When two capacitors are connected in parallel and then connected to a battery, the total stored energy is 6.9 times greater than
when they are connected in series and then connected to the same battery.
What is the ratio of the two capacitances? (Before the battery is connected in each case, the capacitors are fully discharged.)
What is the ratio of the two capacitances? (Before the battery is connected in each case, the capacitors are fully discharged.)
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:
Vo = 4.5 [m/s]
Explanation:
In order to solve this problem, we must use the following equation of kinematics.
where:
Vf = final velocity = 12 [m/s]
Vo = initial velocity [m/s]
a = acceleration = 1.5 [m/s²]
t = time = 5 [s]
Now replacing: