HiggsHunter Posted April 30, 2012 Report Share Posted April 30, 2012 Erm... how on earth is it sensible to calculate the initial acceleration on a body from the resistive force acting on it!? Ridiculous! Could you give the context of that one? Reply Link to post Share on other sites More sharing options...
DropBoite Posted April 30, 2012 Report Share Posted April 30, 2012 (edited) Erm... how on earth is it sensible to calculate the initial acceleration on a body from the resistive force acting on it!? Ridiculous! Could you give the context of that one? Apologies, Ignore that. Higgs, I don't understand this question: Calculate the Weight down the plane of a triangle. Working is W-sintheta. Triangle concerned is attached. I'm sorry, but what the heck does w-sintheta represent!? I makes no sense since that side would be the height which is already given? I'm sorry but this isn't the right image, which is corrupt for some reason now only change is the hypotenuse is 6.4m and the triangle is reducing toward the right, sorry! Edited April 30, 2012 by DropBoite Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted April 30, 2012 Report Share Posted April 30, 2012 Ahaha! I know that, I'm not a fool! But on a serious note, it doesn't seem enough to use those, these IB people are twisted as are their equations... >.> Well the equations of motion describe how some things in the physical world behave, and they are not the invention of the IB! Before Newton, great minds struggled for centuries to understand the phenomena that are described so simply by these equations, and the inventions of the modern world use them in every direction you look. But I agree that some of the physics exam questions seem a little convoluted, although many of them are nice examples to allow you to show that you have mastered the material. Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted April 30, 2012 Report Share Posted April 30, 2012 Higgs, I don't understand this question: Calculate the Weight down the plane of a triangle. Working is W-sintheta. Triangle concerned is attached. I'm sorry, but what the heck does w-sintheta represent!? I makes no sense since that side would be the height which is already given? I'm sorry but this isn't the right image, which is corrupt for some reason now only change is the hypotenuse is 6.4m and the triangle is reducing toward the right, sorry! I presume that the hypotenuse is 6.4km, not 6.4m. The weight W is a gravitational force acting vertically downwards. So the component of this force that acts in the direction down the slope is W x sin (theta), where theta is the angle of the slope. If you are given that the height is 0.3km and the hypotenuse is 6.4km, then sin (theta) = 0.3/6.4 = 0.047 and the component of the force acting down the slope is 0.047 x W. Reply Link to post Share on other sites More sharing options...
DeBrogliez Posted May 6, 2012 Report Share Posted May 6, 2012 The rms current rating of an electric heater is 4 A. What direct current would produce the same power dissipation in the electric heater?A. 4/sqrt(2) AB. 4 AC. sqrt(2)*4 AD. 8 AWhy is the answer BI thought of C because:I/sqrt(2) = Irms Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted May 6, 2012 Report Share Posted May 6, 2012 Yes, answer B is correct. With an AC supply, the instantaneous power varies throughout each cycle. The rms current rating is the constant DC current that would cause the same average power dissipation as the AC current. As you have noted, the peak AC current is sqrt(2) times higher than the rms current. The peak AC voltage is also sqrt(2) times higher than the rms voltage. 2 Reply Link to post Share on other sites More sharing options...
DeBrogliez Posted May 6, 2012 Report Share Posted May 6, 2012 Hmm I get it now, thank you Maybe it would be better if they word it like this: The rms current rating of an electric heater is 4 A. What direct current would produce the same average power dissipation in the electric heater? Because theoritically speaking the electric heater could have a higher max power if provided with with rms current of 4A rather than direct current of 4A, am I right? Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted May 6, 2012 Report Share Posted May 6, 2012 Yes, it is assumed that for a heater it is the average power that is of interest, not the instantaneous peak power that occurs twice per cycle with an AC supply. The higher peak voltage (1.414 x the rms voltage) would be significant for the insulation of the electrical wiring, and the voltage rating of any auxiliary components in the heater. Reply Link to post Share on other sites More sharing options...
DeBrogliez Posted May 6, 2012 Report Share Posted May 6, 2012 Thank You! I have another question if you don't mind It is about Topic 5: Since, Power = I^2 * R and Power = V^2 / R This makes me wonder whether power and resistance are directly or inversely proportional? Because according to the first equation they seem to be directly proportional, but not on the second! I'm confused Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted May 6, 2012 Report Share Posted May 6, 2012 These equations describe how the power dissipated in a fixed resistance R is related to the current I through it or the voltage V across it. If you were to ask the question "How does the power dissipated in a resistance vary with the value of that resistance?", the answer would depend on the internal resistance of the source that was supplying it. Reply Link to post Share on other sites More sharing options...
DeBrogliez Posted May 6, 2012 Report Share Posted May 6, 2012 Let's say the internal resistance in negligibleIf a resistor of resistance 12 ohms is connected in series with the source has power P dissipated in it.What would a 3 ohm resistor have? Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted May 6, 2012 Report Share Posted May 6, 2012 Let's say the internal resistance in negligibleIf a resistor of resistance 12 ohms is connected in series with the source has power P dissipated in it.What would a 3 ohm resistor have?The power dissipated in the 12 ohm resistor P = (Vˆ2)/12.Because the supply has negligible internal resistance, the voltage V remains the same when the resistor is replaced by a 3 ohm one.Hence the power dissipated in the 3 ohm resistor = (Vˆ2)/3 = 4P.Now suppose that the power supply has an EMF of E volts and an internal resistance of 2 ohms.Current with 12 ohm resistor = E/(2 + 12) = E/14Power in 12 ohm resistor = (Eˆ2)12/(14ˆ2)Current with 3 ohm resistor = E/(2 + 3) = E/5Power in 3 ohm resistor = (Eˆ2)3/(5ˆ2)So in this case the power dissipated in the 3 ohm resistor is (3 x 196)/(25 x 12) = 1.96 times the power in the 12 ohm resistor.In general, the power dissipated in a load is greatest when the resistance of the load is equal to the internal resistance of the source. Note that this does not correspond to the condition for maximum efficiency, though! 1 Reply Link to post Share on other sites More sharing options...
DeBrogliez Posted May 7, 2012 Report Share Posted May 7, 2012 Thank you, this was really helpful Can I borrow your mind for the exam day? hahaha Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted May 7, 2012 Report Share Posted May 7, 2012 Thank you, this was really helpful Can I borrow your mind for the exam day? hahaha I'm sure that your young mind has many more lively neurons than my old one, DeBrogliez... Good luck! Reply Link to post Share on other sites More sharing options...
IB Smacher Glau Posted May 8, 2012 Report Share Posted May 8, 2012 I need help .. I dont know but I got I feeling that we will get this question in Physics Exams this yearQ) An object with initial velocity 20 ms^-1 and initial displacement of -75 m experience an acceleration of -2 ms^-2. Draw the displacment time graph of this motion for the first 20 s ? Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted May 8, 2012 Report Share Posted May 8, 2012 The displacement of the object s = -75 +20t -tˆ2, which you can easily plot for t = 0 to 20.The displacement/time graph is a parabola with:maximum s=25 at t=10s=0 at t=5 and t=15and s=-75 at t=0 and t=20(After 20 sec, the object is back where it started) Reply Link to post Share on other sites More sharing options...
IB Smacher Glau Posted May 8, 2012 Report Share Posted May 8, 2012 I am annoyed with the Physics Paper 1 Question the exmainer comment say " wording in the question is important " this question >>The fundamental (first harmonic) frequency for a particular organ pipe is 330 Hz. The pipe is closed at one end but open at the other. What is the frequency of its second harmonic?A. 110 HzB. 165 HzC. 660 HzD. 990 Hzwe should use the equation wavelengh= 4*L right ? I coudn't get the correct answer which is D Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted May 8, 2012 Report Share Posted May 8, 2012 Since the organ pipe is closed at one end and open at the other, the harmonics are odd integer multiples of the fundamental frequency. (There must be a node at the closed end and an antinode at the open end). So the next harmonic after the fundamental is indeed 990 Hz (answer D).But I would still call that the third harmonic, not the second harmonic! 1 Reply Link to post Share on other sites More sharing options...
IB Smacher Glau Posted May 8, 2012 Report Share Posted May 8, 2012 Thanks Man .. The examiner comment included answer C and D as correct ! another one if you dont mind Electrons are accelerated from rest through a potential difference V. Their de Broglie wavelength is λ. The accelerating potential difference is increased to 2 V. Which of the following gives the new de Broglie wavelength? A. 2λ B.SquarRoot λ C. λ / SquarRoot 2 D. λ / 2 again C Reply Link to post Share on other sites More sharing options...
DeBrogliez Posted May 8, 2012 Report Share Posted May 8, 2012 Energy = P^2/2mEnergy = qVSo, qV = P^2/2mSo, P = sqrt(2*m*q*V)Wavelength = h/p= h/(sqrt(2*m*q*V))So the answer C is correct! Reply Link to post Share on other sites More sharing options...
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