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TSmaximR
post May 28 2013, 11:28 AM

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QUOTE(manutd96 @ May 27 2013, 08:31 PM)
We have to find the resistance of every bulb right?
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Hint : Power dissipated = V^2/R
TSmaximR
post May 28 2013, 11:31 AM

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Thank you for your explanation and for your wonderful share ! biggrin.gif
TSmaximR
post May 28 2013, 01:46 PM

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QUOTE(manutd96 @ May 28 2013, 12:13 PM)
Yeah I think so, ans should be D after finding the voltage which is 12 then power which is 144/6 right?
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rclxms.gif

Congratulations . I await Critical_Fallacy's more detailed approach if there is any , if he has the time to do so .

A question :

Are you still in Electricity ? Or have you touched Electromagnetism ?

This post has been edited by maximR: May 28 2013, 01:47 PM
TSmaximR
post May 29 2013, 03:46 PM

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There are so many trigonometric identities to remember if you're just starting to play with them , and deriving some of them is a tedious job .
TSmaximR
post May 30 2013, 04:04 PM

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user posted image

Help please ? I'm not that sure . Thank you .

TSmaximR
post May 30 2013, 04:06 PM

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My gut feeling here is to find the acceleration of the blocks first , then using Block Y's mass and using F = ma , calculate the force acting on it .

Which results in 20N . But I don't have the answer so I need verification .

This post has been edited by maximR: May 30 2013, 04:07 PM
TSmaximR
post May 30 2013, 09:01 PM

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QUOTE(Critical_Fallacy @ May 30 2013, 05:09 PM)
Newton’s second law for a system of particles states that the net external force on a system of particles equals the total mass of the system multiplied by the acceleration of the center of mass.

So, Newton’s 2nd law gives you

Σm*a = Σf = N

(mx + my)*a = fx + fy

Therefore, for Block Y

my*a = fy

where

a = N / (mx + my)

P.S. It should be mentioned that the horizontal surface is frictionless, not just smooth only.
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QUOTE(work_tgr @ May 30 2013, 08:07 PM)
is 20N
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Thank you .

I have several more questions , if they are resolved , I'll be a happy man .

First one :

user posted image

If F is constant , then the trolley's velocity will increase uniformly ( uniform acceleration ) . Since E = 1/2 mv^2 , the graph would be parabolic ( positive parabola ) .

An important one :

In Chapter 2 , Electricity , we have learned that the power lost as heat can be calculated using P = I^2R or P = V^2/R

So when choosing an electrical appliance ( heater , hairdryer ) , in SPM , students are taught to choose an electrical appliance which has the highest resistance ( either the thinnest or longest nichrome wire in the case of a hairdryer ) . Because P = I^2R . When I ask why P = V^2 / R cannot be used , the mark scheme in the past year book is silent , it doesn't provide an answer based on the second equation .

But I learn that if resistance increases , power dissipated will decrease and this holds true for both equations if one takes V = IR into account . ( There is a major error in SPM books )

Today , I learnt about the mechanism of a transformer . One thing that I don't understand is hysteresis .
The other thing is , to reduce power loss in a transformer , resistance of wire should be reduced , so thicker wires should be used as the coils , since P = I^2R .

Why can't P = V^2/R be used ? But if it is used , it'll yield different results .

That's it . Thanks !







TSmaximR
post May 30 2013, 09:41 PM

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QUOTE(Critical_Fallacy @ May 30 2013, 09:21 PM)
Can you give a working example of that? sweat.gif
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I can only give you a conceptual question since SPM questions are mostly conceptual . This is an essay SPM 2009 question . The way to answer is to choose the characteristic [ 1m ] and justify it with an explanation [ 1m ] . Overall 10 marks , 4 characteristics / aspects , plus one final answer about your choice .

user posted image

Are the reasonings correct ? This is provided in past year books , which reflects the SPM's mark scheme .

Meanwhile , about the energy question , is my reasoning correct ?

This post has been edited by maximR: May 30 2013, 09:41 PM
TSmaximR
post May 31 2013, 02:17 PM

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I have two questions :

1 . Based on P = I²R , a higher resistance would imply lower power delivered , if voltage is held constant , because a higher resistance will result in a lower current flow . All of the aspects you've justified are to increase the resistance , but to me , that would decrease the power .

This is the only part in Electricity which I haven't fully understood . It seems like since the power equations are related to Ohm's Law , you can't really see a direct relationship using only one equation . Can you explain ?

2 . Since voltage is assumed to be constant , is that why we use P = I²R instead of P = V²/R ?

TSmaximR
post May 31 2013, 02:19 PM

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QUOTE(Critical_Fallacy @ May 31 2013, 12:50 AM)
By Newtonian mechanics, the mass (m) of the trolley remains unchanged.

constant Force (F) → constant acceleration (a)

constant acceleration → the velocity (v) is increasing linearly

Since the kinetic energy E ∝ v², therefore the graph should be parabolic. Answer: (A)

P.S. Graph C is a damping harmonic response, specifically the overdamped case. The function of this graph is typically f(t) = c*[1 - e^-(σt)], where σ is the resonant angular frequency and c is the steady-state value.
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QUOTE(work_tgr @ May 31 2013, 12:10 PM)
About the trolley, since force is constant, acceleration is constant too.

v=u+at , where u = 0

v= at

and

E=0.5mv^2 = 0.5m(at)^2 = 0.5ma^2 t^2

so

the relationship E and t is same to quadratic function which is a U shape graph .

Answer = A
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Thank you both ! By the way , these are questions from Negeri Sembilan Trial STPM 2009 . I'm happy to have used some Physics knowledge I've gained on some easier STPM questions . smile.gif

TSmaximR
post May 31 2013, 04:40 PM

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Thank you . Sorry for providing the question with the mark scheme though , I had a feeling that it would mislead people into making wrong explanations , but the scanned version of the question in a site has been brought down .

As for the inert gas part , is this conceptual explanation accurate [ the bold part ] ?

The problem with this approach was the evaporation of the tungsten atoms. At such extreme temperatures, the occasional tungsten atom vibrates enough to detach from the atoms around it and flies into the air. In a vacuum bulb, free tungsten atoms shoot out in a straight line and collect on the inside of the glass. As more and more atoms evaporate, the filament starts to disintegrate, and the glass starts to get darker. This reduces the life of the bulb considerably.

In a modern light bulb, inert gases, typically argon, greatly reduce this loss of tungsten. When a tungsten atom evaporates, chances are it will collide with an argon atom and bounce right back toward the filament, where it will rejoin the solid structure. Since inert gases normally don't react with other elements, there is no chance of the elements combining in a combustion reaction.


Based on PV = nkT , a higher pressure of gas in the lamp would result in a higher temperature to heat up the filament , and thus causing an increased evaporation of the filament . A lamp filled with low pressure inert gas would result in a lower temperature for the filament to heat up and give off light and reduce evaporation , is my understanding correct ?

This post has been edited by maximR: May 31 2013, 04:42 PM
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post Jun 3 2013, 04:44 PM

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Google Buletin Sejarah . Download the first and only edition , 2013 .

Seems like the Education Ministry is trying to help us white mice with Sejarah Kertas Tiga .
TSmaximR
post Jun 4 2013, 12:11 PM

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QUOTE(Critical_Fallacy @ May 31 2013, 05:03 PM)
I'm afraid I'm reaching my limit in physics knowledge because I don't conduct inert gas experiment with gas-filled bulbs. Perhaps you can find more information in the following webpages: blush.gif

(1) Operating Principle

(2) Gas Filling Effects

(3) Filament Coiling Effects
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Thank you ! Spent some time going through those and I'm satisfied . smile.gif
TSmaximR
post Jun 4 2013, 12:12 PM

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QUOTE(Boy96 @ Jun 3 2013, 09:51 PM)
Official dates are out. Sorry for spoiling everyone's holiday

http://www.moe.gov.my/lp/files/spm/2013/3_...0SPM%202013.pdf
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The official dates were announced several months ago , you're not spoiling anyone's holidays . sweat.gif
TSmaximR
post Jun 4 2013, 11:10 PM

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QUOTE(RED-HAIR-SHANKS @ Jun 4 2013, 03:38 PM)
rclxms.gif  Yeah,finally complete chapter 4 thermochemistry,looks like I only got 1 chapter left before I can complete the entire syllabus of Form 5 Chemistry. I really hope to finish the whole thing by the end of this weekends, though I lack in doing exercises and questions regarding the Form 5 Chemistry,which will do nothing good to me even if I'm able to finish the whole syllabus. And not to mention that I studied really really slow,which irritates me  doh.gif  Currently I'm entering chapter 5 (Chemicals for Consumers) in the topic of preparation of sodium alkyl sulphate and sodium alkylbenzene sulphonate(it really gives me a big headache just to fathom its formation of long-chain hydrocarbon),by the way I just wanna know how're you guys progress during this holidays? Anyways, make sure you guys have fun and relish this holidays,though  thumbup.gif
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I'm slow when it comes to Chemistry . I'm 3/5 through Oxidation and Reduction , barely touched Carbon Compounds , didn't take a look at Chemical for Consumers . sad.gif

Typing this , I feel a bit bad and worried . I've got Basic Economics to cover , and a lot of gaps in Sejarah .

I'm nearly done with Maths and Physics though .

This post has been edited by maximR: Jun 4 2013, 11:12 PM
TSmaximR
post Jun 5 2013, 02:19 PM

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For future reference . What does this remind you of ?
TSmaximR
post Jun 5 2013, 03:30 PM

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QUOTE(Krevaki @ Jun 5 2013, 02:58 PM)
Why don't you contact schools directly?
Umm, chemistry? PMR science?
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This is actually extracted from a past year GCSE . GCSE and SPM are 99.9% similar , including the mark schemes .

Oh , this is an organic chemistry question .

Any useful guidelines on Carbon Compounds ? I think it's the longest chapter , longer than Salts , Salts is easier .

This post has been edited by maximR: Jun 5 2013, 03:30 PM
TSmaximR
post Jun 5 2013, 03:49 PM

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I don't agree with N.S's decision not to release their Add Maths project question during the holidays . Now all I can do is to watch you all complete your projects while I wait for mine to be announced .
TSmaximR
post Jun 5 2013, 03:58 PM

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QUOTE(Critical_Fallacy @ Jun 5 2013, 03:52 PM)
That's Hydrocarbon Cracking. sweat.gif


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Precisely ! Unfortunately , my school lab doesn't have the necessary materials to conduct the experiment .
The interesting thing is , why , a basic material like broken pieces of pottery can be used to catalyse the cracking process ?
TSmaximR
post Jun 9 2013, 03:18 PM

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QUOTE(1insane @ Jun 9 2013, 02:08 PM)
Is this year's trial set by each school or is every school is gonna sit for the same state paper? Also when are the trials? Aug or early Sep?

Want to get ready for it and dont repeat the same mistakes for this exam. Thanks
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Set by schools . sad.gif

Around August , depends on your state / school .

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