Physics Exam Thoughts (1 Viewer)

[trea99]

That's what I did. If you didn't assume it was a circle, how would you calculate area?

 

[Sp3ctre]

Was it wrong to for the flux density to use the max flux and divide it by the area of the circle loop (It didn't say the loop was circular nor that the given length was the diameter, so idk if what i did was right)

Pretty sure you had to assume that was true

 

[NinjaDatHSC]

I have no clue by my friend was confident to say that you had to find the "area under the curve" from the flux vs time graph - which idk if that made sense

 

[Commando007]

lmfao i did pi(r) ^2 xD

 

[trea99]

Magnetic field strength is flux per unit area. The maximum flux on the flux vs time graph was 0.6Wb I think so the maximum magnetic field strength over the 12 seconds would be 0.6/Area of the loop. Then for part b), the rate of change of flux (flux vs time) is the gradient of the curve which is voltage and then you use the right hand grip rule or something else to find direction. I don't think you had to find the area under the flux vs time curve for anything and you really couldn't do so either because part of it was curved if I remember correctly.

 

[NinjaDatHSC]

Magnetic field strength is flux per unit area. The maximum flux on the flux vs time graph was 0.6Wb I think so the maximum magnetic field strength over the 12 seconds would be 0.6/Area of the loop. Then for part b), the rate of change of flux (flux vs time) is the gradient of the curve which is voltage and then you use the right hand grip rule or something else to find direction. I don't think you had to find the area under the flux vs time curve for anything and you really couldn't do so either because part of it was curved if I remember correctly.

Yeah i did exactly what you said. Yeah i don't think any of the graph was curved but to begin with I'm not sure if area under curve = flux density is a true statement to begin with

 

[trea99]

Yeah i did exactly what you said. Yeah i don't think any of the graph was curved but to begin with I'm not sure if area under curve = flux density is a true statement to begin with

Yea area under Flux/time curve isn't magnetic field strength pretty sure we're correct.

 

[pikachu975]

Magnetic field strength is flux per unit area. The maximum flux on the flux vs time graph was 0.6Wb I think so the maximum magnetic field strength over the 12 seconds would be 0.6/Area of the loop. Then for part b), the rate of change of flux (flux vs time) is the gradient of the curve which is voltage and then you use the right hand grip rule or something else to find direction. I don't think you had to find the area under the flux vs time curve for anything and you really couldn't do so either because part of it was curved if I remember correctly.

Yep finding area under the curve would mean finding the integral of the curve which is clearly wrong. 0.6/Area is correct and for part b yep you had to use faraday's law and it was 0.3 Volts and P was positive using right hand group rule

 

[blinkandyoullmissit]

did anyone else get 10.9 metres for the range of the projectile?

 

[trea99]

Anyone else think aligning will be better than previous years? MC was definitely harder and core had some pretty weird ones like the blackbody curve.

 

[trea99]

did anyone else get 10.9 metres for the range of the projectile?

I got from memory around 9m

 

[maria1999]

I got from memory around 9m

i got 8.8. is that about right??


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[blinkandyoullmissit]

I found time to be 2.18, then Delta X = ux T

which is 2.18 x 5 = 10.9 metres

 

[NinjaDatHSC]

what did you guys get for question 17 mc:

This is the answer from solutions from competitor website lo:
"17 - B , force on magnet is downwards by the right hand slap rule, so the scale reading will be higher (I'm 90% sure). Since the unit is kilograms, it is measuring mass, so we compensate by dividing the force by 9.8."

I did this but my friends said i was wrong and it was "minus" instead of plus

 

[trea99]

Ok, so we're given an initial velocity of 10ms^-1 and an angle of 60 to the horizontal
So initial horizontal component of velocity is 5ms^-1 (ie 10cos60) and initial vertical is 10sin60 (around 8.66)
At max height, vertical velocity is 0,
so 0 = 10sin60 - 9.8t
t = 10sin60/9.8
This is half the total time of flight
Horizontal displacement = horizontal velocity x time = 5 x 20sin60/9.8 which is around 8.8 so yea maria same answer

 

[camstu]

I found time to be 2.18, then Delta X = ux T

which is 2.18 x 5 = 10.9 metres

was Ux 10 or am I retarded?

 

[trea99]

what did you guys get for question 17 mc:

This is the answer from solutions from competitor website lo:
"17 - B , force on magnet is downwards by the right hand slap rule, so the scale reading will be higher (I'm 90% sure). Since the unit is kilograms, it is measuring mass, so we compensate by dividing the force by 9.8."

I did this but my friends said i was wrong and it was "minus" instead of plus

I agree with you and the solution, the force on the magnet is down so there's an additional force of BIl on top of the weight force and yea we divide by 9.8 for mass

 

[camstu]

what did you guys get for question 17 mc:

This is the answer from solutions from competitor website lo:
"17 - B , force on magnet is downwards by the right hand slap rule, so the scale reading will be higher (I'm 90% sure). Since the unit is kilograms, it is measuring mass, so we compensate by dividing the force by 9.8."

I did this but my friends said i was wrong and it was "minus" instead of plus

Where did you find solutions?

 

[blinkandyoullmissit]

Overall velocity was 10, so Ux was 10Cos60 which was about 8.66

 

[NinjaDatHSC]

Where did you find solutions?

Idk if it will let me say the site name on this without getting banned. It's just in hsc exam discussion 2017 for another site if you look for it you'll find it