unicornication Posted October 22, 2012 Report Share Posted October 22, 2012 (edited) 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 I thought about that, but I thought that'd be a little pointless, considering the only change would be the y-shift, and that isn't very useful, lol Isn't that how it's supposed to work? o.O so as to optimally make the gradients as low and as high as possible (our physics teacher told us to do it this way, haha)Light bulbs don't follow Ohm's law linearly, it's something to do with the excessive heating of the filament, which makes it curve.and thanks for your response :-) Edited October 22, 2012 by unicornication Reply Link to post Share on other sites More sharing options...
eelnedross Posted October 22, 2012 Report Share Posted October 22, 2012 Hey IBS! Another questions for ya!I know how to solve question a). However, i can't seem to solve question b)! Have you guys got any idea?The answer seems to be 250Hz... Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted October 22, 2012 Report Share Posted October 22, 2012 Hey IBS! Another questions for ya!I know how to solve question a). However, i can't seem to solve question b)! Have you guys got any idea?The answer seems to be 250Hz...2 x pi x f x t = 500 x pi x tSo f = (500 x pi x t)/(2 x pi x t) = 250 Hz 1 Reply Link to post Share on other sites More sharing options...
break&burn&end. Posted October 24, 2012 Report Share Posted October 24, 2012 Hellooooo can I please have some help on a Relativity question?"Calculate the total energy of an electron moving with a velocity of 0.9c."I tried using the formula E = (y*mo)c^2 but that didn't give me the right answer... Thanks n_n Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted October 24, 2012 Report Share Posted October 24, 2012 Hellooooo can I please have some help on a Relativity question?"Calculate the total energy of an electron moving with a velocity of 0.9c."I tried using the formula E = (y*mo)c^2 but that didn't give me the right answer... Thanks n_nThe gamma factor is 2.294 and the rest energy of the electron 0.511 MeV.So the total energy of the electron is 2.294 x 0.511 = 1.172 MeV. 1 Reply Link to post Share on other sites More sharing options...
break&burn&end. Posted October 24, 2012 Report Share Posted October 24, 2012 Oh, so I didn't need to multiply it by c^2? Thanks. (: Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted October 24, 2012 Report Share Posted October 24, 2012 Oh, so I didn't need to multiply it by c^2? Thanks. (:Yes, your Physics Data Booklet (which you can consult during the exam) gives the rest mass of the electron as 0.511 MeV/cˆ2. So you just have to multiply by cˆ2 to get the rest energy (0.511 MeV), and then by the gamma factor of 2.294 to get the total energy at 0.9c. Reply Link to post Share on other sites More sharing options...
eelnedross Posted October 25, 2012 Report Share Posted October 25, 2012 A question,is apparent magnitude the same as brightness? Also, for calculating ratios of luminosities, should i use the difference in apparent or absolute magnitudes?On the other hand, say you have a mass of 940 MeV c-2, the energy released in the nuclear reaction is 940 MeV?Thank you very much! My exams are nearly in 2 weeks and i'm worried! Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted October 25, 2012 Report Share Posted October 25, 2012 Also, for calculating ratios of luminosities, should i use the difference in apparent or absolute magnitudes?Use the absolute magnitudes when comparing the luminosities of different stars. The apparent magnitude varies with the distance from the observer. 1 Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted October 25, 2012 Report Share Posted October 25, 2012 On the other hand, say you have a mass of 940 MeV c-2, the energy released in the nuclear reaction is 940 MeV?The energy released would be 940 MeV if the mass *decreases* by 940 MeV*cˆ-2 in the nuclear reaction, assuming that the reaction is spontaneous or that the kinetic energy of any projectile particles is negligible. 2 Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted October 25, 2012 Report Share Posted October 25, 2012 A question,is apparent magnitude the same as brightness?Apparent magnitude m is a non-dimensional number, and the larger the number the dimmer the star.It is related to apparent brightness b by: m = -2.5log(b/(2.52*10ˆ-8)) 1 Reply Link to post Share on other sites More sharing options...
IB Smacher Glau Posted October 25, 2012 Report Share Posted October 25, 2012 http://www.mediafire.com/view/?ejo31ciq71b0d2oHow can I solve CII) Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted October 25, 2012 Report Share Posted October 25, 2012 http://www.mediafire...ejo31ciq71b0d2oHow can I solve CII)Time between two adjacent amplitude maxima = 1/(5*10ˆ3) s = 2*10ˆ-4 s.So period of the carrier wave = (2*10ˆ-4)/(1.8*10ˆ4) = 1.111*10ˆ-8 s.Hence frequency = 1/(1.111*10ˆ-8) Hz = 90 MHz. Reply Link to post Share on other sites More sharing options...
IB-Adam Posted October 26, 2012 Report Share Posted October 26, 2012 Hey IBS! Another questions for ya!I know how to solve question a). However, i can't seem to solve question b)! Have you guys got any idea?The answer seems to be 250Hz...2 x pi x f x t = 500 x pi x tSo f = (500 x pi x t)/(2 x pi x t) = 250 HzThe wavelength is 2*L. The length is found by knowing the first positive X values for which the amplitude is zero, that is when x=0, and x=4. Which means that L = 4 m, and hence the wavelength = 8 m. Now, how did you solve for the velocity in order to get the frequency? Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted October 26, 2012 Report Share Posted October 26, 2012 Hey IBS! Another questions for ya!I know how to solve question a). However, i can't seem to solve question b)! Have you guys got any idea?The answer seems to be 250Hz...2 x pi x f x t = 500 x pi x tSo f = (500 x pi x t)/(2 x pi x t) = 250 HzThe wavelength is 2*L. The length is found by knowing the first positive X values for which the amplitude is zero, that is when x=0, and x=4. Which means that L = 4 m, and hence the wavelength = 8 m. Now, how did you solve for the velocity in order to get the frequency?You are given that y = A*cos(500*pi*t) = A*cos(2 *pi*f*t).Hence f = (500*pi)/(2*pi) = 250 HzA = 0 for x = 0, 2, 4, ... (Not x = 0, 4, ...)Hence L = 2m, not 4m. 1 Reply Link to post Share on other sites More sharing options...
IB Smacher Glau Posted October 27, 2012 Report Share Posted October 27, 2012 (edited) http://www.mediafire...qeziqar8k98dlk0This Question .. Now why the answer is B ? be shoudn't it be C in which induced emf ?Also if they ask for the induced current will the answer be Q to S ? Edited October 27, 2012 by IB Smacher Glau Reply Link to post Share on other sites More sharing options...
IB Smacher Glau Posted October 27, 2012 Report Share Posted October 27, 2012 http://www.mediafire.com/view/?7rubrubbznuc9l5How can this Question be solved without using GDC ? Reply Link to post Share on other sites More sharing options...
Drake Glau Posted October 27, 2012 Author Report Share Posted October 27, 2012 The total watts from the sun is split evenly across the surface area of the orbit. Soooo it'd be (4x10^26)/surface areaSA=4pi(r^2) where r=2x10^11 so r^2=4x10^22Tbh, since it wants an estimate, just dividing the power from the sun by the radius squared gives you 10^4, the extra 4pi in the equation would end up in the denominator which would cause the power to actually be smaller so your best estimate would be less than 10^4 which is 10^3.This is how I would do it since you have nothing but paper 2 Reply Link to post Share on other sites More sharing options...
HiggsHunter Posted October 27, 2012 Report Share Posted October 27, 2012 http://www.mediafire...qeziqar8k98dlk0This Question .. Now why the answer is B ? be shoudn't it be C in which induced emf ?Also if they ask for the induced current will the answer be Q to S ?As the free electrons in the conducting wing move from left to right through the magnetic field, they experience a force that causes some migration in the direction from P to R. As a result there is an induced emf from R to P. (Answer B)This migration stops once the electric field has been established in the wing, and there is no induced current during steady flight. 1 Reply Link to post Share on other sites More sharing options...
eelnedross Posted October 28, 2012 Report Share Posted October 28, 2012 Another question for ya IBS!I thought that a pipe closed at one end only had the first harmonic, third harmonic, fifth harmonic and so on... How would you do this one? Reply Link to post Share on other sites More sharing options...
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