HSC Physics Marathon 2013-2015 Archive (3 Viewers)

[Drsoccerball]

re: HSC Physics Marathon Archive

They will affect the equation, won't they? Because we're saying (and similarly for , and then dividing these equations gives the equation .

But terminal voltages V_p and V_s can only be assumed to equal their respective emf's if we assume the internal resistance is 0. The real equation should be: . So we do need to assume there's no energy losses by assuming no internal resistance, e.g. from the wires etc.

(But then again, I don't think internal resistance is dealt with in the HSC, and we just assume V = emf.)

Do we need to know the derivation that you've done for the HSC?

 

[PhysicsMaths]

re: HSC Physics Marathon Archive

As in, it doesn't make THAT much difference.

True, pretty much 99% efficient

 

[InteGrand]

re: HSC Physics Marathon Archive

Do we need to know the derivation that you've done for the HSC?

No. But it's good to know where it comes from (or I find it is, anyway, so the formula is understood better).

 

[Fizzy_Cyst]

re: HSC Physics Marathon Archive

They will affect the equation, won't they? Because we're saying and similarly for , and then dividing these equations gives the equation .

But terminal voltages V_p and V_s can only be assumed to equal their respective emf's if we assume the internal resistance is 0. The real equation should be: . So we do need to assume there's no energy losses by assuming no internal resistance, e.g. from the wires etc.

(But then again, I don't think internal resistance is dealt with in the HSC, and we just assume V = emf.)

Therein lies the assumption! That the rate of change of flux experienced by both coils is the same, i.e., no flux leakage.

Forget about internal resistance! LoL.

 

[Fizzy_Cyst]

re: HSC Physics Marathon Archive

Do we need to know the derivation that you've done for the HSC?

Without understanding where the equation comes from, you would be unable to answer the question which I posed. Most students would cite energy losses due to eddy currents etc.. But that affects the current ratio, not the voltage ratio.

Transformers operate on the principle of mutual induction. We assume that there is no leakage (which is untrue, although the presence of a soft iron core does nearly make this true)

 

[Kaido]

re: HSC Physics Marathon Archive

Well, i guess i got a partially correct answer

 

[Fizzy_Cyst]

re: HSC Physics Marathon Archive

2005HSC Q24
marking criteria insists that we identify the role oh phonons
i'm interested in your answers

and will i get penalised (or not given marks) if i recall information that is beyond the syllabus?

The marking criteria has changed for superconductivity, especially due to recent advancements in understanding (i.e., flux pinning or quantum flux trapping) I really doubt that big questions of superconductivity will arise.

They USED to want you to talk specifically about the role of phonons, but now you do not have to. But, who knows.. This year you may have to :\

Well, i guess i got a partially correct answer

Yep, definitely on the right track re: what you said about flux

 

[jackleung34]

re: HSC Physics Marathon Archive

Have a go at this one

In which case is an astronaut closest to "true" weightlessness

A)Floating around in a cabin of a spaceship orbiting earth
B)Walking on the moon
C)Falling back to Earth during re-entry in a space shuttle
D)Halfway between the moon & Earth

Answer with an explanation.

 

[Thunderstorm]

re: HSC Physics Marathon Archive

Have a go at this one

In which case is an astronaut closest to "true" weightlessness

A)Floating around in a cabin of a spaceship orbiting earth
B)Walking on the moon
C)Falling back to Earth during re-entry in a space shuttle
D)Halfway between the moon & Earth

Answer with an explanation.

D due to the weaker gravitational forces. While every object in the universe in theory is exerting a gravitational force on every other object in the universe, in between the moon and the Earth the gravitational field is undefined. Might also possibly be C haha

 

[Drsoccerball]

re: HSC Physics Marathon Archive

Have a go at this one

In which case is an astronaut closest to "true" weightlessness

A)Floating around in a cabin of a spaceship orbiting earth
B)Walking on the moon
C)Falling back to Earth during re-entry in a space shuttle
D)Halfway between the moon & Earth

Answer with an explanation.

LOL youre in my class was this a moodle test that your teacher gave

 

[PhysicsMaths]

re: HSC Physics Marathon Archive

D due to the weaker gravitational forces. While every object in the universe in theory is exerting a gravitational force on every other object in the universe, in between the moon and the Earth the gravitational field is undefined. Might also possibly be C haha

The answer is D
To feel the sensation of apparent weightlessness, C may be a viable choice. This is because the atoms within your body are accelerating towards the ground at the same rate as the spaceship would be accelerating due to gravity towards Earth. But to experience "true" weightlessness, one must be entirely out of the gravitational field of a planet, or experience no net force due to gravity. When placed halfway between the moon and Earth, the gravitational pull applied by both planets are essentially equal and opposite in direction, so the net force that the astronaut experiences is essentially zero and therefore has "true" weightlessness.

*definition of weight is the force experienced by a mass within a gravitational field

 

[Kaido]

re: HSC Physics Marathon Archive

What are you thoughts on this Q: http://imgur.com/uuZqLkY

 

[jackleung34]

re: HSC Physics Marathon Archive

What are you thoughts on this Q: http://imgur.com/uuZqLkY

plug into F=BILsin theta which should be A

 

[Thunderstorm]

re: HSC Physics Marathon Archive

The answer is D
To feel the sensation of apparent weightlessness, C may be a viable choice. This is because the atoms within your body are accelerating towards the ground at the same rate as the spaceship would be accelerating due to gravity towards Earth. But to experience "true" weightlessness, one must be entirely out of the gravitational field of a planet, or experience no net force due to gravity. When placed halfway between the moon and Earth, the gravitational pull applied by both planets are essentially equal and opposite in direction, so the net force that the astronaut experiences is essentially zero and therefore has "true" weightlessness.

*definition of weight is the force experienced by a mass within a gravitational field

Well said!

 

[Drsoccerball]

re: HSC Physics Marathon Archive

What are you thoughts on this Q: http://imgur.com/uuZqLkY

angle is only not included if m field is in or out of the page

 

[shyjohn]

re: HSC Physics Marathon Archive

Have a go at this one

In which case is an astronaut closest to "true" weightlessness

A)Floating around in a cabin of a spaceship orbiting earth
B)Walking on the moon
C)Falling back to Earth during re-entry in a space shuttle
D)Halfway between the moon & Earth

Answer with an explanation.

If you learn IB Physics (which I didn't either), you will know that the gradient of the gravitational potential (just think of a map with hills on it and you are going downhill, the GPE is how you fall from the potential) is $g$ for gravity. Which (you may assume) that the situation in D is zero because the slope of a tip is zero. Although the actual reason is because at that point the walls of the cabin cannot provide any normal force (which is weight here), hence weightlessness.

The answer is D, but just want to give a technical answer as a complement.

John

 

[someth1ng]

re: HSC Physics Marathon Archive

Have a go at this one

In which case is an astronaut closest to "true" weightlessness

A)Floating around in a cabin of a spaceship orbiting earth
B)Walking on the moon
C)Falling back to Earth during re-entry in a space shuttle
D)Halfway between the moon & Earth

Answer with an explanation.

I would argue that A and D are both correct.

 

[Drsoccerball]

re: HSC Physics Marathon Archive

I would argue that A and D are both correct.

But you still have apparent weight in space

 

[Kaido]

re: HSC Physics Marathon Archive

I think the question is looking for 'closest' - none of the answers define 'true' weightlessness. The definition of apparent weightlessness is seen in (A). Whereas (D) is likely the 'closest' to 'true' weightlessness

 

[someth1ng]

re: HSC Physics Marathon Archive

But you still have apparent weight in space

lol, no - you don't have apparent weight in space.

The question is worded really badly thought - it could mean "which one is the best description for weightlessness" in which A and D both qualify as weightlessness. If you are accelerating at the rate of all acceleration of all weight forces combined, you get weightlessness.

If the question wants to ask for the "best description of the least weight force", D is the better answer.

The thing is, weightlessness is NOT the absence of a weight force - it's the absence of something that opposes gravity (ie usually a normal force).