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
Answer:
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Home Business Cottage and Small Industries Development Committee to organize National Industrial Goods and...
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Cottage and Small Industries Development Committee to organize National Industrial Goods and Technology Expo
By Glocal Khabar - 2368 0
National Industrial Goods and Technology Expo- Glocal Khabar
Kathmandu, February 19, 2018: Cottage and Small Industries Development Committee is set to organize the twenty-ninth edition of National Industrial Goods and Technology Expo from March 23, 2018 at Bhrikutimandap, Kathmandu.
The motto of the five-day event is ‘Let’s use home-made products, move ahead towards prosperity.’ The committee has shared that handicrafts, wool and bamboo products, goods made from handmade papers, different types of pickles and Palpali Dhaka would be put on display in the expo.
Answer:
The answer is
<h2>28 kg</h2>
Explanation:
The mass of an object given it's momentum and velocity / speed can be found by using the formula

where
m is the mass
p is the momentum
v is the speed or velocity
From the question
p = 280 kg/ms
v = 10 m/s
The mass of the object is

We have the final answer as
<h3>28 kg</h3>
Hope this helps you
Answer:
A rocket taking off from earth which pushes gasses in one direction and the rocket in
the other
Answer:
W = 9.93 10² N
Explanation:
To solve this exercise we must use the concept of density
ρ = m / V
the tabulated density of copper is rho = 8966 kg / m³
let's find the volume of the cylindrical tube
V = A L
V = π (R_ext ² - R_int ²) L
let's calculate
V = π (4² - 2²) 10⁻⁴ 3
V = 1.13 10⁻² m³
m = ρ V
m = 8966 1.13 10⁻²
m = 1.01 10² kg
the weight of the tube
W = mg
W = 1.01 10² 9.8
W = 9.93 10² N