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:
Δe=0.578 kJ/kg
Explanation:
Given data
Velocity v₁=0 m/s
Velocity v₂=34 m/s
to find
Specific energy change Δe
Solution
The specific energy change is simply determined from change in velocity
Δe=(v₂²-v₁²)/2
Put the given values to find the specific energy change
Δe=0.578 kJ/kg
Answer:
0.001 s
Explanation:
The force applied on an object is equal to the rate of change of momentum of the object:
where
F is the force applied
is the change in momentum
is the time interval
The change in momentum can be written as
where
m is the mass
v is the final velocity
u is the initial velocity
So the original equation can be written as
In this problem:
m = 5 kg is the mass of the fist
u = 9 m/s is the initial velocity
v = 0 is the final velocity
F = -45,000 N is the force applied (negative because its direction is opposite to the motion)
Therefore, we can re-arrange the equation to solve for the time: