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koban [17]
2 years ago
6

The process of combining two small nuclei into one nucleus of larger mass is called _____.

Physics
2 answers:
ladessa [460]2 years ago
6 0

The correct answer to the question is Fusion i.e the process of combining two small nuclei into one nucleus of larger mass is called fusion.

EXPLANATION:

Before answering this question, first we have to understand nuclear fission and nuclear fusion.

Nuclear fission is the thermo nuclear process in which a heavy unstable nucleus will split into two light stable nuclei with the release of energy.

For instance, the splitting of Uranium-235 to Krypton-92 and Barium-141.

Nuclear fusion is just the opposite process of nuclear fission. It is the thermo nuclear process in which two light unstable nuclei will combine with each other at  a high temperature equal to the temperature of sun to form a large stable nuclei with the release of energy.

For instance, we may take hydrogen bomb i.e the energy formation in stars. Here, four hydrogen nuclei will combine with each other form a helium nuclei with the release of huge amount of energy.

Hence, the correct answer to the question is nuclear fusion.

fiasKO [112]2 years ago
3 0
The process of splitting one large nucleus into
smaller ones is nuclear fission.

The process of combining two small nuclei into
one larger one is nuclear fusion.
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Answer:

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Explanation:

We will separate the problem in 3 parts, part A when there were no coals on the car, part B when there is 1 coal on the car and part C when there are 2 coals on the car. Inertia is the mass in this case.

For each part, and since the coals are thrown vertically, the horizontal linear momentum p=mv must be conserved, that is, p=m_Av_A=m_Bv_B=m_Cv_C, were each velocity refers to the one of the car (with the eventual coals on it) for each part, and each mass the mass of the car (with the eventual coals on it) also for each part. We will write the mass of the hopper car as m_h, and the mass of the first and second coals as m_1 and m_2 respectively

We start with the transition between parts A and B, so we have:

m_Av_A=m_Bv_B

Which means

m_hv_A=(m_h+m_1)v_B

And since we want the mass of the first coal thrown (m_1) we do:

m_hv_A=m_hv_B+m_1v_B

m_hv_A-m_hv_B=m_1v_B

m_1=\frac{m_hv_A-m_hv_B}{v_B}=\frac{m_h(v_A-v_B)}{v_B}

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For the transition between parts B and C, we can write:

m_Bv_B=m_Cv_C

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(m_h+m_1)v_B=(m_h+m_1+m_2)v_C

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v_C=\frac{(3\times10^4Kg+2\times10^4Kg)(0.3m/s)}{(3\times10^4Kg+2\times10^4Kg+2\times10^4Kg)}=0.214m/s

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