The solution for this problem is through this formula:Ø = w1 t + 1/2 ã t^2
where:Ø - angular displacement w1 - initial angular velocity t - time ã - angular acceleration
128 = w1 x 4 + ½ x 4.5 x 5^2 128 = 4w1 + 56.254w1 = -128 + 56.25 4w1 = 71.75w1 = 71.75/4
w1 = 17.94 or 18 rad s^-1
w1 = wo + ãt
w1 - final angular velocity
wo - initial angular velocity
18 = 0 + 4.5t t = 4 s
All of the orbitals in a given subshell have the same value of the "<span>magnetic and principal" quantum number
Hope this helps!</span>
Answer:
The correct answer is option D
KE = 1/2mv^2
KE= 1/2(2)(5)^2
KE= 25 J
