Question

11. Suppose a force F varies as the product of two masses, m and M, inversely as

the square of the distance r between them. If F= 675 when m = 125, M = 225,
and r=0.0530, find the constant of proportionality k.
A. 61.48 X 10-10
B. 1.27X10-
C. 6.74 x 10-5
D. 6.76X10°
E. None

Answer 1

`k = 6.74 * 10^(-5)`

Explanation:

Given

`F\ \alpha\ (m * M)/(r^2)` --- The variation

`F = 675; m = 125; M = 225; r = 0.0530`

Required

Determine the constant of proportionality (k)

`F\ \alpha\ (m * M)/(r^2)`

Express as an equation

`F = k(m * M)/(r^2)`

Multiply both sides by
`r^2`

`Fr^2 = km * M`

Make k the subject

`k = (Fr^2)/(m * M)`

Given that:
`F = 675; m = 125; M = 225; r = 0.0530`

We have:

`k = (675* 0.0530^2)/(125 * 225)`

`k = (1.896075)/(28125)`

`k = 0.000067416`

This can be represented as:

`k = 6.74 * 10^(-5)`