Answer:
there are 1,000m in a km, so 200km is 200,000m
200,000m/10m/s = 20,000s
Explanation:
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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
Answer: 109.89 Nm
Explanation:
The maximum torque will be calculated as the force multiplied by the perpendicular distance. This will be:
Torque = force × perpendicular distance
torque = 333 × 0.33
= 109.89 Nm
Answer:
1.56 J
Explanation:
The potential energy only depends on the vertical height from the ground level.
We consider the ground level to have zero P.E.
So when it is 2 m above the ground level,
P.E. = mgh
= 0.078×10×2
= 1.56 J