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January 28, 2021, 10:29:55 pm

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#### lzxnl

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##### Re: VCE Physics Question Thread!
« Reply #45 on: May 29, 2013, 04:17:40 pm »
0
1. RMS means root-mean-square. It really should be called "square-mean-root" as you square your function, average it and then square root it. Why do we need it? AC voltage is sinusoidally varying; its average value over a cycle is zero. That's not very useful. However, in power calculations, where P=V^2/R, an average of V^2 is just as good. This time, if we square the voltage, it's always positive and now an average makes sense. I won't go into the calculation details, but the RMS voltage is the peak voltage divided by sqrt 2, for a sinusoidal wave only (which is VCE physics). Just know how to calculate RMS voltages, currents, what they mean and the fact that P=I^2 R or V^2/R refers to RMS currents and voltages.

2. If a photodiode is forward-biased, it is a regular diode. Only when reverse-biased does its behaviour depend on light like in the course. Then, the photocurrent is directly proportional to the intensity of light striking the diode. If the light intensity doubles, the current doubles.

3. I'm not great with transistors, sorry. My electronics knowledge is strictly limited to what's in the course.
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#### Robert123

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##### Re: VCE Physics Question Thread!
« Reply #46 on: May 29, 2013, 07:13:22 pm »
0

Thanks for that, I had questions on those topics for a practice SAC and our class haven't cover much of this stuff in detail if at all.
Another question relating to springs.
Pinball game has a ball with mass 0.15kg which is launch by a compressed spring at 4.2m/s that has been compressed by 20 cm. If the spring is "ideal" what is the spring constant k in N/m?
Since k is N/m, would you use constant acceleration formulas to find the acceleration, multiply it by mass (since F=ma) and divide by distance (k=F/x)? Would you have to worry about the constant acceleration question or are you meant to substitute the final speed for acceleration?

#### availn

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##### Re: VCE Physics Question Thread!
« Reply #47 on: May 29, 2013, 08:02:17 pm »
0
Thanks for that, I had questions on those topics for a practice SAC and our class haven't cover much of this stuff in detail if at all.
Another question relating to springs.
Pinball game has a ball with mass 0.15kg which is launch by a compressed spring at 4.2m/s that has been compressed by 20 cm. If the spring is "ideal" what is the spring constant k in N/m?
Since k is N/m, would you use constant acceleration formulas to find the acceleration, multiply it by mass (since F=ma) and divide by distance (k=F/x)? Would you have to worry about the constant acceleration question or are you meant to substitute the final speed for acceleration?

You cannot do this easily with kinematics, because a spring does not accelerate a mass at a constant rate. You should do this with energy, spring potential energy before equals kinetic energy after.

0.5kx2 = 0.5mv2
k · 0.22 = 0.15 · 4.22
∴ k = 66.15Nm-1
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#### random_person

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##### Re: VCE Physics Question Thread!
« Reply #48 on: May 29, 2013, 08:04:22 pm »
0
Thanks for that, I had questions on those topics for a practice SAC and our class haven't cover much of this stuff in detail if at all.
Another question relating to springs.
Pinball game has a ball with mass 0.15kg which is launch by a compressed spring at 4.2m/s that has been compressed by 20 cm. If the spring is "ideal" what is the spring constant k in N/m?
Since k is N/m, would you use constant acceleration formulas to find the acceleration, multiply it by mass (since F=ma) and divide by distance (k=F/x)? Would you have to worry about the constant acceleration question or are you meant to substitute the final speed for acceleration?

We have m=0.15kg, v=4.2ms-1, s=20cm=0.2m

EK=1/2 mv2
=1/2 0.15 x 4.22
=1.323 J
Us=1/2 ks2
therefore
1.323=1/2 k x 0.22
k=66.15Nm-1

Hope this helps

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#### Sentar

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##### Re: VCE Physics Question Thread!
« Reply #49 on: June 03, 2013, 04:01:23 pm »
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Found an exam question but can't find the answer anywhere, don't have the actual answers, seeing if anyone can help.

Explain how it is possible to transmit the information contained in an electrical signal which has both a positive and negative component, when it is impossible to produce a beam of light with negative intensity?

Cheers
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#### lzxnl

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##### Re: VCE Physics Question Thread!
« Reply #50 on: June 03, 2013, 04:29:53 pm »
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You just need a light signal whose amplitude varies just like your electrical signal/has the same shape. E.g. move the electrical signal up so it's always positive.
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#### ~T

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##### Re: VCE Physics Question Thread!
« Reply #51 on: June 10, 2013, 11:19:35 pm »
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Just looking at the 2012 Exam 2 paper and confused with question 1. Basically, there is a point P and it asks for you to draw an arrow at P indicating the direction of the magnetic field. There is a solenoid which will produce a field (comparable to Earth's) going right, and there is an arrow pointing upwards that says "magnetic north."

The answers say that the arrow drawn should be "to the right and up the page at an angle of approximately 45°." However, in saying that "magnetic north" is upwards, does that not mean that the field lines due to the Earth should be going downwards? If magnetic north is up, magnetic south is down, and field lines go from north to south. I would have drawn an arrow "to the right and *DOWN* the page at an angle of approximately 45°."

Basically, in saying that magnetic north is up, does that not mean that geographical north is downwards?
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#### availn

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##### Re: VCE Physics Question Thread!
« Reply #52 on: June 11, 2013, 02:18:50 pm »
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Just looking at the 2012 Exam 2 paper and confused with question 1. Basically, there is a point P and it asks for you to draw an arrow at P indicating the direction of the magnetic field. There is a solenoid which will produce a field (comparable to Earth's) going right, and there is an arrow pointing upwards that says "magnetic north."

The answers say that the arrow drawn should be "to the right and up the page at an angle of approximately 45°." However, in saying that "magnetic north" is upwards, does that not mean that the field lines due to the Earth should be going downwards? If magnetic north is up, magnetic south is down, and field lines go from north to south. I would have drawn an arrow "to the right and *DOWN* the page at an angle of approximately 45°."

Basically, in saying that magnetic north is up, does that not mean that geographical north is downwards?

Nope. My friend got hit by this too.

Magnetic North and North Magnetic Pole is at the earth's north.
Magnetic North Pole is at the earth's south.
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#### ~T

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##### Re: VCE Physics Question Thread!
« Reply #53 on: June 12, 2013, 06:19:45 pm »
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Oh really? So magnetic north is actually a south pole (in dipole terms)? No wonder >50% got that wrong.
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#### jono88

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##### Re: VCE Physics Question Thread!
« Reply #54 on: June 24, 2013, 07:43:46 pm »
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VCAA 2007 exam 2 question 1, the question asks you to sketch ﬁve magnetic ﬁeld lines around the magnet, yet in the assessors report they have 6 lines? wut?

#### Homer

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##### Re: VCE Physics Question Thread!
« Reply #55 on: June 24, 2013, 07:55:32 pm »
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how would  i find the tension in each cable?
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#### Phy124

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##### Re: VCE Physics Question Thread!
« Reply #56 on: June 24, 2013, 08:18:47 pm »
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how would  i find the tension in each cable?
As the structure is in rotational equilibrium you can sum the moments (I think you refer to it as torque in yr 12 physics?) about the point of rotation and equate them to zero.

Hint: The weight force of the bridge and the two vertical components of the tension forces will be the forces that cause rotation. (the horizontal component of tension does not cause rotation as it acts "through" the point of rotation)

Spoiler
$\sum M = -(5)(700)(9.8) + (2)(10)(T\sin(30^{\circ})) = 0$

$\Rightarrow T=3430 N$
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#### Homer

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##### Re: VCE Physics Question Thread!
« Reply #57 on: June 29, 2013, 11:54:24 am »
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Could someone walk me through this question

Thanks

ANS: a) fh=390N, fv=680N
b) 330N
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#### SocialRhubarb

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##### Re: VCE Physics Question Thread!
« Reply #58 on: June 29, 2013, 01:08:34 pm »
+1
Part A is quite simple:

The horizontal component is equal to $800\cos(60)\ N= 400\ N$

The vertical component is equal to $800\sin(60)\ N= 693\ N$

Umm, unless I've done something stupid I think your answers are slightly off.

Part B is a bit beyond the scope of VCE.

Essentially what you're doing is equating torques since the object isn't spinning off in one direction.

$\tau = r \times F$

The net torque on the fence post is 0, so the torque from one source must be equal to the torque from the other.

$0.85\ m\times 400\ N = 1.0\ m\times Ft$   Technically there should be a negative somewhere, but whatever.

$Ft = 340\ N$

Alternatively, using the answer of 390N,

$0.85\ m \times 390\ N = 1.0\ m\times Ft$

$Ft = 331.5\ N$
Fight me.

#### Homer

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##### Re: VCE Physics Question Thread!
« Reply #59 on: June 29, 2013, 02:14:18 pm »
+1
yeah im not sure why theyve got different component values?
anyways

A 6m ladder of mass 7.5 kg leans against a wall at 60 degrees to the horizontal. What is the torque exerted on the ladder by the weight of the ladder itself?

I know it should be 7.5g x 3 x sin(30), but i dont understand why they take theta as 30 instead of 60?
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