An ion’s position vector is initially LaTeX: \vec{r} = 5.0 \hat i - 6.0 \hat j + 2.0 \hat k, r → = 5.0 i ^ − 6.0 j ^ + 2.0 k ^ , and LaTeX: 10\:\text{s}10 s later it is LaTeX: \vec{r} = -2.0 \hat i + 8.0 \hat j - 2.0 \hat k, r → = − 2.0 i ^ + 8.0 j ^ − 2.0 k ^ , all in meters. In unit vector notation, what is its LaTeX: \vec{v}_{avg}v → a v g during the LaTeX: 10\:\text{s}10 s?

Answer: d. 39 cmin the diagram, the approximate length of the minor arc de is 39 centimetres.step-by-step explanation: the formula for the length of an arc in a circle is as displayed below.let the length of the minor arc de = xx = 2( pi )r × angle of sector / 360°from this, we will substitute the values given from the problem into the formula to obtain the length of the minor arc de.x = 2( pi )r × angle of sector / 360°r = radius = 25cmpi as an approximate value = 3.14angle of sector = 120°x = 2( 3.14 )( 25 ) × 120° / 360°x = 157 × 1 / 4x = 39.25x = 39 ( rounded to the nearest whole number as an approximate value )

Ibelieve it is 5,040