IB Smacher Glau Posted October 4, 2012 Report Share Posted October 4, 2012 M091.H1.08Why the answer is A ? Shouldn't it be answer B because we know the potential gradient is equal in magnitude to the field strength Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted October 4, 2012 Report Share Posted October 4, 2012 M091.H1.08Why the answer is A ? Shouldn't it be answer B because we know the potential gradient is equal in magnitude to the field strengthThe work done by the external force is q times the difference between the electric potential at d=2r and d=r. This potential difference is equal to the integral of the electric field from 2r to r, ie. the shaded area under graph 1. (Answer A) 1 Reply Link to post Share on other sites More sharing options...
IB Smacher Glau Posted October 5, 2012 Report Share Posted October 5, 2012 Riviera radio transmit on carrier frequency of 106.5kHz and plays music with a frequency range from 100Hz to 2000Hz. Find the bandwidth and the max and min frequencies ? Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted October 5, 2012 Report Share Posted October 5, 2012 Riviera radio transmit on carrier frequency of 106.5kHz and plays music with a frequency range from 100Hz to 2000Hz. Find the bandwidth and the max and min frequencies ?Riviera Radio (in Monaco) transmits on 106.5 MHz (not kHz) using FM. If the station is transmitting music, the frequency range must be 100 Hz to 20 kHz, not to 2000 Hz. (= 2 kHz!)Assuming the standard FM broadcast frequency deviation of 75 kHz, the bandwidth of the transmitted signal is then approximately 2(75 + 20) kHz = 190 kHz.The maximum and minimum frequencies are respectively (106.5 + 0.095) = 106.595 MHz and (106.5 - 0.095) = 106.405 MHz.(This is only an approximation. Actually there are sidebands outside this frequency range but they are of a lower order) 1 Reply Link to post Share on other sites More sharing options...
IB Smacher Glau Posted October 5, 2012 Report Share Posted October 5, 2012 Riviera radio transmit on carrier frequency of 106.5kHz and plays music with a frequency range from 100Hz to 2000Hz. Find the bandwidth and the max and min frequencies ?Riviera Radio (in Monaco) transmits on 106.5 MHz (not kHz) using FM. If the station is transmitting music, the frequency range must be 100 Hz to 20 kHz, not to 2000 Hz. (= 2 kHz!)Assuming the standard FM broadcast frequency deviation of 75 kHz, the bandwidth of the transmitted signal is then approximately 2(75 + 20) kHz = 190 kHz.The maximum and minimum frequencies are respectively (106.5 + 0.095) = 106.595 MHz and (106.5 - 0.095) = 106.405 MHz.(This is only an approximation. Actually there are sidebands outside this frequency range but they are of a lower order)Thnxs for answering again I toke this question from OSC IB revision guides (Option F: Communcation )If I do the calculations in kHz the bandwidth = 4kHz since Bandwidth = 2Fs( frequency signal ) which will equal to 2 *2000=4kHzand similarly for Max & Min Max = ( Fc+Fs) = 106.5kHz + 2000Hz= 108.5kHzand Min = ( Fc-Fs) = 106.5 - 2000Hz = 104.5kHzSo what I did above yields to the answer given in the book Is it correct ? Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted October 5, 2012 Report Share Posted October 5, 2012 Riviera radio transmit on carrier frequency of 106.5kHz and plays music with a frequency range from 100Hz to 2000Hz. Find the bandwidth and the max and min frequencies ?Riviera Radio (in Monaco) transmits on 106.5 MHz (not kHz) using FM. If the station is transmitting music, the frequency range must be 100 Hz to 20 kHz, not to 2000 Hz. (= 2 kHz!)Assuming the standard FM broadcast frequency deviation of 75 kHz, the bandwidth of the transmitted signal is then approximately 2(75 + 20) kHz = 190 kHz.The maximum and minimum frequencies are respectively (106.5 + 0.095) = 106.595 MHz and (106.5 - 0.095) = 106.405 MHz.(This is only an approximation. Actually there are sidebands outside this frequency range but they are of a lower order)Thnxs for answering again I toke this question from OSC IB revision guides (Option F: Communcation )If I do the calculations in kHz the bandwidth = 4kHz since Bandwidth = 2Fs( frequency signal ) which will equal to 2 *2000=4kHzand similarly for Max & MinMax = ( Fc+Fs) = 106.5kHz + 2000Hz= 108.5kHzand Min = ( Fc-Fs) = 106.5 - 2000Hz = 104.5kHzSo what I did above yields to the answer given in the book Is it correct ?This would be correct for a long-wave AM transmitter on 106.5 kHz transmitting a speech signal of maximum frequency 2 kHz. But Riviera Radio is a VHF FM transmitter on 106.5 MHz, and 2 kHz is not high enough for a good reproduction of music! 1 Reply Link to post Share on other sites More sharing options...
eelnedross Posted October 10, 2012 Report Share Posted October 10, 2012 This question was easy, but i'm not sure at the moment with question f)ii). Does anyone know how to solve it? I've tried everything but nothing happens. I only get answers around the 1.0V range, but the answer is around 0.4V. Thanks! Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted October 10, 2012 Report Share Posted October 10, 2012 This question was easy, but i'm not sure at the moment with question f)ii). Does anyone know how to solve it? I've tried everything but nothing happens. I only get answers around the 1.0V range, but the answer is around 0.4V.Thanks!The same current flows through each of the series-connected lamps, and the sum of the voltages across each of them is 12 V. From the V-I characteristic, this occurs at a current of 0.41 A, when the voltage across lamp A is 9.2 V, and that across lamp B is 2.8 V.For question f(iii), the power dissipated by lamp A is (9.2 x 0.41) = 3.77 W, whereas that dissipated by lamp B is (2.8 x 0.41) = 1.15 W. 1 Reply Link to post Share on other sites More sharing options...
IB Smacher Glau Posted October 11, 2012 Report Share Posted October 11, 2012 M112.H2.B1B This question about orbital motion part (b) (bi) Total energy isn't it = -Gmm/2r how do they found it using -Vm/2 ? Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted October 11, 2012 Report Share Posted October 11, 2012 M112.H2.B1BThis question about orbital motion part (b)(bi) Total energy isn't it = -Gmm/2r how do they found it using -Vm/2 ?Gravitational potential V = -GM/rSo total energy = -GMm/2r = Vm/2From the graph, at r = 1.0 x 10ˆ7, V = -4 x 10ˆ7Hence total energy = (-4 x 10ˆ7 x 8.2 x 10ˆ2)/2 = -1.64 x 10ˆ10 J Reply Link to post Share on other sites More sharing options...
IB Smacher Glau Posted October 12, 2012 Report Share Posted October 12, 2012 M091.H2.B3Bbii) and biii) ?? Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted October 12, 2012 Report Share Posted October 12, 2012 M091.H2.B3Bbii) and biii) ??Can you clarify to which questions of which part of the paper you are referring? Reply Link to post Share on other sites More sharing options...
IB Smacher Glau Posted October 13, 2012 Report Share Posted October 13, 2012 M091.H2.B3Bbii) and biii) ??Can you clarify to which questions of which part of the paper you are referring?This question is about gravitational fields and potential part b ) A binary star system consists of two stars with masses M1 and M2 rotating about a common centre. The centres of the two stars are separated by a distance R = 1.2 × 1010 m.bii) (ii) determine the distance x at which the gravitational field strength due to the two stars is zero. Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted October 13, 2012 Report Share Posted October 13, 2012 This question is about gravitational fields and potentialpart b ) A binary star system consists of two stars with masses M1 and M2 rotating about a common centre. The centres of the two stars are separated by a distance R = 1.2 × 1010 m.bii) (ii) determine the distance x at which the gravitational field strength due to the two stars is zero.The gravitational field strength due to the two stars is zero at distance x = 4.8 x 10ˆ9 m, where the rate of change of the gravitational potential with distance is zero. (The separation of the stars is 1.2 x 10ˆ10 m, not 1.2 x 1010 m!) Reply Link to post Share on other sites More sharing options...
IB Smacher Glau Posted October 15, 2012 Report Share Posted October 15, 2012 A proton and an alpha particle have the same de Broglie wavelength.Which of the following is approximately the ratio ( Speed of alpa particle divided by Speed of proton ) a) 1/4b) 1/2c)2d) 4? Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted October 15, 2012 Report Share Posted October 15, 2012 A proton and an alpha particle have the same de Broglie wavelength.Which of the following is approximately the ratio ( Speed of alpa particle divided by Speed of proton )a) 1/4b) 1/2c)2d) 4?Since the two particles have the same de Broglie wavelength, their momenta are equal. As the mass of the alpha particle is 4 times that of the proton, the ratio of the speed of the alpha particle to that of the proton must therefore be = 1/4. (Answer a) 1 Reply Link to post Share on other sites More sharing options...
IB Smacher Glau Posted October 20, 2012 Report Share Posted October 20, 2012 Q) Radioactive element P has a half-life of 30 days and element Q has a half-life of 20 days. Initially a radioactive source contains equal numbers of each element.What is the ratio ( number of atoms of P / number of atoms of Q) after 60 days have elapsed?A. 1/2B. 2/3C. 3/2D. 2Why the answer D ? shouldn't it beB ? Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted October 20, 2012 Report Share Posted October 20, 2012 Q) Radioactive element P has a half-life of 30 days and element Q has a half-life of 20 days. Initially a radioactive source contains equal numbers of each element.What is the ratio ( number of atoms of P / number of atoms of Q) after 60 days have elapsed?A. 1/2B. 2/3C. 3/2D. 2Why the answer D ?shouldn't it beB ?60 days is 2 half-lives for element P, but 3 half-lives for element Q.So after 60 days, the P atoms have been reduced by a factor of 2ˆ2 = 4 and the Q atoms by a factor of 2ˆ3 = 8. So there are twice as many P atoms as Q atoms left. (Answer D) Reply Link to post Share on other sites More sharing options...
unicornication Posted October 22, 2012 Report Share Posted October 22, 2012 For the following graph, would you put minimum maximum lines as polynomials, or as they are shown in orange and grey on the graph below? Or are they even necessary? (context: ohm's law testing on a light bulb) Thanks! Reply Link to post Share on other sites More sharing options...
Drake Glau Posted October 22, 2012 Author Report Share Posted October 22, 2012 For the following graph, would you put minimum maximum lines as polynomials, or as they are shown in orange and grey on the graph below? Or are they even necessary? (context: ohm's law testing on a light bulb)Thanks!I always thought those lines were your best fit line values with the +/- values attached to it. So it would basically be your best fit line +0.2 or -0.2 since it's just a y shift.Also your current min/max lines seem to end up in the wrong spot? Your gray line goes from the max value of v=0 to your min value of 3.6(ish?)Something semi-related, why is your best fit line a quadratic? Ohm's law is linear Reply Link to post Share on other sites More sharing options...
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