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Elden [556K]
3 years ago
14

Design a diagram, sketch, or other model to show changes in the nucleus of an atom during nuclear fission and fusion. Include a

short description and a way to compare the amounts of energy released in each event
Physics
1 answer:
Drupady [299]3 years ago
6 0

Explanation:

Inside the sun, fusion reactions take place at very high temperatures and enormous gravitational pressures

The foundation of nuclear energy is harnessing the power of atoms. Both fission and fusion are nuclear processes by which atoms are altered to create energy, but what is the difference between the two? Simply put, fission is the division of one atom into two, and fusion is the combination of two lighter atoms into a larger one. They are opposing processes, and therefore very different.

The word fission means "a splitting or breaking up into parts" (Merriam-Webster Online, www.m-w.com). Nuclear fission releases heat energy by splitting atoms. The surprising discovery that it was possible to make a nucleus divide was based on Albert Einstein’s prediction that mass could be changed into energy. In 1939, scientist began experiments, and one year later Enrico Fermi built the first nuclear reactor.

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Is there a question?

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A solution is a homogeneous mixture of one or more solutes dissolved in a solvent. A specific volume of solvent is only able to
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<h3><u>Answer;</u></h3>

C. Supersaturated

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4 years ago
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An electric immersion heater is put at the bottom of a large tank of water. The water next to the heater becomes warm
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4 0
3 years ago
During free fall, what happens to the gravitational potential energy of a ball?​
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3 years ago
Calculate the energy in the form of heat (in kJ) required to change 75.0 g of liquid water at 27.0 °C to ice at –20.0 °C. Assume
REY [17]

Explanation:

The given data is as follows.

          mass, m = 75 g

      T_{1} = 0^{o}C

      T_{2} = 27^{o}C

      Specific heat of water = 4.18

First, we will calculate the heat required for water is as follows.

            q = m C \times (T_{1} - T_{2})

               = 75 g \times 4.18 J/g^{o}C \times (0 - 27)^{o}C

               = 8464.5 J/mol

               = 8.46 kJ ......... (1)

Also, it is given that T_{3} = -20^{o}C = (20 + 273) K = 293 K and specific heat of ice is 2.108 kJ/kg K.

Now, we will calculate the heat of fusion as follows.

        q = mC \times (T_{3} - T_{1})

           = 0.075 kg \times 2.108 kJ/kg K \times (-293 - 0) K

           = -46.32 kJ ......... (2)

Now, adding both equations (1) and (2) as follows.

               8.46 kJ - 46.32 kJ

             = -37.86 kJ

Therefore, we can conclude that energy in the form of heat (in kJ) required to change 75.0 g of liquid water at 27.0^{o}C to ice at -20.0^{o}C is -37.86 kJ.

4 0
4 years ago
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