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March 29, 2024, 10:48:07 am

Author Topic: VCE Physics Question Thread!  (Read 603401 times)  Share 

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Bestie

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Re: VCE Physics Question Thread!
« Reply #555 on: August 05, 2014, 10:29:50 pm »
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question: in a DC motor why does the armature continue to spin even when at one point on force (the forces cancel out) are acting on it?
I know it has something to do with momentum but I don't quite get it....

knightrider

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Re: VCE Physics Question Thread!
« Reply #556 on: August 16, 2014, 11:38:14 pm »
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How would you do this question


A man of mass 70 kg steps forward out of a boat and
onto the nearby river bank with a velocity, when he
leaves the boat, of 2.5 m s−1 relative to the ground.
The boat has a mass of 400 kg and was initially at
rest. With what velocity relative to the ground does
the boat begin to move?

lzxnl

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Re: VCE Physics Question Thread!
« Reply #557 on: August 17, 2014, 10:56:07 am »
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How would you do this question


A man of mass 70 kg steps forward out of a boat and
onto the nearby river bank with a velocity, when he
leaves the boat, of 2.5 m s−1 relative to the ground.
The boat has a mass of 400 kg and was initially at
rest. With what velocity relative to the ground does
the boat begin to move?


Conservation of momentum. Initially both the man and the boat are at rest. Then, the man pushes on the boat to start moving. Have a think about it.
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knightrider

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Re: VCE Physics Question Thread!
« Reply #558 on: August 17, 2014, 12:00:57 pm »
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Conservation of momentum. Initially both the man and the boat are at rest. Then, the man pushes on the boat to start moving. Have a think about it.

Thanks would this be right :)

using m1u1+m2u2=m1v1+m2v2

(70*0)+(400*0)=(70*2.5)+(400*v2)
0=175+400v2
v2=-0.44metres per second what does the negative mean in the answer


Zealous

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Re: VCE Physics Question Thread!
« Reply #559 on: August 17, 2014, 04:54:10 pm »
+2
Thanks would this be right :)

using m1u1+m2u2=m1v1+m2v2

(70*0)+(400*0)=(70*2.5)+(400*v2)
0=175+400v2
v2=-0.44metres per second what does the negative mean in the answer

The negative simply means the boat is moving in the opposite direction to the man. If you define the man's velocity to be positive 2.5m/s, the boat's velocity will be 0.44m/s in the other direction - this is because the man's movement pushes the two apart, they are not in the same direction.
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Alwin

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Re: VCE Physics Question Thread!
« Reply #560 on: August 17, 2014, 05:42:21 pm »
+2
What accelerating potential difference would be required to give an alpha particle a de Broglie wavelength of 2.0nm?

mass of alpha is 6.67x10^-27kg

I attached the solution and am just unsure why they doubled the coulomb of charge. The way i tried to work it out is find the velocity using de broglie's wavelength formula which gave me 49.7m/s. Then use this to find the kinetic energy which was 8.24x10^-24J. Once i did this i used the formula W=qV and subbed in 8.24x10^-24=(1.6x10^-19)V which gave me the answer of 5.1x10^-5V required.

Think about the charge on an alpha particle ;) It's a bit of a trick question because you have to had remembered what the charge is from year 11, or do it the way you did it with year 12 methods

what is the importance of frequency in determining the non-ideal behaviour of transformers. are there any other parameters?

Hmm interesting question but usually not examinable. A quick run down of things that can make transformers less ideal:
Eddy currents
Non ideal joins (side note is that Wilson Transformers actually makes all their transforms by hand and its an amazing process to get maximum efficiency)
Heat in the coil (usually transformers are oil-convection cooled)
Whether or not the core is laminated (this goes back to the eddy currents issue)
And the last one I can think of is flux leakage, but tbh I don't really get this either

But basically in essence the most important one as you've pointed how is the frequency as this will have an effect on eddy currents (imagine it as a sort of like a residue current in the coil)

question: in a DC motor why does the armature continue to spin even when at one point on force (the forces cancel out) are acting on it?
I know it has something to do with momentum but I don't quite get it....

I assume you mean something like this :)


I'll start with B first, because it's the most obvious: Force on side is going up and the force on the other is going down, so the coil spins, seems simple enough :)
But then when it's in points A and C, the force on the top is going upwards and the force on the bottom is acting downwards so I get there can be some confusion as how the coil keeps on turning. In fact, it's because of the commutator. At that instant the coil briefly loses connection with the batter (the 'split' part of the split ring commutator) so there is no current flowing and thus no force acting on the coil. Now, the coil was originally spinning (see point B) so it will continue to spin as there is no braking or frictional force acting on the coil. The terminals of the coil then touch the other side of the split ring commutator (reversing the current) and now the forces are reversed as well meaning the coil keeps on spinning.

So in essence, the coil's spinning momentum just before the perpendicular position is what carries to through the vertical position and allows the direction of the current to be changed via the commutator
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Stick

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Re: VCE Physics Question Thread!
« Reply #561 on: August 18, 2014, 12:08:26 pm »
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Lol, I know I'm not doing VCE anymore but I'm doing a VCE equivalent (kinda) Physics subject at university and I'm getting confused on the question attached. My answer for the second situation doesn't make sense so I'm assuming I'm doing it wrong. Thanks for the help. :)
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Re: VCE Physics Question Thread!
« Reply #562 on: August 18, 2014, 05:08:56 pm »
+2
Lol, I know I'm not doing VCE anymore but I'm doing a VCE equivalent (kinda) Physics subject at university and I'm getting confused on the question attached. My answer for the second situation doesn't make sense so I'm assuming I'm doing it wrong. Thanks for the help. :)

Hey Stick,

From what I've calculated, Rachel would be 6m from the centre pivot point, which is not actually on the see-saw. Yeah - doesn't make sense to me also.
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Re: VCE Physics Question Thread!
« Reply #563 on: August 18, 2014, 05:11:51 pm »
+1
I got the same answer, so it must be a dodgy question lol. Thanks for checking that. :)
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knightrider

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Re: VCE Physics Question Thread!
« Reply #564 on: August 18, 2014, 09:05:34 pm »
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The negative simply means the boat is moving in the opposite direction to the man. If you define the man's velocity to be positive 2.5m/s, the boat's velocity will be 0.44m/s in the other direction - this is because the man's movement pushes the two apart, they are not in the same direction.

Thanks Zealous :)
Can you help with these questions?

A small research rocket of mass 250 kg is launched
vertically as part of a weather study. It sends out
50 kg of burnt fuel and exhaust gases with a velocity
of 180 m s−1 in a 2 s initial acceleration period.

a)What is the velocity of the rocket after this initial
acceleration?
b)What upward force does this apply to the rocket?
c)What is the net upward acceleration acting on the
rocket?

Alwin

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Re: VCE Physics Question Thread!
« Reply #565 on: August 19, 2014, 12:46:36 pm »
+2
Thanks Zealous :)
Can you help with these questions?

A small research rocket of mass 250 kg is launched
vertically as part of a weather study. It sends out
50 kg of burnt fuel and exhaust gases with a velocity
of 180 m s−1 in a 2 s initial acceleration period.

a)What is the velocity of the rocket after this initial
acceleration?
b)What upward force does this apply to the rocket?
c)What is the net upward acceleration acting on the
rocket?


Sorry, I'm not Zealous but I hope you'll accept my help too :P

To me, the trick to this question is that the mass of the rocket changes from 250 to 200kg (50 kg of fuel is sent out)
So, I would use the 'average' mass of 225kg in my calculations. (there are more scientific methods to find the average mass than just choosing the middle value, but they're too bothersome and not required for VCE)

Part a
Conservation of momentum tells us that the momentum before the launch is equal to that after the launch. In maths form,


Now initially we know that the rocket was stationary and we're given all the information in the question (just don't forget to use the average mass!). Note that I take up as positive


Part b
To find the upwards force, we can use the impulse as we know the momentum and we are given the time


Part c
To find the net acceleration we just use Newton's 2nd law remembering to use the average mass again


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A pessimist says a glass is half empty, an optimist says a glass is half full.
An engineer says the glass has a safety factor of 2.0

knightrider

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Re: VCE Physics Question Thread!
« Reply #566 on: August 19, 2014, 06:13:39 pm »
+1
Sorry, I'm not Zealous but I hope you'll accept my help too :P

To me, the trick to this question is that the mass of the rocket changes from 250 to 200kg (50 kg of fuel is sent out)
So, I would use the 'average' mass of 225kg in my calculations. (there are more scientific methods to find the average mass than just choosing the middle value, but they're too bothersome and not required for VCE)

Part a
Conservation of momentum tells us that the momentum before the launch is equal to that after the launch. In maths form,


Now initially we know that the rocket was stationary and we're given all the information in the question (just don't forget to use the average mass!). Note that I take up as positive


Part b
To find the upwards force, we can use the impulse as we know the momentum and we are given the time


Part c
To find the net acceleration we just use Newton's 2nd law remembering to use the average mass again






Thank you so much Alwin i really appreciate your help, you are very nice :)

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Re: VCE Physics Question Thread!
« Reply #567 on: August 20, 2014, 10:44:45 pm »
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how would I show acceleration in the opposite direction on a velocity time graph?
thanks

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Re: VCE Physics Question Thread!
« Reply #568 on: August 21, 2014, 11:38:07 pm »
+2
Acceleration is the slope of a velocity time graph, as , so if the velocity is positive, you need a negative gradient for the acceleration to be in the opposite direction of the velocity. If the velocity is negative, then the gradient must be positive.
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Re: VCE Physics Question Thread!
« Reply #569 on: August 22, 2014, 02:28:18 am »
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Cold-blooded animals (ectotherms) such as reptiles are able to sustain a higher population density than similarly sized warm-blooded animals (endotherms) on the same terrain. Can you
explain this observation in terms of energy and heat flow? (