Answer:
elastic partial width is 2.49 eV
Explanation:
given data
ER E = 250 eV
spin J = 0
cross-section magnitude σ = 1300 barns
peak P = 20ev
to find out
elastic partial width W
solution
we know here that
σ = λ²× W / ( E × π × P ) ...................1
put here all value
σ = (0.286)² × W × / ( 250 × π × 20 )
1300 × = (0.286)² × W × / ( 250 × π × 20 )
solve it and we get W
W = 249.56 ×
so elastic partial width is 2.49 eV
Answer:
Fnet = 12 N
Explanation:
Force on a point charge due to another point charge = kq1q2 / d^2
Force on +32uC = due to + 20uC + due to -60uC
where uC = 1 x 10^-6 C and k = 9 x 10^9 N m^2 / C^2
Net Force =
Fnet = 12 N
Answer:
Centre of mass of any body is a point where all mass of a body is supposed to be concentrated
it lies in geometrical centre....