the gravitational force of attraction between two bodies at a certain distance is 70 N if the distance of separation between them is doubled then find the percentage change in force of attraction
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Answered by
12
F=G.m1.m2/r^2
F'=G.m1.m2/(2r)^2
So, F'=F/4
Change in force=F-F/4
=3F/4
percentage change = (3F/4)/F*100
=75%
F'=G.m1.m2/(2r)^2
So, F'=F/4
Change in force=F-F/4
=3F/4
percentage change = (3F/4)/F*100
=75%
Incredible29:
n as i said that F=x
so F' =F/4
did u get it???
ohhh got the point thanks for spending your valuable time in my question bye
its my pleasure...byee
have a good day
yeah...same 2 u
can u mark my answer as brainliest plz???
oh yeah
thnx
Answered by
4
F=G.m1.m2/r^2
F'=G.m1.m2/(2r)^2
So, F'=F/4
Change in force=F-F/4
=3F/4
percentage change = (3F/4)/F*100
=75%
F'=G.m1.m2/(2r)^2
So, F'=F/4
Change in force=F-F/4
=3F/4
percentage change = (3F/4)/F*100
=75%
thanks paaji
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