The attractive force (F) between two objects varies directly as the product of their masses (M1 and M2) and inversely as the square of their distance (d). If M1 and M2 are both halved and d is doubled , find the ratio of the new value of F to its original value.
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2009-05-29 2:28 am
回答 (2)
2009-05-29 3:25 am
✔ 最佳答案
The answer is as follows:圖片參考：http://i320.photobucket.com/albums/nn347/old-master/090528_2.jpg?t=1243509916
2009-05-29 3:09 am
Let F= q(M1)x(M2) + p/(d^2)--------------------q&p is a integer number
M1&M2both halved , d is doubled
so, newF=q(1/2)(M1)x(1/2)(M2)+p/(2d)^2
=(q/4)(M1)x(M2)+p/(4d^2)
=(1/4)(q(M1)x(M2) + p/(d^2))
=(1/4)F
so newF=(1/4)F
newF/F=1/4
newF:F=1:4
M1&M2both halved , d is doubled
so, newF=q(1/2)(M1)x(1/2)(M2)+p/(2d)^2
=(q/4)(M1)x(M2)+p/(4d^2)
=(1/4)(q(M1)x(M2) + p/(d^2))
=(1/4)F
so newF=(1/4)F
newF/F=1/4
newF:F=1:4
參考: ME
收錄日期: 2021-04-29 18:22:08
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