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s2008m [1.1K]
2 years ago
9

(a) Consider a fusion generator built to create 3.00 GW of power. Determine the rate of fuel burning in grams per hour if the DT

reaction is used.
Physics
1 answer:
Stels [109]2 years ago
4 0

The rate of fuel burning in grams per hour if the DT reaction is used is 1.08 ×10^1^3 J/g per hour

<h3>How is the rate of fuel burning in grams per hour calculated when the D-T reaction is used?</h3>
  • D + T → He + n
  • The D-T fusion reaction results in a Helium (He) and  neutron (n)

E = 17.59 MeV

Mass = 2.014u + 3.016u

= 5.030u

Energy per Kg = (17.59×10^6×1.6×10^-^1^9) ÷ ( 5.030×1.66×10^-^2^7)

= 3.37×10^1^4 J/Kg

= 3.0× 10^9 J/g

Rate of fuel burning in grams per hour = 3.0× 10^9 ×  3600

= 3.6×3.0×10^1^2

= 1.08 ×10^1^3 J/g per hour

To learn more about fusion reactor and energy production, refer

brainly.com/question/13399644

#SPJ4

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Two parallel disks of diameter D 5 0.6 m separated by L 5 0.4 m are located directly on top of each other. Both disks are black
oksian1 [2.3K]

To solve this problem it is necessary to apply the concepts related to the Stefan-Boltzmann law which establishes that a black body emits thermal radiation with a total hemispheric emissive power (W / m²) proportional to the fourth power of its temperature.

Heat flow is obtained as follows:

Q = FA\sigma\Delta T^4

Where,

F =View Factor

A = Cross sectional Area

\sigma = Stefan-Boltzmann constant

T= Temperature

Our values are given as

D = 0.6m

L = 0.4m\\T_1 = 450K\\T_2 = 450K\\T_3 = 300K

The view factor between two coaxial parallel disks would be

\frac{L}{r_1} = \frac{0.4}{0.3}= 1.33

\frac{r_2}{L} = \frac{0.3}{0.4} = 0.75

Then the view factor between base to top surface of the cylinder becomes F_{12} = 0.26. From the summation rule

F_{13} = 1-0.26

F_{13} = 0.74

Then the net rate of radiation heat transfer from the disks to the environment is calculated as

\dot{Q_3} = \dot{Q_{13}}+\dot{Q_{23}}

\dot{Q_3} = 2\dot{Q_{13}}

\dot{Q_3} = 2F_{13}A_1 \sigma (T_1^4-T_3^4)

\dot{Q_3} = 2(0.74)(\pi*0.3^2)(5.67*10^{-8})(450^4-300^4)

\dot{Q_3} = 780.76W

Therefore the rate heat radiation is 780.76W

5 0
2 years ago
Give an example of a force applied to an object that does not change the object's velocity. Why does the object's velocity not c
Xelga [282]

Answer:

A chair at rest on the floor has two forces acting on it its own weight that pulls it downward and the floor pushing upward on the chair, both of these forces are acting on it but the net force is 0, so the chair remains at rest and its velocity stays at 0.

8 0
1 year ago
Can someone help me ​
Julli [10]
The answer is the first one
6 0
3 years ago
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When the core of a star like the sun uses up its supply of hydrogen for fusion, the core begins to ________?
pav-90 [236]

When the core of a star like the sun uses up its supply of hydrogen for fusion, the core begins to contract in size. The gravitational forces becomes strong and cause the star to reduce in size. as the star contracts, the temperature of the star also rises since the heat is distributed inside smaller Volume now.

4 0
3 years ago
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A 10 kg box hangs from a rope. What is the tension in the rope (in Newtons) if the box is stationary
Archy [21]

Answer:

T = 98 N

Explanation:

The gravity of the earth is known to be 9.8 m/s²

Data:

  • m = 10 kg
  • g = 9.8 m/s²
  • T = ?

Use formula:

  • \boxed{\bold{T=m*g}}

Replace and solve:

  • \boxed{\bold{T=10\ kg*9.8\frac{m}{s^{2}}}}
  • \boxed{\boxed{\bold{T=98\ N}}}

The tension in the rope is <u>98 Newtons.</u>

Greetings.

5 0
2 years ago
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