00:00Question is, compute the gravitational attraction between two college students of mass 80 and
00:1650 kg respectively 2 meters apart from each other.
00:26Is this for worth worrying about?
00:30First of all, we will make a data is the mass of first student is m1 which is 80 kg and the
00:53mass of second student is m2 which is 50 kg and the distance between them is 2 meters and
01:08the value of capital G is 6.67 into 10 raised power minus 11 Newton into meter square per
01:18kilogram square.
01:20So, we have to find the gravitational force of attraction between them.
01:27So, for solution, the gravitational force of attraction will be equal to G into m1 m2 upon
01:40R square.
01:42So, F will be equal to as the value of G is 6.67 into 10 raised power minus 11 into the
01:52mass of m1 is 80 kg into the mass of m2 which is 50 kg divided by R square as R is equal to
02:052 meters.
02:06So, square on 2 is the gravitational force will be obtained as 6.67 into 10 raised power minus
02:1611 into 80 into 50.
02:20It will be 4000 divided by the square of 2 is 4.
02:28So, the table of 4 on 4 is 1 times and on 4000 is 1000 times.
02:38So, the gravitational force will be obtained as 6.67 into 10 raised power minus 11 is 1000
02:45can be written as 10 cubed Newton.
02:51So, in multiplication, when the bases are same, then we will add the powers.
02:58So, gravitational force will be 6.67 into 10 raised power minus 11 plus 3 Newton.
03:07So, it will be obtained as 6.67 into 10 raised power minus 11 plus 3.
03:14So, it will be minus 8 Newton.
03:17So, this is our required condition is this force is very small.
03:29So, it is not worth worrying about.
03:33It will be pieces of 2-1 and it will adjust to the Manhattan子 numbers from
03:528 to 10.
03:53So, if the rel sink runs away from 120 to 15 years, we will review the
04:00ằngli Giovanna.
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