HSC Physics Marathon 2013-2015 Archive (6 Viewers)

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yasminee96

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re: HSC Physics Marathon Archive

I'm not too sure about my answer.

Let y=0 be ground level
y=1/2 a(y)t^2 + u(y) +100
Uy=0, as it is released horizontally---> So y=1/2 a(y)t^2 + 100 (1)
x= U(x)t, Ux=10---> x=10t (2)
tan45=y/x, so y=x
Equating (1) and (2), we get a quadratic equation in terms of t---> solving I got t=3.6 and t=-5.65(t>=0, so ignore this value)
Sub t=3.6s into equation (1) we find a value for y (y= 36.5m) (how far it falls above the ground), so 100-36.5=63.5m is the distance in drops?

Sorry when I say 'y' and 'x' I mean 'delta y' and 'delta x'

Yeah i did this too, however when using exact values (such as t=3.61.... rather than just t=3.6) i got 63.9.
Are we supposed to use exact values or rounded throughout are workings? which is more accepted?
 
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JJ345

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re: HSC Physics Marathon Archive

Yeah i did this too, however when using exact values (such as t=3.61.... rather than just t=6) i got 63.9.
Are we supposed to use exact values or rounded throughout are workings? which is more accepted?
I think we use exact values throughout working(stored in our calculators), then round our final answer...but I was just rushing through the question so wasn't paying much attention to decimal places and stuff.
 

Fizzy_Cyst

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re: HSC Physics Marathon Archive

Awesome stuff you two!

Answer is indeed 64m :)
 

Fizzy_Cyst

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re: HSC Physics Marathon Archive

Yeah i did this too, however when using exact values (such as t=3.61.... rather than just t=3.6) i got 63.9.
Are we supposed to use exact values or rounded throughout are workings? which is more accepted?
Use exact values throughout working. Only round off your final answer to appropriate number of sig figs
 

Menomaths

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re: HSC Physics Marathon Archive

That's so confusing =/
 

someth1ng

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re: HSC Physics Marathon Archive

It doesn't work because that's for the angle between the conductor and the field lines.

In Q8, the conductor and field likes are perpendicular.
 

JJ345

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Menomaths

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re: HSC Physics Marathon Archive

You apply F(x)/F(y) =[GmM/r(x)^2]/ [GmM/r(y)^2]--> The G,m,M will cancel leaving you with r(y)^2/r(x)^2--> Measure these radii directly from the diagram since it is to scale and you should get your answer :)
Ah okay same procedure. My rulers playing tricks lol
 

Menomaths

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re: HSC Physics Marathon Archive

Why doesn't DC work for transformers? (Good question, and I can't seem to understand why it doesn't work)
 

ROTUS

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re: HSC Physics Marathon Archive

Why doesn't DC work for transformers? (Good question, and I can't seem to understand why it doesn't work)
transformers function on the basis of a change in flux
DC will not produce a change in flux (unlike AC)
 

someth1ng

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re: HSC Physics Marathon Archive

transformers function on the basis of a change in flux
DC will not produce a change in flux (unlike AC)
Just a note, be careful with what you say.

Saying that DC does not produce a change in flux is technically incorrect in some cases because a DC signal, if it appears like below will produce a small amount of change of flux but the transfer of that energy would largely be inefficient.

 

Menomaths

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re: HSC Physics Marathon Archive

So DC CAN be used, but it's not used since it's change in flux is very minute?
 

Menomaths

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re: HSC Physics Marathon Archive

Ohhhh...Never mind just understood why it can't be used. Blanked out when I was reading about transformers
 
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