WEBVTT 00:00:00.000 --> 00:00:03.240 This is College Physics Answers with Shaun Dychko. 00:00:03.640 --> 00:00:05.960 The critical mass density needed to 00:00:05.960 --> 00:00:07.680 stop the expansion of the universe 00:00:07.680 --> 00:00:11.320 is about 10 to the minus 26 kilograms per cubic meter. 00:00:11.700 --> 00:00:16.700 We will convert this into electron volts per c squared per cubic meter 00:00:16.780 --> 00:00:19.780 by multiplying 10 to the minus 26 kilograms per cubic meter by 00:00:19.780 --> 00:00:24.320 1 atomic mass for every 1.6605 times 10 to the minus 27 kilograms 00:00:24.480 --> 00:00:28.200 and that gives atomic mass units per cubic meter 00:00:28.520 --> 00:00:31.480 and then convert the atomic mass units into 00:00:31.520 --> 00:00:33.740 megaelectron volts per c squared by multiplying by 00:00:33.740 --> 00:00:38.500 931.5 megaelectron volts per c squared for every atomic mass unit 00:00:38.800 --> 00:00:40.240 and then convert this 00:00:40.280 --> 00:00:43.020 megaelectron volts into electron volts by multiplying by 00:00:43.020 --> 00:00:45.780 10 to the 6 electron volts for every megaelectron volt. 00:00:46.060 --> 00:00:49.640 So now we have electron volts per c squared per cubic meter 00:00:49.720 --> 00:00:50.360 and that is 00:00:50.360 --> 00:00:54.700 6 times 10 to the 9 electron volts per c squared per cubic meter. 00:00:55.120 --> 00:00:58.080 Part (b) asks how many neutrinos would you need 00:00:58.260 --> 00:01:00.500 to have this density? 00:01:01.740 --> 00:01:06.400 So it would be one neutrino for every 7 electron volts per c squared 00:01:06.400 --> 00:01:10.100 because we are told this is the supposed mass of a neutrino 00:01:10.320 --> 00:01:15.240 times 5.61 times 10 to the 9 electron volts per c squared for every cubic meter 00:01:15.420 --> 00:01:21.800 and this works out to 8 times 10 to the 8 neutrinos per cubic meter.