the ans is 4 times but how? can anyone explain..?
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initially, let the masses b m1 & m2 and separation dist. be 'r' . A/Q 'r' is unchanged .. initial force of attraction = Gm1m2/r² .. now if both masses are doubled. then F'= G(2m1)(2m2)/r²=4Gm1m2/r²=4F
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Heya...!!
initially
gravitational force = Gm1m2/r^2
It is given that both mass are doubled
i.e = m1 = 2 times m1
m2 = 2 times m2
so ,
new force F' = G×4×m1m2/r^2
so new force F' will be 4 times the original Force F .
i.e F'= 4F
HOPE IT HELPS U
initially
gravitational force = Gm1m2/r^2
It is given that both mass are doubled
i.e = m1 = 2 times m1
m2 = 2 times m2
so ,
new force F' = G×4×m1m2/r^2
so new force F' will be 4 times the original Force F .
i.e F'= 4F
HOPE IT HELPS U
rohit710:
pls mark my answer as brainliest
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