The structure of a water molecule is shown in figure (9-E1). Find the distance of the center mass of the molecule from the center of the oxygen atom.
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ANSWER::
Given:-
m₁=1 gm m₂=1 gm
x₁= - (0.96 x 10⁻¹⁰) sin 52° x₂= - (0.96 x 10⁻¹⁰)sin52° x₃= 0
y₁=0 y₂=0 y₃=(0.96 x 10⁻¹⁰) cos52°
So , the position of centre of mass is =
[(m₁x₁+m₂x₂+m₃x₃)/(m₁+m₂+m₃) , (m₁y₁+m₂y₂+m₃y₃)/(m₁+m₂+m₃)]
=[(- (0.96 x 10⁻¹⁰) x sin 52°+ (0.96 x 10⁻¹⁰) sin 52°+16 x 0) / (1+1+16) , (0+0+16(0.96 x 10⁻¹⁰) cos 52°)/(18)]
=(0,(8/9)0.96 x 10⁻¹⁰cos 52°)
Hope it helps!
ANSWER::
Given:-
m₁=1 gm m₂=1 gm
x₁= - (0.96 x 10⁻¹⁰) sin 52° x₂= - (0.96 x 10⁻¹⁰)sin52° x₃= 0
y₁=0 y₂=0 y₃=(0.96 x 10⁻¹⁰) cos52°
So , the position of centre of mass is =
[(m₁x₁+m₂x₂+m₃x₃)/(m₁+m₂+m₃) , (m₁y₁+m₂y₂+m₃y₃)/(m₁+m₂+m₃)]
=[(- (0.96 x 10⁻¹⁰) x sin 52°+ (0.96 x 10⁻¹⁰) sin 52°+16 x 0) / (1+1+16) , (0+0+16(0.96 x 10⁻¹⁰) cos 52°)/(18)]
=(0,(8/9)0.96 x 10⁻¹⁰cos 52°)
Hope it helps!
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19
Answer:
6.6x10^-12 m
Explanation:
Step-by-step explanation is shown in the above photo....
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