Je2
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Doctor_Einstein
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(Original post by Je2)
I'll give you a hint. Part ii) uses a temperature of 3.7x10^9 K, but fusion occurs at 7.4x10^9 K. Therefore you would expect KE to be double for part iii). Secondly PE = 2KE, so therefore since KE in part iii) is double that for part ii), then PE in part iii) will be 4 times the value of KE you calculated in part ii).
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(Original post by Doctor_Einstein)
I'll give you a hint. Part ii) uses a temperature of 3.7x10^9 K, but fusion occurs at 7.4x10^9 K. Therefore you would expect KE to be double for part iii). Secondly PE = 2KE, so therefore since KE in part iii) is double that for part ii), then PE in part iii) will be 4 times the value of KE you calculated in part ii).
Thanks, i think I get it but why is PE = 2KE?
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astro67
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Thanks, i think I get it but why is PE = 2KE?
The kinetic energy calculated is the KE of one of the particles involved in the interaction. The potential energy is the energy of the two-particle system, so the max potential energy of the system is the sum of the kinetic energy of each of the two particles.
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