Physics, 27.07.2019 17:10 masonbitterman7488

Aneutron in a reactor makes an elastic headon collision with the nucleus of an atom initially at rest. assume: the mass of the atomic nucleus is about 14.9 the mass of the neutron. what fraction of the neutron's kinetic energy is transferred to the atomic nucleus?

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Answer from: rayne40

The fraction of the neutron's kinetic energy transferred to the atomic nucleus is 1.87.

Explanation:

Use the conservation of momentum in head-on collision.

$mv_{i}=mv_{f}+MV_{f}$

$v_{i}-v_{f}=\frac{M}{m}V_{f}$

Put

$\frac{M}{m}=14.9$

$v_{i}-v_{f}=14.9V_{f}$ ....... (1)

Here, m, M are the masses of the neutron and atom, $v_{i}$ is the initial velocity of the neutron, $v_{f}$ is the final velocity of the neutron and $V_{f}$ is the final velocity of the atom.

Use the conservation of the kinetic energy.

$\frac{1}{2}mv_{i}^{2}=\frac{1}{2}mv_{f}^{2}+\frac{1}{2}MV_{f}^{2}$

$v_{i}^{2}-v_{f}^{2}=\frac{M}{m}V_{f}^{2}$

Put

$\frac{M}{m}=14.9$ .

$v_{i}^{2}-v_{f}^{2}=14.9V_{f}^{2}$ ...... (2)

Divide (2) by (1).

$\frac{v_{i}^{2}-v_{f}^{2}}{v_{i}-v_{f}}=\frac{14.9V_{f}^{2}}{14.9V_{f}}$

$\frac{v_{i}^{2}-v_{f}^{2}}{v_{i}-v_{f}}=V_{f}$

$v_{i}+v_{f}=V_{f}$ ...... (3)

Solve equation (1) and (3).

$v_{i}=7.9V_{f}$

Calculate the fraction of the neutron's kinetic energy transferred to the atomic nucleus.

$Fraction\ transferred\ to\ atom =\frac{\frac{1}{2}MV_{f}^{2}}{\frac{1}{2}mv_{i}^{2}}$

Put

$\frac{M}{m}=14.9$

$Fraction\ transferred\ to\ atom=\frac{14.9}{7.95}$

$Fraction\ transferred\ to\ atom=1.87$

Therefore, the fraction of the neutron's kinetic energy transferred to the atomic nucleus is 1.87.

Answer from: Quest

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.

Answer from: Quest

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j

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