Calculate the energy released by the electron-capture decay of 5727Co. Consider only the energy of the nuclei (ignore the energy of the surrounding electrons). The following masses are given:5727Co: 56.936296u5726Fe: 56.935399uExpress your answer in millions of electron volts (1u=931.5MeV/c2) to three significant figures.

Explanation:

The mass difference is

Δm = m2 - m1

56.936296 u - 56.935399 u

= 0.000897‬ u

The energy released by the electron-capture decay

E = Δmc²

  ( 0.000897‬ u) c² ( 931. 5 MeV /c²÷ 1 u)

= 0.8355555 MeV

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards

amplitude

hope this