Answer:
39.40 MeV
Explanation:
<u>Determine the minimum possible Kinetic energy </u>
width of region = 5 fm
From Heisenberg's uncertainty relation below
ΔxΔp ≥ h/2 , where : 2Δx = 5fm , Δpc = hc/2Δx = 39.4 MeV
when we apply this values using the relativistic energy-momentum relation
E^2 = ( mc^2)^2 + ( pc )^2 = 39.4 MeV ( right answer ) because the energy grows quadratically in nonrelativistic approximation,
Also in a nuclear confinement ( E, P >> mc )
while The large value will portray a Non-relativistic limit as calculated below
K = h^2 / 2ma^2 = 1.52 GeV
Answer:
I believe the answer is A and D.
I am unsure of C.
The answer to this statement is true!
The potential difference between points a and b is zero.
<h3>Total emf of the series circuit</h3>
The total emf in the circuit is the sum of all the emf in the circuit.
emf(total) = 1.5 + 1.5 = 3.0 V
<h3>Potential difference</h3>
The potential difference between two points, a and b is calculated as follows;
V(ab) = Va - Vb
V(ab) = 1.5 - 1.5
V(ab) = 0
Thus, the potential difference between points a and b is zero.
Learn more about potential difference here: brainly.com/question/3406867
