Login

Welcome, Guest. Please login or register.

March 29, 2024, 08:47:29 am

Author Topic: VCE Physics Question Thread!  (Read 603380 times)  Share 

0 Members and 3 Guests are viewing this topic.

lzxnl

  • Victorian
  • ATAR Notes Legend
  • *******
  • Posts: 3432
  • Respect: +215
Re: VCE Physics Question Thread!
« Reply #885 on: March 03, 2015, 09:49:44 pm »
0
Could someone please explain to me how the equation for resistance total for parallel circuits proves the validity of Kirchoff's Law?

I'm not sure if the resistance equation establishes Kirchhoff's law or the other way around...but anyway

If you have two resistors in parallel, the total resistance satisfies 1/Reff = 1/R1 + 1/R2 where R1, R2 are the resistances of each parallel part.
The total current is therefore V/Reff = V/R1 + V/R2 = I1 + I2
This shows that it's valid to say that the current after the junction is the sum of the currents before the junction. Also, I've used the same letter V for both resistors; it's clearly also valid to say that both parts of the circuit are at the same voltage.

When finding the kinetic energy of a projectile in motion do we use the horizontal component, verticle component or the resultant vector component?

KE = 1/2*mass*(speed)^2. Find the speed. It's not a vector.
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

paper-back

  • Victorian
  • Forum Obsessive
  • ***
  • Posts: 340
  • "I must govern the clock, not be governed by it"
  • Respect: +7
Re: VCE Physics Question Thread!
« Reply #886 on: March 03, 2015, 10:27:16 pm »
0
So is the speed the resultant component?

lzxnl

  • Victorian
  • ATAR Notes Legend
  • *******
  • Posts: 3432
  • Respect: +215
Re: VCE Physics Question Thread!
« Reply #887 on: March 03, 2015, 11:01:10 pm »
0
Speed = size of velocity
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

orgekas

  • Victorian
  • Adventurer
  • *
  • Posts: 15
  • Respect: 0
  • School: Cheltenham Secondary College
Re: VCE Physics Question Thread!
« Reply #888 on: March 04, 2015, 12:55:44 pm »
0
Hi guys,

So we did this prac in physics for circular motion. The aim was to derive the formula F=mv^2/r. The apparatus we used contained paper clip, tube, fishing line, rubber stoper and slotted masses. So what we did with my partner was  measure the time and record it for the rubber stopper to make 10 revolutions and then keep radius constant change mass and calculate the velocity. Then, we repeated the same thing instead we changed the radius and kept the mass constant.
And now we have to use graphs to show that F=mv^2/r. We know that Fc=ma=F by the hanging masses but we dont know what graphs to draw in order to prove the formula.

Please help,
Thanks.

Stevensmay

  • Guest
Re: VCE Physics Question Thread!
« Reply #889 on: March 04, 2015, 01:29:23 pm »
0
Hi guys,

So we did this prac in physics for circular motion. The aim was to derive the formula F=mv^2/r. The apparatus we used contained paper clip, tube, fishing line, rubber stoper and slotted masses. So what we did with my partner was  measure the time and record it for the rubber stopper to make 10 revolutions and then keep radius constant change mass and calculate the velocity. Then, we repeated the same thing instead we changed the radius and kept the mass constant.
And now we have to use graphs to show that F=mv^2/r. We know that Fc=ma=F by the hanging masses but we dont know what graphs to draw in order to prove the formula.

Please help,
Thanks.

I'm not too sure how to prove it with a graph (I've never seen anything proved like that before).

I'm happy to try writing up a derivation but it involves angular velocity/acceleration and I can't remember whether this is in the course. And a bit of calculus (basic).

lzxnl

  • Victorian
  • ATAR Notes Legend
  • *******
  • Posts: 3432
  • Respect: +215
Re: VCE Physics Question Thread!
« Reply #890 on: March 04, 2015, 02:49:30 pm »
0
Hi guys,

So we did this prac in physics for circular motion. The aim was to derive the formula F=mv^2/r. The apparatus we used contained paper clip, tube, fishing line, rubber stoper and slotted masses. So what we did with my partner was  measure the time and record it for the rubber stopper to make 10 revolutions and then keep radius constant change mass and calculate the velocity. Then, we repeated the same thing instead we changed the radius and kept the mass constant.
And now we have to use graphs to show that F=mv^2/r. We know that Fc=ma=F by the hanging masses but we dont know what graphs to draw in order to prove the formula.

Please help,
Thanks.

Graph F against v^2. You should get a straight line of slope m/r
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

orgekas

  • Victorian
  • Adventurer
  • *
  • Posts: 15
  • Respect: 0
  • School: Cheltenham Secondary College
Re: VCE Physics Question Thread!
« Reply #891 on: March 04, 2015, 03:03:49 pm »
0
Graph F against v^2. You should get a straight line of slope m/r

If you can see the attachment those are the results that I have. And the graph that Excel produces. Is that right? If yes then what?

JackSonSmith

  • Victorian
  • Forum Obsessive
  • ***
  • Posts: 288
  • "Failure is part of nature, success is man-made"
  • Respect: +4
  • School Grad Year: 2015
Re: VCE Physics Question Thread!
« Reply #892 on: March 04, 2015, 04:08:44 pm »
0
We did a prac about friction on inclined planes. We got the length of hypotenuse (1m), 3 angles (0,12.7 and 24.8 degrees) and 3 times (0.81,1.07 and 1.17) respectively.

A 600g trolley was pulled by a 400g weight.

My calculation for 0 degrees, was friction = 0.95N.
 
I used constant acceleration formula to calculate acceleration to be 3.05.
Then
Fnet = Fapplied - Fopposing
3.05 = 4 - friction.
Friction = .95N

I have had trouble finding the friction when the plane was inclined. Do I have to somehow factor in the force down the plane due to the component of weight down? Could someone help please?
2014: Psychology
2015: English | Methods | Chinese SL | Specialist | Physics 

2016: Bachelor of Commerce - The University of Melbourne

Start where you are. Use what you have.  Do what you can. – Arthur Ashe

Kel9901

  • Victorian
  • Trendsetter
  • **
  • Posts: 158
  • Respect: +2
  • School: Kardinia International College
  • School Grad Year: 2015
Re: VCE Physics Question Thread!
« Reply #893 on: March 04, 2015, 05:03:58 pm »
+1
We did a prac about friction on inclined planes. We got the length of hypotenuse (1m), 3 angles (0,12.7 and 24.8 degrees) and 3 times (0.81,1.07 and 1.17) respectively.

A 600g trolley was pulled by a 400g weight.

My calculation for 0 degrees, was friction = 0.95N.
 
I used constant acceleration formula to calculate acceleration to be 3.05.
Then
Fnet = Fapplied - Fopposing
3.05 = 4 - friction.
Friction = .95N

I have had trouble finding the friction when the plane was inclined. Do I have to somehow factor in the force down the plane due to the component of weight down? Could someone help please?

You need to take into account the force down the plane due to, as you said, the weight force, and that's the only difference. What you've done so far is correct. I won't do the calculations for you (since it's your prac), but expect lower friction forces
s=change in displacement for physics
2011: Methods [47]
2012: Spesh [42] Further [47]
2013: UMEP Maths [4.5]
2014: Chem [47] Physics [48] Music Performance [43]
2015: Spesh [redo] English Accounting Music Investigation

lzxnl

  • Victorian
  • ATAR Notes Legend
  • *******
  • Posts: 3432
  • Respect: +215
Re: VCE Physics Question Thread!
« Reply #894 on: March 04, 2015, 09:22:06 pm »
0
If you can see the attachment those are the results that I have. And the graph that Excel produces. Is that right? If yes then what?

Something like that. Get Excel to display the line equation to get the slope. Right click the trendline.
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

chekside

  • Victorian
  • Trailblazer
  • *
  • Posts: 39
  • Respect: 0
  • School: Melbourne High School
Re: VCE Physics Question Thread!
« Reply #895 on: March 06, 2015, 10:40:22 pm »
+1
I found a question that dealt with an IR emitter and an IR phototransistor receiver. The question asked if the voltage drop across the receiver would increase or decrease when the IR emitter was on. The answer stated that it would decrease, but didn't give an explanation, could someone please explain why the voltage drop would decrease when the IR began hitting the receiver?
Thanks

JackSonSmith

  • Victorian
  • Forum Obsessive
  • ***
  • Posts: 288
  • "Failure is part of nature, success is man-made"
  • Respect: +4
  • School Grad Year: 2015
Re: VCE Physics Question Thread!
« Reply #896 on: March 07, 2015, 10:14:54 pm »
0
A geosynchronous satellite is one with a period of 24h positioned exactly above the equator. It appears motionless viewed from the surface of Earth. Explain why it must be in an orbit above the equator.

Two questions. 1: How important is knowing this theory. 2: Could someone explain in layman's terms what is going on.

Thanks.
2014: Psychology
2015: English | Methods | Chinese SL | Specialist | Physics 

2016: Bachelor of Commerce - The University of Melbourne

Start where you are. Use what you have.  Do what you can. – Arthur Ashe

lzxnl

  • Victorian
  • ATAR Notes Legend
  • *******
  • Posts: 3432
  • Respect: +215
Re: VCE Physics Question Thread!
« Reply #897 on: March 07, 2015, 11:47:40 pm »
+1
I found a question that dealt with an IR emitter and an IR phototransistor receiver. The question asked if the voltage drop across the receiver would increase or decrease when the IR emitter was on. The answer stated that it would decrease, but didn't give an explanation, could someone please explain why the voltage drop would decrease when the IR began hitting the receiver?
Thanks

Not part of the course I'm sure. Don't need to know about transistors

A geosynchronous satellite is one with a period of 24h positioned exactly above the equator. It appears motionless viewed from the surface of Earth. Explain why it must be in an orbit above the equator.

Two questions. 1: How important is knowing this theory. 2: Could someone explain in layman's terms what is going on.

Thanks.
1. I guess it's unlikely VCAA will test this
2. Any orbit must have the centre of the Earth as the centre of the orbit. If you're not over the equator, you will still orbit around the centre of the Earth so your orbit will be lopsided.
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

stockstamp

  • Victorian
  • Trailblazer
  • *
  • Posts: 33
  • Respect: 0
  • School Grad Year: 2015
Re: VCE Physics Question Thread!
« Reply #898 on: March 08, 2015, 12:18:00 am »
0
Hoping someone can clarify this:

If I drop an object why does it stop when it hits the ground? I'm not challenging whether or not it should, but If I was asked this question in a test I doubt I could answer it satisfactorily.

I know that every action creates and equal and opposite reaction, but I don't understand how the size of the normal force can change during the contact time, and how equal and opposite reactions can cause a decelleration as a result of that.

Not sure if I've expressed my self properly here, but I hope someone follows my train of thought...

Thanks
2014: Methods [40] (neck myself)

I'm here to avoid homework

http://i.imgur.com/1tf3V.gif

lzxnl

  • Victorian
  • ATAR Notes Legend
  • *******
  • Posts: 3432
  • Respect: +215
Re: VCE Physics Question Thread!
« Reply #899 on: March 08, 2015, 10:52:12 am »
0
Hoping someone can clarify this:

If I drop an object why does it stop when it hits the ground? I'm not challenging whether or not it should, but If I was asked this question in a test I doubt I could answer it satisfactorily.

I know that every action creates and equal and opposite reaction, but I don't understand how the size of the normal force can change during the contact time, and how equal and opposite reactions can cause a decelleration as a result of that.

Not sure if I've expressed my self properly here, but I hope someone follows my train of thought...

Thanks

Equal and opposite forces act on DIFFERENT OBJECTS. That's why a deceleration can result. If you had the equal but opposite forces acting on the same object, then of course no net force would result.

As for why objects stop when they hit the ground, it's because collisions with the ground are inelastic and the object tends to lose its kinetic energy thermally.
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