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#### lzxnl

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##### Re: VCE Physics Question Thread!
« Reply #60 on: June 29, 2013, 03:13:09 pm »
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It depends on how the angle is defined. The formula $\tau = rF sin{\theta}$ is only used when the angle is measured between the heads of the vectors r and F (it's a vector equation). In this case, 60 degrees is the angle between the ladder and the horizontal, while here theta is the angle between r and mg, in this case the ladder and the VERTICAL.

This is why I like thinking that torque = moment arm * force where the moment arm is the component of the radius that is perpendicular to the force. You'll never get it wrong this way.
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#### Homer

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##### Re: VCE Physics Question Thread!
« Reply #61 on: July 02, 2013, 11:41:55 am »
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The young's modulus for a rope of length 30m, cross-sectional area 30cm2, which stretches 2m under a load of 1000N is? ANS: 5 x 10^6 Nm-2

« Last Edit: July 02, 2013, 05:00:14 pm by 2/cos(c) »
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#### sin0001

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##### Re: VCE Physics Question Thread!
« Reply #62 on: July 02, 2013, 11:57:53 am »
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It depends on how the angle is defined. The formula $\tau = rF sin{\theta}$ is only used when the angle is measured between the heads of the vectors r and F (it's a vector equation). In this case, 60 degrees is the angle between the ladder and the horizontal, while here theta is the angle between r and mg, in this case the ladder and the VERTICAL.
Is the torque formula examinable material for our core topics, or is it only required for a particular detailed study because I haven't seen it in the Heinemann textbook :/
Also, are you guys gonna reinforce the Unit 3 knowledge-through some mid-year exams- or continue on with the course, as revision?
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#### Phy124

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##### Re: VCE Physics Question Thread!
« Reply #63 on: July 02, 2013, 04:59:53 pm »
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The young's modulus for a rope of length 30m, cross-sectional area 30cm2, which stretches 2m under a load of 1000N is? ANS: 5 x 10^6 Nm-2

$E = \frac{\sigma}{\varepsilon }}$

$\sigma = \frac{F}{A}$

$\varepsilon = \frac{\Delta L}{L}$

$\Rightarrow E = \frac{FL}{A\Delta L} = \frac{(1000)(30)}{(0.003)(2)} = 5 \times 10^6 \ Nm^{-2}$

Is the torque formula examinable material for our core topics, or is it only required for a particular detailed study because I haven't seen it in the Heinemann textbook :/
Also, are you guys gonna reinforce the Unit 3 knowledge-through some mid-year exams- or continue on with the course, as revision?

I would think that torque is only relevant to the "Materials and their use in structures" detailed study.

Although your question wasn't directed at me, I would advise you to do some mid-year practice exams during these holidays to both reinforce key concepts and get a feel for VCAA physics exams.

« Last Edit: July 02, 2013, 10:21:51 pm by 2/cos(c) »
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#### Homer

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##### Re: VCE Physics Question Thread!
« Reply #64 on: July 03, 2013, 10:11:24 am »
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$\Rightarrow E = \frac{FL}{A\Delta L} = \frac{(1000)(30)}{(0.003)(2)} = 5 \times 10^6 \ Nm^{-2}$

hey just a question why is it 0.003 wouldnt it be $(\frac{30}{100})^2$ so  $\frac{(1000)(30)}{(0.09)(2)}$ ?

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#### Homer

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##### Re: VCE Physics Question Thread!
« Reply #65 on: July 03, 2013, 11:45:34 am »
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Also, What would be the tension in the cables of the cranes? Thankyou
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#### Alwin

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##### Re: VCE Physics Question Thread!
« Reply #66 on: July 03, 2013, 12:58:15 pm »
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Also, What would be the tension in the cables of the cranes? Thankyou

Let the tensions of the cables be T1 and T2, corresponding the the crane numbers Crane 1 and Crane 2.
Note I have included units in my working, just a small personal habit.

Using Crane 2 as the pivot,
$\Rightarrow 25m \times T_1 = 20m \times 300,000 kg \times 10 m/s^{2}$
$\therefore T_1 = 2,400,000 N$

You can now either say,
$T_1 + T_2 = 300,000 kg \times 10 m/s^{2}$

or use Crane 1 as a pivot,
$25m \times T_2 = 5m \times 300,000 kg \times 10 m/s^{2}$

Either way, it works out that
$T_2 = 600,000 N$

So, then tensions on cables of the cranes 1 and 2 are 2,400,000N an 600,000N in that order

$\Rightarrow E = \frac{FL}{A\Delta L} = \frac{(1000)(30)}{(0.003)(2)} = 5 \times 10^6 \ Nm^{-2}$

hey just a question why is it 0.003 wouldnt it be $(\frac{30}{100})^2$ so  $\frac{(1000)(30)}{(0.09)(2)}$ ?

30cm2 is not the same as 30cm x 30cm = (0.3m)^2 = 0.09 m2

30cm2 is the same as 30cm x 1cm = (0.3m) x (0.01m) = 0.003 m2

The general rule is that 1cm2 = 0.0001m2

EDIT: answered both of homer's questions, rather than double post
« Last Edit: July 03, 2013, 01:08:24 pm by Alwin »
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#### Homer

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##### Re: VCE Physics Question Thread!
« Reply #67 on: July 05, 2013, 02:05:26 pm »
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Having trouble with 3. I dont know what they are asking for and how to work it out.
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#### SocialRhubarb

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##### Re: VCE Physics Question Thread!
« Reply #68 on: July 05, 2013, 03:00:17 pm »
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They've given you a compass on the side, and the earth's magnetic field generally speaking runs south to north. Near the poles it's actually more into the ground, but usually it's mostly running south to north. Using right hand rule, if magnetic field is running from south to north, and current is going into the page, the field must be towards the right, or towards B.
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#### Nato

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##### Re: VCE Physics Question Thread!
« Reply #69 on: July 06, 2013, 11:16:12 pm »
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Hey guys,

i am having a little trouble getting my head around Newton's third law. so it's the whole equal and opposite reaction thing.
so if an object exerts a force on the another, that object while exert the same force of same magnitude in the opposite direct. what i don't understand is how object are still able to move (i understand how the forces are acting on different and can't be cancelled out).

for example if you kick a ball with 100N, the ball exerts 100N back on you. Where does that *extra* force to make the ball go flying??

thank you guys
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#### sin0001

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##### Re: VCE Physics Question Thread!
« Reply #70 on: July 07, 2013, 12:09:18 am »
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Hey guys,

i am having a little trouble getting my head around Newton's third law. so it's the whole equal and opposite reaction thing.
so if an object exerts a force on the another, that object while exert the same force of same magnitude in the opposite direct. what i don't understand is how object are still able to move (i understand how the forces are acting on different and can't be cancelled out).

for example if you kick a ball with 100N, the ball exerts 100N back on you. Where does that *extra* force to make the ball go flying??

thank you guys
Okay so the forces are identical, but what is the difference here? Something has to be different between you and the ball, because clearly one is moving while the other is stable. The difference is mass; you weigh a lot more than the ball you've kicked and according to the formula: a = F/m, you are going to experience negligible acceleration while the ball is going to experience much greater acceleration, because if 'm' increases, then acceleration of an object will decrease. Therefore, you are going to 'absorb' a reaction force of 100 N by only experiencing negligible acceleration, but this force will cause the ball to go 'flying'.
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#### SocialRhubarb

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##### Re: VCE Physics Question Thread!
« Reply #71 on: July 07, 2013, 05:30:34 pm »
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You exert a force on the ball of 100N. The 'reaction force' which Newton's third law talks about is the ball's force applied to you. That force doesn't affect the ball's acceleration at all, because it doesn't act on the ball. It acts on you. So the original 100N of force which you apply isn't cancelled out by the reaction force from the ball because they act on different objects.
« Last Edit: July 07, 2013, 11:16:08 pm by SocialRhubarb »
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#### Homer

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##### Re: VCE Physics Question Thread!
« Reply #72 on: July 08, 2013, 02:45:18 pm »
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What would be the direction of the magnetic force and how do we work it out?
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#### jssantucci

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##### Re: VCE Physics Question Thread!
« Reply #73 on: July 08, 2013, 03:58:21 pm »
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Could someone please explain why when the secondary coil in a transformer is connected to a circuit with an open switch (i.e. one with no load) the energy used by the primary coil is zero? Wouldn't the power being dissipated by the primary coil always just be IxV regardless of what's going on in the secondary circuit? This is in relation to question 9 of chapter 10.6 in Heinemann.

Thanks

#### SocialRhubarb

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##### Re: VCE Physics Question Thread!
« Reply #74 on: July 08, 2013, 05:07:45 pm »
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Unfortunately I don't really understand the physics behind this but at the very least I can show you with the formulas.

Let the primary current be $I_P$, the secondary current be$I_S$, the turns on the primary coil be $N_P$ and the turns on the secondary coil be $N_S$.

We know that $I_S=0$, because the resistance across an open circuit is infinite, and $I=\frac{V}{R}$.

We also know that $\frac{I_S}{I_P}=\frac{N_P}{N_S}$.

Therefore, $I_P=\frac{N_S}{N_P}I_S=0$.

Thus, since $I_P=0$ and $P=VI$, the power dissipated in the primary circuit must be 0.

Like I said before, I don't actually understand the physics behind this, but somehow the secondary circuit must somehow influence the primary circuit? Maybe someone else can help with that.
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