All of them
2006-07-28 09:17:36
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answer #1
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answered by rules27 6
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The issue is not so much how many split but how fast the reaction grows. If the reaction does not grow fast enough then the fuel in the bomb when heat up, expand, deform, etc... killing the reaction. The period T measures how fast the power level can increase by a factor of e. Thus the power P at time t is P=Po*exp(t/T). For a bomb to work the period needs to be very very small, 10^-6 s or less. For a period of 10^-6 the power can increase by a factor of 3*10^43 in a 10000th of a second.
2006-07-31 18:24:42
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answer #2
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answered by sparrowhawk 4
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Most nuclear bombs uses a fussion of atoms instead of fission in a atom bomb but depending on how much plutonium238 or uranium 235 is used in the bomb, after the loose neutron is sent out and a atom in the plutonium or uranium accepts it it splits and sends of 2 or 3 neutrons hitting other atoms this is called fission. After every generation of fission more and more atoms are being split real easing massive amouts of energy and radiation in a chain reaction until the fission is in a controled state, and eventually ripps apart the bomb and sets off the explosives realesing about 12,500 pounds of tnt realesing a heat wave disinigrating anything flamable ten a blast wave realesees energy that puts out most fire folled by the radiation created by the fission.
2006-07-29 08:25:13
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answer #3
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answered by jacob m 1
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Only one or two to start. Each releases several neutrons which split other U235 atoms, and so on, extremely rapidly - chain reaction.
2006-07-28 16:17:21
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answer #4
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answered by helixburger 6
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The number is not so much important as the process by which the reaction starts.
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Watch Pay It Forward or Einstein's Letter to get an idea of exponential distribution!
2006-07-28 16:18:33
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answer #5
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answered by Anonymous
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It starts with one. then the rest is a controlled chain reaction
2006-07-28 16:16:11
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answer #6
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answered by Donald S 3
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one
2006-07-28 16:16:45
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answer #7
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answered by lilmama 4
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y u wanna
2006-07-28 16:17:47
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answer #8
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answered by sweets 2
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