Hydropower _____.
1 answer:
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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:
The speed of the block when it has returned to the bottom of the ramp is 6.56 m/s.
Explanation:
Given;
mass of block, m = 4 kg
coefficient of kinetic friction, μk = 0.25
angle of inclination, θ = 30°
initial speed of the block, u = 5 m/s
From Newton's second law of motion;
F = ma
a = F/m
Net horizontal force;
∑F = mgsinθ + μkmgcosθ
At the top of the ramp, energy is conserved;
Kinetic energy = potential energy
¹/₂mv² = mgh
¹/₂ v² = gh
¹/₂ x 5² = 9.8h
12.5 = 9.8h
h = 12.5/9.8
h = 1.28 m
Height of the ramp is 1.28 m
Now, calculate the speed of the block (in m/s) when it has returned to the bottom of the ramp;
v² = u² + 2ah
v² = 5² + 2 x 7.022 x 1.28
v² = 25 + 17.976
v² = 42.976
v = √42.976
v = 6.56 m/s
Therefore, the speed of the block when it has returned to the bottom of the ramp is 6.56 m/s.