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V125BC [204]
3 years ago
13

The core of a certain reflected reactor consist of a cylinder 10 ft high 10 ft in diameter The measured maximum-to-average flux

is 1.5. When the reactor is operated at a power level of 835 MW. what is the maximum power density in the reactor in kW/liter?
Physics
1 answer:
Westkost [7]3 years ago
6 0

Answer:

The maximum power density in the reactor is 37.562 KW/L.

Explanation:

Given that,

Height = 10 ft = 3.048 m

Diameter = 10 ft = 3.048 m

Flux = 1.5

Power = 835 MW

We need to calculate the volume of cylinder

Using formula of volume

V =\pi r^2 h

Put the value into the formula

V=\pi\times(1.524)^2\times 3.048

V= 22.23\m^3

V = 22.23\times10^{3}\ Liter

We need to calculate the maximum power density in the reactor

Using formula of power density

P=\dfrac{E}{V}

Where, P = power density

E = energy

V = volume

Put the value into the formula

P=\dfrac{835\times10^{6}}{22.23\times10^{3}}

P=37561.85 = 37.562\times KW/L

Hence, The maximum power density in the reactor is 37.562 KW/L.

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Explanation:

Effective nuclear charge is defined as he net positive charge experienced by an electron in an atom. It is termed "effective" because the shielding effect of electrons prevents higher orbital electrons from experiencing the full nuclear charge of the nucleus due to the repelling effect of inner-layer electrons.

The 1s is the closest shell to the nucleus of an therefore maximum nuclear charge is experienced. The formula for effective nuclear charge is:

Zeff = Z – S

where

Z = the number of protons in the nucleus, and

S = the shielding constant, the average number of electrons between the nucleus and the electron.

Hence, the energy required to remove an electron from the 1s orbital is the strongest.

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4 years ago
How much work is done against gravity when lifting a 2-kg sack of groceries a distance of 2.5 meters?
Crazy boy [7]
W = F x d/x = (m x Ag) x h, therefore, mass (2kg x 9.8) x 2.5m = 49J
8 0
3 years ago
Fossils are generally found inside of what type of rocks?
dusya [7]

Answer:

C sedimentary rocks is the answer.

Explanation:

8 0
4 years ago
Read 2 more answers
Depict the following path: You drive 14 blocks East, 7 blocks North, and 2 blocks West. Use the vertex of the graph (coordinates
horrorfan [7]

Answer:

Displacement = 12\hat{i} +7\hat{j}

Magnitude of displacement = 13.89 units

Angle of displacement = 30.26°

Explanation:

Mark the co-ordinates of each of the end points of the vectors.

The marked co-ordinates are shown below.

Displacement is the shortest distance between two points. Here, displacement between the starting and end points is the vector \overrightarrow{OA}.

Displacement vector for a point A(x,y) from origin O is given as:

\overrightarrow{OA}=(x-0)\hat{i}+(y-0)\hat{j}

Magnitude is given as:

|\overrightarrow{OA}|=\sqrt{x^{2}+y^{2}}

Direction of the vector is given as:

\theta=tan^{-1}\frac{y}{x}

Therefore, displacement vector is:

\overrightarrow{OA}=(12-0)\hat{i}+(7-0)\hat{j}\\ \overrightarrow{OA}=12\hat{i}+7\hat{j}

Magnitude of displacement is given as:

|\overrightarrow{OA}|=\sqrt{x^{2}+y^{2}}\\ |\overrightarrow{OA}|=\sqrt{12^{2}+7^{2}}\\ |\overrightarrow{OA}|=\sqrt{144+49}\\\\ |\overrightarrow{OA}|=13.89

Angle of displacement is:

tan\theta=\frac{y}{x} =\frac{7}{12}\\\theta=tan^{-1}(\frac{7}{12})

\theta=30.26°

8 0
4 years ago
The perimeter of a sector of a circle is the sum of the two sides formed by the radii and the length of the included arc. A sect
snow_tiger [21]

Answer:

Length of the arc of this sector, l = 14 cm

Explanation:

It is given that, the perimeter of a sector of a circle is the sum of the two sides formed by the radii and the length of the included arc.

Perimeter of sector, P = 28 cm

Area of sector, A=49\ cm^2

According to figure,

2r + l = 28 ............(1)

Area of sector, A=\dfrac{\theta}{360}\times \pi r^2

Where, \theta is in radian and \theta=\dfrac{l}{r}

Since, 1^{\circ}=\dfrac{\pi}{180}\ radian

A=\dfrac{l}{2\pi r}\times \pi r^2

r=\dfrac{98}{l}

Put the value of r in equation (1) so,

2\times (\dfrac{98}{l})+l=28

l^2-28l+196=0

On solving above equation for l we get, l = 14 cm. So, the length of the arc of this sector is 14 cm. Hence, this is the required solution.

5 0
3 years ago
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