Question

An apparatus like the one cavendish used to find g has a large lead ball that is 7.7 kg in mass and a small one that is 0.061 kg. their centers are separated by 0.05 m. find the force of attraction between them. the value of the universal gravitational constant is 6.67259 × 10−11 n · m2 /kg2 . answer in units of n.

Answer 1

An apparatus like the one cavendish used to find g has a large lead ball that is 7.7 kg in mass and a small one that is 0.061 kg. their centers are separated by 0.05 m. find the force of attraction between them. the value of the universal gravitational constant is 6.67259 × 10−11 n · m2 /kg2 . answer in units of n.

Answer 2

The force of attraction between them is
`(1.2536).10^(-8)N`

Step-by-step explanation:

We are going to use the Newton's Gravitational Law to solve this exercise.

The law states that :

The gravitational force between two objects is proportional to their masses and inversely proportional to the square of the distance between their centers.

The equation is :

`F_(G)=(G.m_(1).m_(2))/(d^(2))`

Where ''
`G`'' is the universal gravitational constant.

Where ''
`m_(1)`'' and ''
`m_(2)`'' are the masses of the objects.

Where ''
`d`'' is the distance between the centers of the objects.

If we replace all the data in the equation :

`F_(G)=((6.67259).10^(-11)(N.m^(2))/(kg^(2)).(7.7kg).(0.061kg))/((0.05m)^(2))`

`F_(G)=(1.2536).10^(-8)N`