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:
velocity of the object
Explanation:
For an object moving at a constant acceleration, we would expect to see a position graph with a curved shape and a velocity graph with a straight shape.
Incompletevquestion. However, I inferred from a general perspective about perpendicular lines.
<u>Explanation:</u>
Put simply, <u>perpendicular lines</u> are lines that are at right angles (90°) to each other. Thus, we could say based on this definition that for lines lll and mmm to be perpendicular they intersect and be at right angles (90°) to each other as <u>found on the attached image.</u>
Answer:

Explanation:
Impulse on an object is given by
.
However, it's also given as change in momentum (impulse-momentum theorem).
Therefore, we can set the change in momentum equal to the former formula for impulse:
.
Momentum is given by
. Because the truck's mass is maintained, only it's velocity is changing. Since the truck is being slowed from 26.0 m/s to 18.0 m/s, it's change in velocity is 8.0 m/s. Therefore, it's change in momentum is:
.
Now we plug in our values and solve:
(two significant figures).
A nuclear particle with no charge is a neutron