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11 months ago

If a critical mass of fissionable material in a spherical shape is flattened like a hamburger, it will be?

1 answer:

8 0

A subcritical state will result by flattening a critical mass of fissionable material into the shape of a hamburger.

To find the answer, we need to know about the critical mass.

<h3>What is critical mass?</h3>

A material's nuclear characteristics (particularly, its nuclear fission cross-section), density, shape, enrichment, purity, temperature, and environment all affect its critical mass.

A mass of fissile material is considered to be in a critical state when a nuclear chain reaction in the mass is self-sustaining and there is no change in power, temperature, or neutron population.
At a specific temperature, a mass might be precisely critical.

A subcritical state will result by flattening a critical mass of fissionable material into the shape of a hamburger.

The cross sections for fission and absorption grow as the relative neutron velocity drops.

Thus, we can conclude that, a subcritical state will result by flattening a critical mass of fissionable material into the shape of a hamburger.

Learn more about the critical mass here:

brainly.com/question/12545809

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Read 2 more answers

QUESTION 10

Elena L [17]

The maximum value of θ of such the ropes (with a maximum tension of 5,479 N) will be able to support the beam without snapping is:

$\theta =37.01^{\circ}$

We can apply the first Newton's law in x and y-direction.

If we do a free body diagram of the system we will have:

x-direction

All the forces acting in this direction are:

$T_{1}sin(\theta)-T_{2}sin(\theta)=0$ (1)

Where:

T(1) is the tension due to the rope 1
T(2) is the tension due to the rope 2

Here we just conclude that T(1) = T(2)

y-direction

The forces in this direction are:

$T_{1}cos(\theta)+T_{2}cos(\theta)-W=0$ (2)

Here W is the weight of the steel beam.

We equal it to zero because we need to find the maximum angle at which the ropes will be able to support the beam without snapping.

Knowing that T(1) = T(2) and W = mg, we have:

$T_{1}cos(\theta)+T_{1}cos(\theta)-m_{steel}g=0$

$2T_{1}cos(\theta)-m_{steel}g=0$

$2T_{1}cos(\theta)=m_{steel}g$

T(1) must be equal to 5479 N, so we have:

$cos(\theta)=\frac{m_{steel}g}{2T_{1}}$

$cos(\theta)=\frac{892*9.81}{2*5479}$

$cos(\theta)=0.80$

Therefore, the maximum angle allowed is θ = 37.01°.

You can learn more about tension here:

brainly.com/question/12797227

I hope it helps you!

8 0

3 years ago

If the mass of a material is 50 grams and the volume of the material is 12 cm^3, what would the density of the material be?

AlekseyPX

The density of the material would be
25/6 grams per cm^3.
to obtain the result above this is what we do:
density is calculated as: (the mass of the given material or object) / volume of the material
which leads us to 50grams /12cm^3

4 0

3 years ago