Compare the minimum energies required to remove a neutron from Ca, zzCa and 23Ca, using m40=37225.15 MeV/c2, m41=38156.36 MeV/c?, m42=39084.46 MeV/c?, m43=40016.10 MeV/c and mn=939.57 MeV/c2. Why there is a distinct increase of this energy for Ca?

Explanation:

From the given information, we are to compare the minimum energies required to remove a neutron from  ,    and  

To start with   ; the minimum  energy required to remove a neutron from is :

=

= (38156.36 - 37225.15 -939.57) MeV

= -8.36 MeV   Since energy is being given out

Thus, E = 8.36 MeV

the minimum  energy required to remove a neutron from is :

=

= ( 39084.46 - 38156.36 - 939.57) MeV

= -11.47 MeV     Since energy is being given out

Thus, E = 11.47 MeV  

the minimum  energy required to remove a neutron from is :

=

= (40016.10 -39084.46 -939.57) MeV

=  - 7.93 MeV     Since energy is being given out

Thus, E =7.93  MeV

Why there is a distinct increase of this energy for ?

This is as a result of electronic configuration of which posses the same number of proton and neutron, as such, tends to acquire more stability. For this reason, it will be difficult to remove a neutron from .

i don't understand. explain it clearly

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer