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This is an example question in TPRH Physics Ch 3 on center of mass
Example 3-3: An ammonia molecule (NH3) contains 3 hydrogen atoms that are positioned at the vertices of an equilateral triangle. The nitrogen atom lies 38pm (1pm= 1 picometer = 10E-12m) directly above the center of this triangle. If the N:H ratio is 14:1, how far below the N atom is the center of mass of the molecule?
They give you their work showing Ycm= [(14)(ocm) + (3)(38pm)]/(14+3)=7pm
Their answer is 7/38 or 1/6 of the way down from the nitrogen atom towards the plane of the hydrogen.
Im confused as to why the answer just wasnt 7pm?
Example 3-3: An ammonia molecule (NH3) contains 3 hydrogen atoms that are positioned at the vertices of an equilateral triangle. The nitrogen atom lies 38pm (1pm= 1 picometer = 10E-12m) directly above the center of this triangle. If the N:H ratio is 14:1, how far below the N atom is the center of mass of the molecule?
They give you their work showing Ycm= [(14)(ocm) + (3)(38pm)]/(14+3)=7pm
Their answer is 7/38 or 1/6 of the way down from the nitrogen atom towards the plane of the hydrogen.
Im confused as to why the answer just wasnt 7pm?
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