Taking a shot in the dark here as i mostly deal with atomic structure and the fact i haven't handled a diverging mirror ut would it be a virtual image because that needs parallel normal lines which cannot occur
Answer:
2.857 gm
Explanation:
<u>Step 1: Energy from uranium fission
</u>
The ratio of energy reeled from uranium to coal= 2.5 million times=2.5*10⁶
<u>step2: mass of uranium 235 required
</u>
To get the same energy that of 1 ton coal, the mass reuied will be = (1/2.5) * 10⁻⁶ ton=0.4 * 10⁻³ Kg
where we have taken 1 ton = about 1000Kg
<u>step3: mass of uranium
</u>
mass of u-235 = 70% of natural Uranium=0.7 Mu
So Mu= (1/0.7 )* mass of U-235=(0.4 * 10⁻³ Kg)/0.7=0.571 gm
<u>Step 4: Mass of ore
</u>
mass of Mu = 20% of ore=0.2 M
So, mass of ore= (1/0.2 )* mass of MU-=(0.571)/0.2 gm=2.857 gm
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Since v = dx/dt = 200t
I took the integral of dx = integral of 200t dt
and I got x = 100t^2 + C
So I plugged in 0.5 as t and got x = 25
So x = 25 miles
Answer:
There are three main cloud types.
The foundation consists of 10 major cloud types. In addition to cirrus, stratus, cumulus, and nimbus clouds, there are cirrostratus, cirrocumulus, altostratus, altocumulus, stratocumulus, nimbostratus, and cumulonimbus clouds. The following table places these cloud types into the four major cloud groups.
Explanation:
So false, depends what you have learned and your grade level ig
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
