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
I will say increasing temperature but you dont have that option on your list, so I would take B. Increasing Concentration.
When v= 4.8 m/s:
E.K=mv²/2
=3,1(4.8)²/2
=35.712J
when v= 2(4.8):
E.K=mv²/2
=3.1(9.6)²/2
=142.848J
Answer:
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Explanation:
When object density is lesser than liquid density, it floats....BUT when it's density is greater than the liquid density, then it sinks...
