plzz solve Q v . a. b. c. in the following image
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The original force.
F = m₁m₂/r²
a) If r becomes 2r,
F = m₁m₂/(2r)² = m₁m₂/4r²
Here, the difference between the earlier and new force = 1/4
So,the force decrease by 1/4 times
b) If m₁ and m₂ becomes 2m₁ and 2m₂
F = 2m₁ x 2m₂/r² = 4m₁m₂/r²
Here the difference between earlier and new force = 4
So,the force increases by 4 times
c) Newtons' Law of Universal Gravitation is used. It states that
Every object attracts every other object with a force that is
a) directly proportional to product of their masses
b) inversely proportional to square of distance between their centers.
F = m₁m₂/r²
a) If r becomes 2r,
F = m₁m₂/(2r)² = m₁m₂/4r²
Here, the difference between the earlier and new force = 1/4
So,the force decrease by 1/4 times
b) If m₁ and m₂ becomes 2m₁ and 2m₂
F = 2m₁ x 2m₂/r² = 4m₁m₂/r²
Here the difference between earlier and new force = 4
So,the force increases by 4 times
c) Newtons' Law of Universal Gravitation is used. It states that
Every object attracts every other object with a force that is
a) directly proportional to product of their masses
b) inversely proportional to square of distance between their centers.
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G = m1 . M2/r^2
so u can see that here,
g is directly proportional to masses and inversely proportional to the radius
therefore when 1) radius is doubled force of gravitation will become 1/4 of the first value
2)m1 and m1 are double force of gravitation will become 4 times the first value
3) law used is law of universal gravitation
so u can see that here,
g is directly proportional to masses and inversely proportional to the radius
therefore when 1) radius is doubled force of gravitation will become 1/4 of the first value
2)m1 and m1 are double force of gravitation will become 4 times the first value
3) law used is law of universal gravitation
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