Question

The attractive force between any two masses is the Gravitational Constant 6.67 x 10-11 N kg2/m2. How would this look in standard notation? 66,700,000,000,000 N kg2/m2 667,000,000,000 N kg2/m2 N kg2/m2 .00000000667 N kg2/m2 .0000000000667 N kg2/m2

Answer 1

`G=6.67* 10^(-11)\ Nkg^2/m^2=0.0000000000667\ Nkg^2/m^2`

Step-by-step explanation:

The attractive force acting between any tow masses is given by :

`F=G(m_1m_2)/(r^2)`

G is the universal gravitational constant

The value of G is,
`G=6.67* 10^(-11)\ Nkg^2/m^2`

We need to write the value of G in standard notation. A number is written in scientific notation as :

`N=a* 10^b`

The given value of G is in scientific notation. Its standard notation is given by :

`G=6.67* 10^(-11)\ Nkg^2/m^2=0.0000000000667\ Nkg^2/m^2`

So, the correct option is (d). Hence, this is the required solution.

Answer 2

Answer:
`0.0000000000667 N (kg^(2))/(m^(2))`

Step-by-step explanation:

According to the Universal Law of Gravitation:

The force
`F` exerted between two bodies of masses
`m1` and
`m2` and separated by a distance
`r` is equal to the product of their masses and inversely proportional to the square of the distance.

Written in a mathematicall form is:

`F=G((m1)(m2))/(r^2)`

If we rewrite this formula:

`G=(Fr^2)/((m1)(m2))`

Where
`G=6.67(10)^(-11)N (kg^(2))/(m^(2))` is the gravitational constant, which in standard notation is:

`0.0000000000667 N (kg^(2))/(m^(2))`