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
if you're shopping for a rack switch, the component on the switch that tells you it can be mounted to a rack is the:
<h3>What are rack ears?</h3>
Rack ears are L-shaped objects that can be used to hold a rack switch firmly to the support walls. Rack ears are often found at the front panel of the rack which is to be mounted.
When a person purchases a rack switch and finds the rack ears there, it is a signal that the item can be secured firmly to rails. Sometimes, the rack ears also appear as extensions.
Learn more about rack ears here:
brainly.com/question/13318148
#SPJ4
Answer:
λ = 5940 Angstroms
Explanation:
This is an exercise of the relativistic Doppler effect
f’= f √((1- v / c) / (1 + v / c))
Where the speed in between the strr and the observer is positive if they move away
Let's use the relationship
c = λ f
f = c /λ
We replace
c /λ’ = c /λ √ ((1- v / c) / (1 + v / c))
λ = λ’ √ ((1- v / c) / (1 + v / c))
Let's calculate
v = 0.01 c
v = 0.01 3 10⁸
v= 3 10⁶ m / s
λ = 6000 √ [(1- 3 10⁶/3 10⁸) / (1+ 3 10⁶/3 10⁸)]
λ = 6000 √ [0.99 / 1.01]
λ = 5940 Angstroms
