HiggsHunter

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  1. The specific heat capacity of water is 4200 J kgˆ-1 Kˆ-1 and 3.0 dm3 of water weighs 3 kg. So the thermal energy lost in cooling from 45 to 38 C = (45 - 38)*3*4200 = 88,200 J. Hence the average rate is 88,200/(8*60*60) = 3 J sˆ-1.
  2. Voltage across R = (1.44*10ˆ-18)/(1.6*10ˆ-19) = 9V Hence R = 9/((12-9)/5) = 15 ohms
  3. The absorption lines in the spectrum of light from a star can indicate the elements present in its atmosphere, providing the temperature of the star is taken into account. The lines are red shifted if the star is receding, and because of its rotation the Doppler shifts can be different for the part moving towards us and the part that is moving away. But since there are many absorption lines, and their patterns are characteristic, these spectral shifts do not normally hinder their identification.
  4. Pairs of binary stars orbit their common centre of mass. The absorption lines in the spectrum of the light received from spectroscopic binaries are red-shifted when a star is moving away from the earth, and blue-shifted when it is moving towards us. The orientation of the orbital plane of eclipsing binaries results in their periodically blocking each other when viewed from earth, modulating their apparent brightness.
  5. Could you give details of the context of the question? Many transformers (in modulators or amplifiers, for example) are not limited to a single constant frequency.
  6. I would assume that by "second harmonic frequency" the question is referring to the next higher resonant frequency, which would indeed be three times that of the fundamental. So the third harmonic would have a resonant frequency five times that of the fundamental? And what does it mean exactly with "resonant frequency"? The natural frequency of oscillation which gives the highest amplitude for the oscillation, if i'm not mistaken? Thank you as always Mr. Higgs! Yes, it is much more common practice to number the harmonics by the corresponding multiple of the fundamental frequency. But the question appears to adopt the less usual nomenclature of Tsokos (see Q3 on page 254) in which the "second harmonic" means the next after the fundamental. So for this question we would have to draw 1N - 1A - 1N - 1A in the second diagram? And in a pipe closed at one end, say, if there is a N - A (fundamental frequency), the second harmonic would be the one that has another N - A and so on? In contrast, in a pipe open at both ends, say, if there is a N - N (fundamental frequency), the second harmonic would be the one that has another N - N and so on? I'm not so sure if you understand me. N: Node A: Antinode Yes, except that the fundamental in the pipe open at both ends would be A - N - A (not N - N) and each higher harmonic would have an additional N - A.
  7. I would assume that by "second harmonic frequency" the question is referring to the next higher resonant frequency, which would indeed be three times that of the fundamental. So the third harmonic would have a resonant frequency five times that of the fundamental? And what does it mean exactly with "resonant frequency"? The natural frequency of oscillation which gives the highest amplitude for the oscillation, if i'm not mistaken? Thank you as always Mr. Higgs! Yes, it is much more common practice to number the harmonics by the corresponding multiple of the fundamental frequency. But the question appears to adopt the less usual nomenclature of Tsokos (see Q3 on page 254) in which the "second harmonic" means the next after the fundamental.
  8. I would assume that by "second harmonic frequency" the question is referring to the next higher resonant frequency, which would indeed be three times that of the fundamental.
  9. 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.
  10. 2 x pi x f x t = 500 x pi x t So f = (500 x pi x t)/(2 x pi x t) = 250 Hz The 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 Hz A = 0 for x = 0, 2, 4, ... (Not x = 0, 4, ...) Hence L = 2m, not 4m.
  11. 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.
  12. 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))
  13. 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.
  14. Use the absolute magnitudes when comparing the luminosities of different stars. The apparent magnitude varies with the distance from the observer.
  15. 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.