ATAR Notes: Forum
VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: Wingtips on June 11, 2012, 10:21:02 am
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Is there some leeway with reading graphs? For example I had to calculate the resistance from an I vs V graph. My answer was 1250 ohms, compared to 1200 ohms.
Would they accept this?
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I'm sure that in the case that the graph was one were multiple answers could be taken from it then there would probably be an acceptable range, although i doubt the VCAA would have a graph were reading of it took "guessing" they would almost always have the value you need to read of the graph right on a line that corresponds to a direct value so there can be no mistake with reading it.
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Oh okay. Sweet, that makes sense. Thanks
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Yes, there is leeway with graphs and their values. VCAA allow gaps, but within reason
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Cool. Another question. In VCAA 2010, question 12 of motion. The formula is: net force x change in time = (net) impulse.
In that case the net average force was equal to the average force exerted on the ball by the racquet. Why is that? Is it because the ball is hit when is momentarily stationary, so at rest?
What would be some cases where the net average force doesn't equal the force of object A on B or vice versa?
For example, if a ball hit the ground, the average force of the ground on the ball would not be equal to the net average force. Cause you would have to add the weight force.
Cheers!
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In that case the net average force was equal to the average force exerted on the ball by the racquet. Why is that? Is it because the ball is hit when is momentarily stationary, so at rest?
Wouldn't you have to account for the Weight Force regardless of whether its stationary or not when calculating the average Net Force on the ball, so it wouldn't be equal to the average Force that the racquet exerted on the ball?
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In that case the net average force was equal to the average force exerted on the ball by the racquet. Why is that? Is it because the ball is hit when is momentarily stationary, so at rest?
Wouldn't you have to account for the Weight Force regardless of whether its stationary or not when calculating the average Net Force on the ball, so it wouldn't be equal to the average Force that the racquet exerted on the ball?
I'm not sure cause all the solutions have is "Substituting into the relationship I = F x t, the force was 212.5 N" But it is the net average force right? not just force? Maybe it's because it is being hit horizontally, so the Weight Force doesn't matter?
On another note, is this the correct energy transfer process for a opto-electronic system:
X (i.e sound) -> electrical -> light -> electrical -> X (i.e sound etc.)
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I'm not sure cause all the solutions have is "Substituting into the relationship I = F x t, the force was 212.5 N" But it is the net average force right? not just force? Maybe it's because it is being hit horizontally, so the Weight Force doesn't matter?
On another note, is this the correct energy transfer process for a opto-electronic system:
X (i.e sound) -> electrical -> light -> electrical -> X (i.e sound etc.)
With the formulae I=Fnet x Δt this only applies to collisions, so the Net Force is only talking about the Net for of object A on object B and does not include Weight Force because that is an external force to the collision.
Yes that is correct for an opto-electronic system, you could also say:
Microphone ⇒ electric signal ⇒ LED ⇒ opticle fibre ⇒ photodiode ⇒ electrical signal ⇒ speaker
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Thank you so much man. So basically the net force is the sum of the forces acting ON the ball and not of the whole system?
So with the ball colliding with the ground again. If the impulse acting On ball is say 4.2 Ns. Then the net force calculated using that equation is the net force acting ON the ball. So net force wouldn't equal Force of ball on ground - Force of ground on ball. Because that's acting on two different objects?
I really hope I'm not confusing you. :S
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However, if the cars collide on an icy, horizontal road, the collision can be considered to take place in an isolated system. The sum of external forces (including the force of gravity and the normal reaction force) acting on the system of the cars would be negligible compared with the forces that each car applies to the other.
Thats an extract from my textbook, which has got be doubting myself and i am quite honestly not sure at all anymore.
maybe someone else can chime in and clear things up?
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Yep, I'm confused as well. Sorry that I have got you doubting yourself. >< Yeh, hopefully someone else will be able to explain it.
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Yep, I'm confused as well. Sorry that I have got you doubting yourself. >< Yeh, hopefully someone else will be able to explain it.
Ahh I think I figured it out as I was just re-doing the VCAA 2010 paper.
The reason you only take the horizontal force and you do not include the weight force is because in the question it states that the ball is given a horizontal impulse of 1.7kgms-1 so the average force will be the average horizontal force on the ball which can only be exerted by the racquet so therefore it is equal to the average force exerted by the racquet on the ball. :D
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Ah yup. It makes sense now :D Cheers dude.
And another question. What's the difference between Amplitude modulation and light intensity modulation?
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I'm not 100% as my teacher never actually taught modulation and i never really bothered with it too much but,
these are the definitions from my textbook, so hope they help.
Amplitude modulation is the changing of the amplitude of the carrier signal by superimposing the waveform the information signal that you want to be sent (music or voice).
Optical intensity modulation in photonic communications systems involves varying the output intensity of a carrier light source (for example, an LED or laser diode) by using the electric signal from a sound to electricity transducer such as a microphone, or from a sound playback device such as a tape recorder.
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Cheers! I find it hard to understand the details of modulation. Just gonna chuck those definitions on a cheat sheet and hope for the best Haha
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Don't worry i'm in the exact same boat man, just hoping for the best with those definitions... ;)