Two people, who may be treated as point masses for this problem, have masses of
ID: 2010369 • Letter: T
Question
Two people, who may be treated as point masses for this problem, have masses of 66 kg and 84 kg. They carry equal but opposite charges. The negative charge is due to 3.5×1010 excess electrons. The individuals are ten m apart. Assume that the people are mostly water for which 6.02×1023 molecules has a mass of 18 g.1.How many molecules of water are there in the more massive person?
2.What is the magnitude of the gravitational force of attraction between the two people?
3.What is the magnitude of the electrical force of attraction between these two people?
4.What is the ratio of the magnitudes of the gravitational force to the electric force? (Take note of the number of charges and the number of water molecules; the result emphasizes the weakness of gravity.)
Explanation / Answer
Assume that the people are mostly water for which 6.02×1023 molecules has a mass of 18 g.1 gram of mass has 6.023*1023 molecues / 18 = 0.33461*1023 molecules 1) molecules of water are there in the more massive person is 84 * 0.33461 = 2.8107*1027 molecules 2) givne mass of two persons m1 = 66kg m2 = 84kg seperation between two persons is r = 10 m gravitational force of attraction between the two people F g = G m1 m2 / r 2 = (6.67*10-11 ) ( 66)(84) / ( 102 ) = 3.6978*10-9 N 3) electrical force of attraction between these two people F = k q1 q2 / r 2 here q1 = q2 = q F = k q 2 / r 2 = ( 9*109 ) ( 3.5*1010 *1.6*10-19 ) 2 / 100 = 2.8224*10-9 N 4)ratio of the magnitudes of the gravitational force to the electric force Fg / F = 3.6978*10-9 / 2.8224*10-9 N = 1.310
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