iTute 2010 Question

[kyzoo]

I just did the exam and there one question I don't understand

Question 6 For the material which of the following alternatives is correct?
A. The stress at which the material changes from elastic to plastic behaviour is easily detected.
B. The material is elastic because it absorbs a large amount of energy before it breaks.
C. The material is tough because it has a high fracture stress.
D. The material is ductile because its stress-strain curve is not linear.

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I chose D but the answer is A =/. I don't see what's wrong with D.

[olly_s15]

poorly written question

[Akirus]

I suppose it's one of those questions where you have to choose the answer that is 'most right', although it is pretty shitty.

[olly_s15]

well it is not B nor C - I'm leaning towards D but who knows it could be A? Does it mean easily detectable via the graph or what i dunno what option A is implying

[superflya]

yes that question was weird I think it was A

[kyzoo]

well it is not B nor C - I'm leaning towards D but who knows it could be A? Does it mean easily detectable via the graph or what i dunno what option A is implying

I think a large difference between the gradient of the linear region, and the gradient of the strain-stress graph immediately after the elastic limit, enables easy detection of plasticity

[Akirus]

A makes sense; there is a distinct point at which the graph changes from elastic deformation to plastic. If I had to hazard a guess, it is "more right" than D because the material is still PARTLY linear, making it a somewhat erroneous statement?

[olly_s15]

A makes sense; there is a distinct point at which the graph changes from elastic deformation to plastic. If I had to hazard a guess, it is "more right" than D because the material is still PARTLY linear, making it a somewhat erroneous statement?

this is true - although i interpreted A in a slightly different way

if A said:

The stress at which the material changes from elastic to plastic behaviour is easily detected from the stress-strain graph.

then it is much clearer - but i guess it is A

my teacher said the itute exams are usually poorly written since they are free

[Akirus]

My teacher told my class not to touch iTute because they're unnecessarily difficult; going to do them anyway, though ;p

You would assume it meant 'from the graph', given that the graph was specifically provided.

[kyzoo]

iTute is not unneccesarily difficult. I got 84/90 on this one whereas I got 85/90 for VCAA 2009 >.<.

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Another question which I dispute the solution to

Question 7.

If gravity is the only force acting on the skier (i.e. he/she is in freefall), then he undergoes no horizontal acceleration. Hence horizontal velocity at R is the same as the horizontal velocity at Q.

Q is the top of an inverted parabola, which is flat. Therefore the velocity at Q is entirely horizontal. Hence horizontal velocity at Q = V.

Now at R, the skier's velocity is 45 degrees to the horizontal. Hence at R, the skier's vertical velocity = his horizontal veloicty. Hence at R, his vertical velocity = V.

...

Is it wrong to assume that Q is the top of the inverted parabola?

Btw the solution's answer involves conservation of energy

...

[kyzoo]

Another question lol

Question 12 Which one or more of the above graphs best represent the signal in the fibre optic cable when the
signal generator is turned on?

Answer is B and D.

I thought that light intensity is the amplitude of the light wave? (So I chose A and B).

[Akirus]

iTute is not unneccesarily difficult. I got 84/90 on this one whereas I got 85/90 for VCAA 2009 >.<.

Yeah, but consider why you got those grades. You're far from an average student so your results don't reflect the actual difficulty of the paper.

[googoo]

iTute is not unneccesarily difficult. I got 84/90 on this one whereas I got 85/90 for VCAA 2009 >.<.

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Another question which I dispute the solution to

Question 7.

If gravity is the only force acting on the skier (i.e. he/she is in freefall), then he undergoes no horizontal acceleration. Hence horizontal velocity at R is the same as the horizontal velocity at Q.

Q is the top of an inverted parabola, which is flat. Therefore the velocity at Q is entirely horizontal. Hence horizontal velocity at Q = V.

Now at R, the skier's velocity is 45 degrees to the horizontal. Hence at R, the skier's vertical velocity = his horizontal veloicty. Hence at R, his vertical velocity = V.

...

Is it wrong to assume that Q is the top of the inverted parabola?

Btw the solution's answer involves conservation of energy

...

Both methods are right. yours used the properties of projectile motions, itutes used energy conservation. V= sqrt(V^2/2 +40), .: V=sqrt(40).

[Akirus]

Another question lol

Question 12 Which one or more of the above graphs best represent the signal in the fibre optic cable when the
signal generator is turned on?

Answer is B and D.

I thought that light intensity is the amplitude of the light wave? (So I chose A and B).

Shouldn't the signal in the optical fibre be modulated? I don't see how it's D either.

[googoo]

Another question lol

Question 12 Which one or more of the above graphs best represent the signal in the fibre optic cable when the
signal generator is turned on?

Answer is B and D.

I thought that light intensity is the amplitude of the light wave? (So I chose A and B).

light intensity is proportional to amplitude^2