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Kruka [31]
3 years ago
9

How do products of a fusion reaction differ from the products of a fission reaction

Physics
2 answers:
Sloan [31]3 years ago
6 0
Both fission<span> and </span>fusion<span> are nuclear </span>reactions<span> that produce energy, but the applications are not the same. </span>Fission<span> is the splitting of a heavy, unstable nucleus into two lighter nuclei, and </span>fusion<span> is the process where two light nuclei combine together releasing vast amounts of energy.
Hope this helps! =)</span>
Romashka [77]3 years ago
3 0

Answer:

C.) Fission reactions are used to generate electricity for consumers, but fusion reactions are not.

Please give brainliest!!

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Y_Kistochka [10]

Answer:

It's A & C

Explanation:

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7 0
3 years ago
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Work may be measured using units of <br> A.watts<br> B.newton<br> C.joules<br> D.newtons per second
DochEvi [55]

If you are working with electricity most people will measure it in watts.

So, A. watts is your best answer.

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4 years ago
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qaws [65]
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3 years ago
What is the heat extracted from the cold reservoir for the refrigerator?
zaharov [31]
What is the heat extracted from the cold reservoir for the refrigerator shown in(Figure 1) ? Assume that W1 = -123J and W2 = 88J . 

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K=105J
7 0
3 years ago
A +12 μC charge and -8 μC charge are 4 cm apart. Find the magnitude and direction of the E-field at the point midway between t
Natasha_Volkova [10]

Answer:

Explanation:

Given

Charge of first Particle q_1=+12\ \mu C

Charge of second Particle q_2=-8\ \mu C

distance between them d=4\ cm

k=9\times 10^{9}

magnetic field due to first charge at mid-way between two charged particles is

E_1=\frac{kq_1}{r^2}

r=\frac{d}{2}=\frac{4}{2}=2\ cm

E_1=\frac{9\times 10^9\times 12\times 10^{-6}}{(2\times 10^{-2})^2}

E_1=27\times 10^7\ N/C (away from it)

Electric field due to q_2=-8\ \mu C

E_2=\frac{kq_2}{r^2}

E_2=-\frac{9\times 10^9\times 8\times 10^{-6}}{(2\times 10^{-2})^2}

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E_{net}=E_1+E_2

E_{net}=9\times 10^7\ N/C(away from first charge)        

8 0
3 years ago
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