Look first for the relation between deBroglie wavelength (λ) and kinetic energy (K):
K = ½mv²
v = √(2K/m)
λ = h/(mv)
= h/(m√(2K/m))
= h/√(2Km)
So λ is proportional to 1/√K.
in the potential well the potential energy is zero, so completely the electron's energy is in the shape of kinetic energy:
K = 6U₀
Outer the potential well the potential energy is U₀, so
K = 5U₀
(because kinetic and potential energies add up to 6U₀)
Therefore, the ratio of the de Broglie wavelength of the electron in the region x>L (outside the well) to the wavelength for 0<x<L (inside the well) is:
1/√(5U₀) : 1/√(6U₀)
= √6 : √5
<span>The
answer is Local Area Network (LAN). Examples of LANs, are those in workplaces,
libraries, and universities. The nodes are
connected and share common resources over
a WiFi, Ethernet network, Token rings, and
etcetera. The opposite of LAN is a Wide Area Network (WAN) that covers a
wide geographical area.</span>
Answer:
You will reach both your arms out to break your fall and save your head.
Explanation:
It common sense you don't want your head injured. Do you?
"Frequency decreases" is the one way among the following choices given in the question that <span>frequency change as wavelength increases. The correct option among all the options that are given in the question is the second option. I hope that this is the answer that has actually come to your desired help.</span>
The final velocity of the red barge in the collision elastic is 0.311 m/s when it collides with blue barge pf mass 1000000 kg.
Final velocity(v3) of the red barge is calculated by following formula
m1×v1+ m2×v2= (m1+m2)v3
Substituting the value of m1= 150000 kg, v1= 0.25 m/s, m2= 1000000 kg, v2= 0.32 m/s
150000 × 0.25+ 1000000×0.32= (150000+1000000)×v3
37500+ 320000= 1150000×v3
357500= 1150000×v3
v3= 0.311 m/s
<h3>What is elastic collision velocity? </h3>
- The velocity of the target particle after a head-on elastic impact in which the projectile is significantly more massive than the target will be roughly double that of the projectile, but the projectile velocity will remain virtually unaltered.
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