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3 years ago

How does critical mass play a role in nuclear reactions?

Physics

2 answers:

malfutka [58]3 years ago

8 0

Answer;

C.It is the minimum amount of material needed to sustain a fission reaction.

Explanation;

-A critical mass is the smallest or the minimum amount of fissile material needed for a sustained nuclear chain reaction. A critical mass must be achieved in order for the chain reaction to continue and release the atomic energy. A critical mass is needed for both a nuclear reactor and an atomic or hydrogen bomb.

-The critical mass of a fissionable material depends upon its nuclear properties (specifically, the nuclear fission cross section), its density, its shape, its enrichment, its purity, its temperature, and its surroundings.

ryzh [129]3 years ago

3 0

The correct answer for above statement is:

c.It is the minimum amount of material needed to sustain a fission reaction.

Explanation:

A nuclear reaction is taken into account to be the method during which 2 nuclear paticles (two nuclei or a nucleus and a nucleon) move to supply 2 or a lot of nuclear particles or gamma rays.

Thus, a natural action should cause a change of a minimum of one nuclide to a different. typically if a nucleus interacts with another nucleus or particle while not dynamic the mature of any nuclide, the method is mentioned a nuclear scattering.

The important mass is that the smallest amount of fissile material required to sustain nuclear reactions below expressed conditions

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When a coil is carrying a current of 25.0 A that is increasing at 145 A/s the induced emf in the coil has magnitude 3.70 mV.

Aleks04 [339]

Answer:

a) 25.5 µH

b) 22.95 mV

Explanation:

Induced emf in a inductor is given by

E = L * di/dt, where

E is the voltage of the circuit

L is the inductance of the circuit

di/dt if the rate of inductance

So we have

0.0037 = L * 145

L = 0.0037 / 145

L = 0.0000255

L = 25.5 µH

i(t) = 225t²

Recall that

E = L * di/dt, so that

E = 25.5 µH * |225t²|

Differentiating with respect to t, we have

E = 25.5 * 2 * 225t

E = 25.5 * 450t

Solving for t = 2,we get

E = 25.5 * 450(2)

E = 25.5 * 900

E = 22950 µV or

E = 22.95 mV

6 0

3 years ago

When you push on an object such as a wrench, a steel pry bar, or even the outer edge of a door, you produce a torque equal to th

Korvikt [17]

<span>Since youc oncetrate all your force directly towards the moment arm it means that you push it at an angle of your force is directed to the left or the right and I bet that it must be 90</span> degrees to the bar. Obviuosly, if you are about to push it you will do it straight up but not in a zig zag way. In other words, it should be perpendicular to the arm because the<span> torque can be produced only if force is applied at a constant index (90).
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3 years ago

A star with a mass like the Sun which will soon die is observed to be surrounded by a large amount of dust and gas -- all materi

Dafna11 [192]

Answer: D. Infrared

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2 years ago

A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V?What is the total energy stores in the

Rama09 [41]

1) $1.11\cdot 10^{-7} J$

The capacitance of a parallel-plate capacitor is given by:

$C=\frac{\epsilon_0 A}{d}$

where

$\epsilon_0$ is the vacuum permittivity

A is the area of each plate

d is the distance between the plates

Here, the radius of each plate is

$r=\frac{2.0 cm}{2}=1.0 cm=0.01 m$

so the area is

$A=\pi r^2 = \pi (0.01 m)^2=3.14\cdot 10^{-4} m^2$

While the separation between the plates is

$d=0.50 mm=5\cdot 10^{-4} m$

So the capacitance is

$C=\frac{(8.85\cdot 10^{-12} F/m)(3.14\cdot 10^{-4} m^2)}{5\cdot 10^{-4} m}=5.56\cdot 10^{-12} F$

And now we can find the energy stored,which is given by:

$U=\frac{1}{2}CV^2=\frac{1}{2}(5.56\cdot 10^{-12} F/m)(200 V)^2=1.11\cdot 10^{-7} J$

2) 0.71 J/m^3

The magnitude of the electric field is given by

$E=\frac{V}{d}=\frac{200 V}{5\cdot 10^{-4} m}=4\cdot 10^5 V/m$

and the energy density of the electric field is given by

$u=\frac{1}{2}\epsilon_0 E^2$

and using

$E=4\cdot 10^5 V/m$ , we find

$u=\frac{1}{2}(8.85\cdot 10^{-12} F/m)(4\cdot 10^5 V/m)^2=0.71 J/m^3$

7 0

3 years ago

Two Polaroids are aligned so that the initially unpolarized light passing through them is a maximum. At what angle should one of

steposvetlana [31]

To solve this problem it is necessary to apply the law of Malus which describes the change in the Intensity of Light when it crosses a polarized surface.

Mathematically the expression is given as

$I = I_0 cos^2\theta$