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3 years ago

How do fission nuclear reactions differ from fusion nuclear reactions?

2 answers:

6 0

The path of the raction occurs on the basis of mass of the nuclei involved in reaction.
In case of nuclear fusion, two or more nuclei having less mass fuse(combine, join) together to form a new nuclei(heavier mass but it is relatively stable). During fusion, matter is not conserved because some of the matter is converted into energy(light). This reaction evolves a huge amount of energy and there comes Einstein's famous Energy-mass equivalence formula E=mc^2! :D. The nuclear reaction occuring in stars(including our sun ) is "fusion".

Fission occurs with heavier nuclei such as that of Uranium-235. Which splits into smaller subatomic particles like gamma, neutrons and enormous amount of energy.
Both, Fission and Fusion releases enormous amount of energy and modern nuclear weapons works on the principle of nuclear fission.

Dmitriy789 [7]3 years ago

4 0

The correct answer is C.<span>Fission reactions are used to generate electricity for consumers, but fusion reactions are not. This is according to my online study class I just took. Hopefully i have helped! :)</span>

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3 years ago

In 1976, the SR-71A, flying at 20 km altitude (T = –56 0C), set the official jet-powered aircraft speed record of 3530 km/hr (21

Lapatulllka [165]

To solve this problem we will apply the concepts related to the calculation of the speed of sound, the calculation of the Mach number and finally the calculation of the temperature at the front stagnation point. We will calculate the speed in international units as well as the temperature. With these values we will calculate the speed of the sound and the number of Mach. Finally we will calculate the temperature at the front stagnation point.

The altitude is,

$z = 20km$

And the velocity can be written as,

$V = 3530km/h (\frac{1000m}{1km})(\frac{1h}{3600s})$

$V = 980.55m/s$

From the properties of standard atmosphere at altitude z = 20km temperature is

$T = 216.66K$

$k = 1.4$

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Velocity of sound at this altitude is

$a = \sqrt{kRT}$

$a = \sqrt{(1.4)(287)(216.66)}$

$a = 295.049m/s$

Then the Mach number

$Ma = \frac{V}{a}$

$Ma = \frac{980.55}{296.049}$

$Ma = 3.312$

So front stagnation temperature

$T_0 = T(1+\frac{k-1}{2}Ma^2)$

$T_0 = (216.66)(1+\frac{1.4-1}{2}*3.312^2)$

$T_0 = 689.87K$

Therefore the temperature at its front stagnation point is 689.87K

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3 years ago

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3 years ago

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Answer:

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