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romanna [79]
3 years ago
6

During World War II, it was found that r, the radius of the shockwave produced during an atomic bomb explosion, depends on the e

nergy released, W, the elapsed time, t, and the air density, .
Physics
1 answer:
cupoosta [38]3 years ago
7 0

The atomic bomb explosion, depends on the energy released, W, the elapsed time, t, and the air density which altogether lead to the vaporisation of the contents within the mushroom cloud created.

<h3>What is an atomic bomb?</h3>

An atomic bomb is type of nuclear weapon that releases a vast amount of energy upon explosion in the form of fission reaction.

During an explosion of atomic bomb, a cloud of mushroom fire is formed which vaporises anything found within it. The extent of destruction depends on:

  • energy released, W,

  • the elapsed time, t, and

  • the air density.

Therefore, the atomic bomb explosion, depends on the energy released, W, the elapsed time, t, and the air density which altogether lead to the vaporisation of the contents within the mushroom cloud created.

Learn more about nuclear energy here:

brainly.com/question/24295936

#SPJ1

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A sphere is originally at a temperature of 500°c. The sphere is melted and recast, without loss of mass, into a cube with the sa
aleksandr82 [10.1K]

Answer: The value of the celsius temperature of the cube is 472.2°c.

Explanation:        

The expression for the power radiated is as follows;

P=A\epsilon\sigma T^{4}

Here, A is the area, \sigma is the stefan's constant,\epsilon is the emissivity and T is the temperature.

It is given in the problem that A sphere is originally at a temperature of 500°c. The sphere is melted and recast, without loss of mass, into a cube with the same emissivity as the sphere.

Then the expression for the radiated power for the cube and the sphere can be expressed as;

A_{1}\epsilon \e\sigma T_{1}^{4}=A_{2}\epsilon \e\sigma T_{2}^{4}

Here, A_{1} is the area of the sphere, A_{2} is the area of the cube,T_{1}  is the temperature of the sphere and T_{2}  is the temperature of the cube.

The radiated powers and emissivity of the cube and the sphere are same.

A_{1}T_{1}^{4}=A_{2}T_{2}^{4}

The area of the sphere is A_{1}=4\pi \times r^{2}.

Here, r is the radius of the sphere.

The area of the cube is A_{2}=6\times a^{2}.

Here, a is the edge of the cube.

Put A_{1}=4\pi \times r^{2} and A_{2}=6\times a^{2}.

T_{2}=T_{1}(\frac{2\pi }{3}\times (\frac{r}{a})^{2})^{\frac{1}{4}}  ....(1)

The masses and the densities of the sphere and the cube are same. Then the volumes are also same.

V_{1}=V_{2}

Here,V_{1},V_{1} are the volumes of the sphere and the cube.

\frac{4}{3}\pi r^{3}=a^{3}

\frac{r}{a}=(\frac{3}{4\pi })^{\frac{1}{3}}  

Put this value in the equation (1).

T_{2}=T_{1}(\frac{2\pi }{3}\times (\frac{r}{a})^{2})^{\frac{1}{4}}T_{2}=T_{1}(\frac{2\pi }{3}\times ((\frac{3}{4\pi })^{\frac{1}{3}})^{2})^{\frac{1}{4}}

Put T_{1}=500°c.

T_{2}=(500)(\frac{2\pi }{3}\times (\frac{3}{4\pi })^{\frac{2}{3}})^{\frac{1}{4}}

T_{2}=472.2^{\circ}c

Therefore, the value of the celsius temperature of the cube is 472.7°c.    

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