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
<h3>
Answer:</h3>
49500 kgm/s
<h3>
Explanation:</h3>
Data given;
- First car; Mass = 1100 kg
- Velocity = 30 m/s
- Second car; mass = 1100 kg
- Velocity = 15 m/s
We are required to calculate the total momentum of the system.
- We need to know that momentum is calculated by multiplying the velocity of a body by its mass.
- Therefore;
Momentum of the first car = 1100 kg × 30 m/s
= 33,000 kgm/s
Momentum of the second car = 1100 kg × 15 m/s
= 16,500 kgm/s
Therefore;
Total momentum = 33,000 kgm/s + 16,500 kgm/s
= 49500 kgm/s
Thus, the total momentum of the system is 49500 kgm/s
<h3><u>Answer</u>;</h3>
D. Suspension
<h3><u>Explanation;</u></h3>
- <u>Suspension</u> is a type of construction where the main load-bearing elements are hung from suspension cables that are then suspended from vertical supports anchored into the ground or a riverbed.
- <em><u>Suspension structures are building structures in which the main elements that support the load such as wires, cables, chains, grids, sheet diaphragms,etc are subject only to forces of extension.</u></em>
- Suspension structures may be either plane or spatial. Spatial suspension structures are generally used for large-span roofs of public and industrial buildings.