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The electric potential between the two charges is 91.68 V.
<h3>
Electric potential between the two charges</h3>
The electric potential between the two charges is calculated as follows;
V = Ed
where;
- V is electric potential
- E is electric field
- d is the distance of the charge
Substitute the given parameters and solve for electric potential,
V = 573 N/c x 0.16 m
V = 91.68 V
Thus, the electric potential between the two charges is 91.68 V.
Learn more about electric potential here: brainly.com/question/26978411
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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
Nonmetals often share or gain
electrons. The nonmetals in the periodic table increases as you move to the
right and decreases as you go down. This is because, the smaller the atom, the
reactive it gets due to less electron attached to the orbits of the atom. The
reactivity of nonmetals is arranged in decreasing order.
<span>
Carbon
</span>
Nitrogen
Oxygen
Fluorine
Phosphorus
<span>
Sulfur</span>
Chlorine
<span>
Selenium</span>
<span>
Bromine</span>
<span>
Iodine</span>
