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Ksenya-84 [330]
4 years ago
15

What is the function of concrete shielding in a nuclear reactor?*

Physics
1 answer:
Reptile [31]4 years ago
3 0

The function of concrete shielding in a nuclear reactor is: <u>to contain radiation.</u>

<u>Explanation</u>:

The reactor core in the nuclear reactor is placed in a steel pressure vessel. The steel vessel is then shielded by a thick layer of concrete. The concrete shielding is done to protect the workers from radiation.

The nuclear power station must be built on hard solid rock as the reactor is heavy in structure. The radiation emitted from the reactor may harm the human being. To avoid this, the concrete shield is made to prevent the emission of radiation.

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A cord is wrapped around the outer surface of the 8-kg disk. If a force of F = (1⁄4u2) N, where u is in radians, is applied to t
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A cord is wrapped around the outer surface of the 8-kg disk. If a force of F = (1⁄4θ²) N, where θ is in radians, is applied to the cord, determine the disk’s angular acceleration when it has turned 5 revolutions. The disc has an initial angular velocity \omega _o = 1 \ rad/s and radius from the center of the disc = 300 mm

Answer:

the angular acceleration = 205.706 rad/sec²

Explanation:

GIVEN THAT:

The disc mass = 8 kg

Force = \dfrac{1}{4} \ \ \theta ^2* N

We are told that the given θ is in radians; Therefore; the when it has turned 5 revolutions; we have the θ to be:

\theta = 5 rev * (\dfrac{2 \  \pi  * rad}{1 rev}) \\ \\ \theta = 10 \ \pi \ rad

Also;

the initial angular velocity \omega _o = 1 \ rad/s

radius from the center of the disc = 300 mm = 0.3 m

Thus; the mass moment about the Inertia can be determined via the following expression;

I_o = \dfrac{1}{2}*m*r^2

I_o = \dfrac{1}{2}*8*0.3^2

I_o = 0.36 \ kg/m^3

Now to calculate the angular acceleration; we equate the sum of the moments acting on the Inertia;

SO:

\sum M_o = I_o \alpha

F*0.3 = 0.36* \alpha

\dfrac{1}{2}* \theta^2 *0.3 = 0.36* \alpha

\alpha = 0.20836 \  \theta^2 \ rad/sec^2

\alpha = 0.20836 \  (10 \ \pi )^2 \ rad/sec^2

\alpha = 205.706 \ rad/sec^2

Hence; the angular acceleration = 205.706 rad/sec²

6 0
3 years ago
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