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:
a. A = 0.735 m
b. T = 0.73 s
c. ΔE = 120 J decrease
d. The missing energy has turned into interned energy in the completely inelastic collision
Explanation:
a.
4 kg * 10 m /s + 6 kg * 0 m/s = 10 kg* vmax
vmax = 4.0 m/s
¹/₂ * m * v²max = ¹/₂ * k * A²
m * v² = k * A² ⇒ 10 kg * 4 m/s = 100 N/m * A²
A = √1.6 m ² = 1.26 m
At = 2.0 m - 1.26 m = 0.735 m
b.
T = 2π * √m / k ⇒ T = 2π * √4.0 kg / 100 N/m = 1.26 s
T = 2π *√ 10 / 100 *s² = 1.99 s
T = 1.99 s -1.26 s = 0.73 s
c.
E = ¹/₂ * m * v²max =
E₁ = ¹/₂ * 4.0 kg * 10² m/s = 200 J
E₂ = ¹/₂ * 10 * 4² = 80 J
200 J - 80 J = 120 J decrease
d.
The missing energy has turned into interned energy in the completely inelastic collision
A perfect black body can’t be realized
Explanation:
click the link you can get answer of the question
Answer:
m,lkj,mkn,njn
Explanation:because she is telling you to do a project
