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RED-HAIR-SHANKS
post May 29 2013, 04:01 PM

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QUOTE(Boy96 @ May 29 2013, 02:40 PM)
Why my teacher never told us about this? Even my seniors had no idea about it
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Well, I reckon your teacher keeps it as a secret out of personal reasons or, maybe because you guys haven't even question him/her anything regarding the Add Math folio, so he/she just keeps quiet about it, but sooner or later you will have to do it. Btw, I had no idea that your seniors haven't even got a clue about the Add Math folio. Anyways,FYI, the Add Math folio that we're dealing with in this year kinda resembles with our Kerja Kursus Sejarah Tingkatan 3.

This post has been edited by RED-HAIR-SHANKS: May 29 2013, 04:22 PM
RED-HAIR-SHANKS
post May 29 2013, 04:20 PM

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QUOTE(maximR @ May 29 2013, 03:46 PM)
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 .
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biggrin.gif I'm concur with you biggrin.gif I still remember that I spent for about 20 minutes solving a rather confusing and mind-bending question that involves basic identities, but when I look up on the Internet and found an exactly similar question with the ones that I faced, I realized that, its not that confusing or extremely hard, but its rather very tricky,and all it needs is 5 to 6 simple steps to solve it rclxms.gif I reckoned that basic identities is more of a puzzle-solving question rather than a pure mathematical question.

ZerokillerXX
post May 30 2013, 02:33 AM

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recently I found this youtube channel which is quite useful for SPM students. But it only have Maths and Add Maths.. >.<

Msia Free Education

And they even categorized their Add Maths video chapter by chapter in the facebook page
Y=mx+c

This post has been edited by ZerokillerXX: May 30 2013, 02:37 AM
RED-HAIR-SHANKS
post May 30 2013, 08:00 AM

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QUOTE(ZerokillerXX @ May 30 2013, 02:33 AM)
recently I found this youtube channel which is quite useful for SPM students. But it only have Maths and Add Maths.. >.<

Msia Free Education

And they even categorized their Add Maths video chapter by chapter in the facebook page
Y=mx+c
*
Btw Edunation Malaysia also provide numerous tutoring videos that are quiet helpful too thumbup.gif Unlike Msia Free Education which focus solely on maths and Add Maths(though the tutoring videos are really helpful), Edunation Malaysia covers variety of subjects that range from the pure science subjects(Physic,Biology,Chemistry) to accounting subjects rclxms.gif

This post has been edited by RED-HAIR-SHANKS: May 30 2013, 08:11 AM
Critical_Fallacy
post May 30 2013, 09:53 AM

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QUOTE(ZerokillerXX @ May 30 2013, 02:33 AM)
And they even categorized their Add Maths video chapter by chapter in the facebook page
Y=mx+c
Very good! Please continue to share good stuffs here or B.U.M.P. this link to help the students, especially those weak at Elementary Algebra. icon_rolleyes.gif
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
Critical_Fallacy
post May 30 2013, 05:09 PM

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QUOTE(maximR @ May 30 2013, 04:04 PM)
user posted image
QUOTE(maximR @ May 30 2013, 04:06 PM)
Which results in 20N. But I don't have the answer so I need verification.
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.
work_tgr
post May 30 2013, 08:07 PM

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is 20N
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
*
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 !







Critical_Fallacy
post May 30 2013, 09:21 PM

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QUOTE(maximR @ May 30 2013, 09:01 PM)
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

Why can't P = V^2/R be used ? But if it is used , it'll yield different results .
Can you give a working example of that? sweat.gif
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
*
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
Critical_Fallacy
post May 31 2013, 12:03 AM

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QUOTE(maximR @ May 30 2013, 09:41 PM)
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Assumption: All five filament bulbs are designed to operate on the fixed voltage (V). The brightness of the bulb depends on the power delivered to the filament (P = I²R), which depends mainly on the resistance of the filament. Therefore, our interest is the resistance of the filament that depends primarily on two factors: material it is made of, and its shape.

(1) Filament Material
Copper has higher electrical conductivity than Tungsten. Thus, Copper is a better conductor, and the current would pass through the filament without dissipating much energy, which means nothing would light up. Moreover, copper has lower young's modulus than tungsten, which means it will easily break.

(2) Filament Thickness
By R = ρl/A, a thick tungsten filament has lower resistance than an otherwise identical thin tungsten filament.

(3) Filament Length
Similarly, by R = ρl/A, a straight tungsten filament has lower resistance than an otherwise longer coiled tungsten filament.

(4) Inert Gas Pressure
The main cause of the failure of bulb is the evaporation of tungsten, which causes weak spots in the filament. By PV = nRT, the pressure is proportional to the temperature. As the temperature is raised, the filament evaporation rate rises. The inert gas also has the cooling effect.

Conclusion: To produce the brightest light among five, the characteristics of the bulb have to be (1) tungsten filament, (2) thin filament, (3) coiled filament, and (4) low pressure inert gas. And Bulb R has all these features.

EDIT: Please refer to Post #899.

This post has been edited by Critical_Fallacy: May 31 2013, 04:23 PM
Critical_Fallacy
post May 31 2013, 12:50 AM

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QUOTE(maximR @ May 30 2013, 09:01 PM)
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).
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.
work_tgr
post May 31 2013, 12:10 PM

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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
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
*
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

Critical_Fallacy
post May 31 2013, 04:19 PM

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QUOTE(maximR @ May 31 2013, 02:17 PM)
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 ?
I don't run the experiment to verity that but your point is valid. I made mistakes on Post #894 because I was misled by the marking scheme, and you probably should consider making a experimental validation. It is generally true that the brightness of a light bulb is a function of its filaments. For a given voltage of operation, a less resistive filament will draw more current, run hotter and will glow more brightly, as to say getting more incandescent.

Consider the following power rating of two bulbs:

(1) 120V 60W Bulb: The ideal filament resistance must be R = (120V)² / 60W = 240Ω. The measured rms current shall be approximately I = 60W / 120V = 0.5A.

(2) 120V 100W Bulb: The ideal filament resistance must be R = (120V)² / 100W = 144Ω. The measured rms current shall be approximately I = 100W / 120V = 0.8333A.

For more info, read this article Attached File  Basic_Physics_of_the_Incandescent_Lamp.pdf ( 268.68k ) Number of downloads: 4
, and study the experiment on the last page. icon_idea.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
Critical_Fallacy
post May 31 2013, 05:03 PM

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QUOTE(maximR @ May 31 2013, 04:40 PM)
As for the inert gas part , is this conceptual explanation accurate [ the bold part ] ?

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?
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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