the right, we can see the current will flow X to Y. Therefore, we can deduce that it must rotate the opposite way (anticlockwise).
You can do the same analysis for anticlockwise to confirm it.

Rod

  • Victorian
  • Part of the furniture
  • *****
  • Posts: 1755
  • The harder the battle, the sweeter the victory
  • Respect: +101
Re: VCE Physics Question Thread!
« Reply #381 on: December 17, 2013, 09:43:51 pm »
0
Can the year 11's moving into year 12 use this thread next year? Or do we have to make our own, new thread?
2013-2014:| VCE
2015-2018:| Bachelor of Science (Neuroscience) @ UoM
2019-X:| Doctor of Dental Surgery (discontinued)
2019 -2021:| Master of Physiotherapy

Currently: Physiotherapist working at a musculoskeletal clinic. Back pain, sore neck, headaches or any other pain limiting your study? Give me a PM (although please do see your personal health professional first!)

Any questions related to pathways towards studying dentistry or physiotherapy? Gimmi a PM!

Stevensmay

  • Guest
Re: VCE Physics Question Thread!
« Reply #382 on: December 17, 2013, 09:44:48 pm »
+3
Can the year 11's moving into year 12 use this thread next year? Or do we have to make our own, new thread?

Feel free to post whatever.

Einstein

  • Guest
Re: VCE Physics Question Thread!
« Reply #383 on: March 01, 2014, 11:44:44 am »
0
Can someone explain fissile and fissionable?


katie101

  • Victorian
  • Trailblazer
  • *
  • Posts: 47
  • Katie
  • Respect: 0
  • School Grad Year: 2015
Re: VCE Physics Question Thread!
« Reply #384 on: March 01, 2014, 05:31:53 pm »
0
If a dead battery and a new battery were used in series, would the well used battery have any effect on the output voltage of the new battery? Why/why not?

Thorium

  • Victorian
  • Trendsetter
  • **
  • Posts: 189
  • Why can ants lift 50x their own weight?
  • Respect: +20
  • School Grad Year: 2014
Re: VCE Physics Question Thread!
« Reply #385 on: March 05, 2014, 09:23:45 pm »
0
A question says: "A spring has stiffness 20 N/m and is hang vertically. When a mass is attached, it stretches by 0.2 m. Assuming that the spring has no mass, what is the value of the mass."

What I did was that:
change(Elastic E)=Change(Gravitational E)
1/2kx^2=mgh
1/2(20)(0.2)^2=m(10)(0.2)
                     m=0.2 kg

But the ans says to use kx=mg, and the ans is 0.4 kg.

I am thinking that since the F due to gravity on the mass is constant, the 1/2kx^2 should not be used. Is that the reason why my method gave the wrong ans?
2013: Further | Persian
2014: English (AL) | Methods | Physics | Specialist
ATAR: 96.70

2015: Bachelor of Engineering @ Monash

RKTR

  • Victorian
  • Forum Leader
  • ****
  • Posts: 613
  • Respect: +17
Re: VCE Physics Question Thread!
« Reply #386 on: March 05, 2014, 09:31:18 pm »
0
that day someone in my class asked a similar question. my teacher said maybe there is some energy lost. so it is better to use mg=kx if possible
2015-2017: Bachelor of Biomedicine (Neuroscience)
2018: Doctor of Medicine (Withdrawn)
2019: Bachelor of Commerce (Actuarial Studies?)

lzxnl

  • Victorian
  • ATAR Notes Legend
  • *******
  • Posts: 3432
  • Respect: +215
Re: VCE Physics Question Thread!
« Reply #387 on: March 05, 2014, 09:43:41 pm »
+1
A question says: "A spring has stiffness 20 N/m and is hang vertically. When a mass is attached, it stretches by 0.2 m. Assuming that the spring has no mass, what is the value of the mass."

What I did was that:
change(Elastic E)=Change(Gravitational E)
1/2kx^2=mgh
1/2(20)(0.2)^2=m(10)(0.2)
                     m=0.2 kg

But the ans says to use kx=mg, and the ans is 0.4 kg.

I am thinking that since the F due to gravity on the mass is constant, the 1/2kx^2 should not be used. Is that the reason why my method gave the wrong ans?

Let's think about this reasonably. The only thing you know about the spring when it is still is that the net force is zero, so you use forces.

Now, let's see what happens when you attach a mass to the relaxed spring and let go. Obviously, the spring is going to start moving. Initially, the spring only has gravitational potential energy. By equating mgx with 1/2 kx^2, all you are doing is finding when all of the gravitational potential energy goes completely into spring energy. By conservation of energy, this means the kinetic energy at this point (where all of the potential energy is spring) is still the same at the st art, where the kinetic energy is zero. This is where the object is momentarily stationary. HOWEVER, the forces are clearly not balanced as the spring is still headed upwards.

The point is, if you just let a spring go, it will oscillate forever assuming no resistive drag forces. Note how when you attach a spring, the forces are unbalanced. In order to balance the spring, the person holding the mass must physically move it with the spring to the point of stable equilibrium. This action drains energy from the spring and explains why you cannot just equate the energies.
2012
Mathematical Methods (50) Chinese SL (45~52)

2013
English Language (50) Chemistry (50) Specialist Mathematics (49~54.9) Physics (49) UMEP Physics (96%) ATAR 99.95

2014-2016: University of Melbourne, Bachelor of Science, Diploma in Mathematical Sciences (Applied Maths)

2017-2018: Master of Science (Applied Mathematics)

2019-2024: PhD, MIT (Applied Mathematics)

Accepting students for VCE tutoring in Maths Methods, Specialist Maths and Physics! (and university maths/physics too) PM for more details

Rishi97

  • Victorian
  • Part of the furniture
  • *****
  • Posts: 1042
  • Respect: +40
  • School: The University of Melbourne
  • School Grad Year: 2014
Re: VCE Physics Question Thread!
« Reply #388 on: March 09, 2014, 01:31:37 pm »
0
Can I pls have help with the following questions?
1) An astronaut standing on the moon experiences a gravitational force of attraction of 160N. He moves away from the surface of the moon to an altitude where the gravitational force is 40N.
     a) How far from the centre of the moon is this new location in terms of the radius of the moon?
     b) The astronaut now travels to another location at a height of three moon radii above the surface. Calculate the gravitational force at this altitude


Cheers ;)
2014: VCE completed
2015-2017: BSc at Melb Uni

DREAM, BELIEVE, ACHIEVE!!!

lzxnl

  • Victorian
  • ATAR Notes Legend
  • *******
  • Posts: 3432
  • Respect: +215
Re: VCE Physics Question Thread!
« Reply #389 on: March 09, 2014, 01:48:53 pm »
+1
Can I pls have help with the following questions?
1) An astronaut standing on the moon experiences a gravitational force of attraction of 160N. He moves away from the surface of the moon to an altitude where the gravitational force is 40N.
     a) How far from the centre of the moon is this new location in terms of the radius of the moon?
     b) The astronaut now travels to another location at a height of three moon radii above the surface. Calculate the gravitational force at this altitude


Cheers ;)

So...F = k/r^2 where k is some constant.
If the force quarters, the distance r must double, so this new location is now two moon radii from the centre of the moon.

Now, three moon radii above the surface = four moon radii from centre of moon (r is ALWAYS measured from the centre of the moon), so we've quadrupled the distance r with regards to the initial distance, so the force drops to 1/16, or 10 N.
2012
Mathematical Methods (50) Chinese SL (45~52)

2013
English Language (50) Chemistry (50) Specialist Mathematics (49~54.9) Physics (49) UMEP Physics (96%) ATAR 99.95

2014-2016: University of Melbourne, Bachelor of Science, Diploma in Mathematical Sciences (Applied Maths)

2017-2018: Master of Science (Applied Mathematics)

2019-2024: PhD, MIT (Applied Mathematics)

Accepting students for VCE tutoring in Maths Methods, Specialist Maths and Physics! (and university maths/physics too) PM for more details