Question

(6.67×10−11)(5.97×1024)(6.38×106)2. Determine the values of a and k when the value of this expression is written in scientific notation.

Answer 1

a = 9.78

k = 0

Explanation:

This is the magnitude of the acceleration experienced by a freely falling body near the surface of the earth, expressed in meters per second squared ( m/s2 ). This acceleration is usually denoted by the symbol g . You may learn later that this acceleration is related to the mass Me and radius Re of the earth as g=GMe/R2e , where G=6.67×10−11N⋅m2/kg2 is called the gravitational constant, postulated by Sir Isaac Newton and first measured by Henry Cavendish in 1798.

Answer 2

The general scientific form = a *
`10^(k)`

`(6.67 * 10^(-11) )(5.97 * 10^(24)  )(6.38 * 10^(6) )^(2)`

=>
`(6.67 * 10^(-11) ) (5.97 * 10^(24) )(6.38 * 10^(6)) (6.38 * 10^(6) )`

=
`1620.845138 * 10^(-11+24+6+6)`

=
`1620.845138 * 10^(25)`

=
`1.620 * 10^(28)`

hence , a = 1.620 and k= 28