The federal reserve goes at every 2 percent.
Cobalt has an atomic number (Z) of 27, which means the nuclei of all its isotopes have 27 protons. Cobalt 60 has an atomic mass of 60, so it has 60-27 = 33 neutrons.
The mass of 27 isolated protons plus the mass of 33 isolated neutrons would be:
27*(1.007825 u) + 33*(1.008665 u) = 60.497220 u
The actual mass of the nucleus of 60-Co is 59.933820 u.
Mass defect: 60.497220 u - 59.933820 u = 0.563400 u
The mass defect is equal to the binding energy of a nucleus.
using the fact that 1 u = 931.5 MeV/c^2
(0.563400 u)*(931.5 MeV/u) = 524.807 MeV
Answer:
B
Step-by-step explanation:
7mn³|7m²n³ ---14mn^5
| m---2n²
LCM=7mn³×m×2n²
=14m²n^5
or LCM of 7,14 =14
Highest power of m=2
Highest power of n=5
LCM=14m²n^5
Answer:
Step-by-step explanation:
Given
Required
Evaluate when n = 12
Substitute 12 for n in
Simplify the fraction
So we have:
xk = v + w solve for x
divide both sides by k to isolate the variable x:
x = (v+w)/k
^^This is your answer!
Hope I helped!
