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statuscvo [17]
3 years ago
14

The hydrogen bomb uses the process of _____

Chemistry
2 answers:
Leokris [45]3 years ago
7 0

Answer:-

a) Nuclear Fusion.

Explanation:-

A hydrogen bomb utilizes Nuclear fusion using the isotopes of hydrogen deuterium and tritium to cause explosion. Atom bomb uses nuclear fission.

OLEGan [10]3 years ago
7 0

Answer: Hydrogen bomb uses the process of nuclear fusion reactions.

Explanation: Nuclear fusion reactions are the reactions in which two lighter nuclei fuse together to form a heavier nuclei.

Hydrogen bomb uses the fusion of isotopes of hydrogen. These reactions require a large amount of heat to be initiated.

Reaction for the hydrogen bomb are:

Step 1 :                     _1^1H+_1^1H\rightarrow _1^2H+_{+1}^0e

Step 2:                     _1^1H+_1^2H\rightarrow _2^3He

Step 3:                    _1^1H+_2^3He\rightarrow _2^4He+_{+1}^0e

Overall reaction:    4_1^1H\rightarrow _2^4He+2_{+1}^0e

The hydrogen bomb is also known as Thermonuclear bomb.

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The volume increases when the balloon temperature increases.

<u>Explanation:</u>

-10 F is converted into Kelvin as 249 K.

0°C is nothing but 0+ 273 = 273 K

And the room temperature is 25°C which is converted into Kelvin as 273 + 25 = 298 K.

249 K is below room temperature.

As per the Charles' law volume and temperature are directly proportional to each other, when the pressure of the gas remains constant.

V ∝ T

As the balloon temperature increases, the volume also increases.

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3 years ago
A person's heart beats 72 times a minute. At this rate, how many times does it beat in one hour?
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At one minute, a persons's heart beats 72 times.

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3 years ago
An alloy with an average grain diameter of 35 μm has a yield strength of 163 Mpa, and when it undergoes strain hardening, the gr
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Answer:

\sigma_y\ =210.2\ MPa  

Explanation:

Given that

d= 35 μm ,yield strength = 163 MPa

d= 17 μm ,yield strength = 192 MPa

As we know that relationship between diameter and yield strength

\sigma_y=\sigma_o+\dfrac{K}{\sqrt d}

\sigma_y\ =Yield\ strength

d = diameter

K =Constant

\sigma_o\ =material\ constant

So now by putting the values

d= 35 μm ,yield strength = 163 MPa

163=\sigma_o+\dfrac{K}{\sqrt 35}      ------------1

d= 17 μm ,yield strength = 192 MPa

192=\sigma_o+\dfrac{K}{\sqrt 17}           ------------2

From equation 1 and 2

192-163=\dfrac{K}{\sqrt 17}-\dfrac{K}{\sqrt 35}

K=394.53

By putting the values of K in equation 1

163=\sigma_o+\dfrac{394.53}{\sqrt 35}

\sigma_o\ =96.31\ MPa

\sigma_y=96.31+\dfrac{394.53}{\sqrt d}

Now when d= 12 μm

\sigma_y=96.31+\dfrac{394.53}{\sqrt 12}

\sigma_y\ =210.2\ MPa

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