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HiggsHunter

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HiggsHunter last won the day on October 29 2012

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  1. The displacements of the beads in Diagram 2 from their initial positions in Diagram 1 are: 0cm, 0.5cm, 0.8cm, 0.5cm, 0cm, -0.5cm, etc So the best estimate of the amplitude is 0.8cm. (Answer B)
  2. Since bulb A is open circuit, the current in the circuit flows only through C and D in series. As these bulbs have equal resistance, the potential difference across C is half of 30v, i.e. 15v. Since there is no current through bulb B due to the open circuit in A, this 15v potential difference also appears across the bulb A.
  3. Resistance of lamp in parallel with YZ = (4x12)/(4+12) = 3 ohms. Hence potential difference across lamp = 3(3/(3+12)) v = 0.6v.
  4. The emf generated is equal to the rate of change of the magnetic flux through the coil multiplied by the number of turns in the coil, which is the rate of change of the magnetic flux linkage. (Answer a)
  5. b (i) Resistance between A and B = 6*(6+6)/(6+6+6) = 4 ohms b (ii) Resistance between C and D = 6 + 4*6/(4+6) = 8.4 ohms
  6. Since a = F/m, the acceleration is 1m/s2 for t = 0 to 5s and -1m/s2 after 5s.
  7. For the slower cyclist's speed v, solve v^2 + 10v - 450 = 0 Speed of slower cyclist = 16.8km/h, time 53.6 minutes Speed of faster cyclist = 26.8km/h, time 33.6 minutes how did you get to v^2 + 10v - 450 = 0? v(15/v) + 10(15/v) - v/3 - 10/3 = 15
  8. For the slower cyclist's speed v, solve v^2 + 10v - 450 = 0 Speed of slower cyclist = 16.8km/h, time 53.6 minutes Speed of faster cyclist = 26.8km/h, time 33.6 minutes
  9. Are you referring to the Physics Guide published by the IBO? If so, the March 2007 edition (for first examinations 2009) is still the latest version to have been published. It is available in the Files section: http://www.ibsurviva...1-physics-2009/ He has to be a VIP to DL from file section.. Unless the rules have been changed, he should not have to be a VIP to download an IBO document from that part of the Files section.
  10. Are you referring to the Physics Guide published by the IBO? If so, the March 2007 edition (for first examinations 2009) is still the latest version to have been published. It is available in the Files section: http://www.ibsurviva...1-physics-2009/
  11. 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.
  12. Voltage across R = (1.44*10ˆ-18)/(1.6*10ˆ-19) = 9V Hence R = 9/((12-9)/5) = 15 ohms
  13. 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.
  14. 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.
  15. 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.
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