<span>59.027m/s^2
Centripetal force is give by F = mv^2/r
Since
F = ma
The acceleation accoiated with this centripetal force is
a=v^2/r
the radius is found by r=d/2=5.75m
velocity is found by
v=d/t
distance is the circumference of the circle
c=pi*d=11.5*3.14=36.11m
time=60s/30.6rpm=1.96s
v=36.11/1.96=18.423m/s
plugging this back in to
a=v^2/r=(18.423m/s)^2/5.75m=59.027m/s^2</span>
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
The data must be reproducible: it means that the data will be reliable and representative and not that this data would lead to one conclusion and another data would lead to another.
Data being varied, unique or surprising does not have anything to do with acceptability of the conclusion.