WEBVTT
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This is College Physics Answers
with Shaun Dychko.
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The critical mass density needed to
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stop the expansion of the universe
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is about 10 to the minus 26 kilograms
per cubic meter.
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We will convert this into electron volts
per *c squared* per cubic meter
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by multiplying 10 to the minus 26 kilograms
per cubic meter by
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1 atomic mass for every 1.6605 times
10 to the minus 27 kilograms
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and that gives atomic mass units
per cubic meter
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and then convert the atomic
mass units into
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megaelectron volts per *c squared* by
multiplying by
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931.5 megaelectron volts per *c squared*
for every atomic mass unit
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and then convert this
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megaelectron volts into electron volts by
multiplying by
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10 to the 6 electron volts for
every megaelectron volt.
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So now we have electron volts per
*c squared* per cubic meter
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and that is
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6 times 10 to the 9 electron volts
per *c squared* per cubic meter.
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Part (b) asks how many neutrinos
would you need
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to have this density?
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So it would be one neutrino for every
7 electron volts per *c squared*
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because we are told this is the supposed
mass of a neutrino
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times 5.61 times 10 to the 9 electron volts
per *c squared* for every cubic meter
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and this works out to 8 times 10 to
the 8 neutrinos per cubic meter.