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ExtremeBDS [4]
2 years ago
11

the atomic number of uranium-235 is 92, its half-life is 704 million years, and the radioactive decay of 1 kg of 235u releases 6

.7 × 1013 j. radioactive material must be stored in a safe container or buried deep underground until its radiation output drops to a safe level. generally, it is considered "safe" after 10 half-lives.
Physics
1 answer:
wel2 years ago
7 0

The overall efficiency of the conversion is 6.93%.

<h3>What is the overall efficiency?</h3>

We know that generation of electricity from radioactive fuels is one of the major ways by which we can generate electricity in many countries. Now the efficiency of each of the steps have been shown in the question;

  • Conversion efficiency = 35%
  • Transmission efficiency = 90%
  • Light  efficiency = 22%

Overall efficiency = (0.35 * 0.90 * 0.22) * 100 = 6.93%

The energy is obtained from;

6.7X10^13 J * 6.93/100

= 4.6 * 10^12 J

Learn more about nuclear energy:brainly.com/question/12793906

#SPJ4

Missing parts;

The atomic number of uranium-235 is 92, its half-life is 704 million years and the radioactive decay of l kg of 235U releases 6.7X10^13 J. Generally, radioactive material is considered safe after 10 half-lives.

Assume that a nuclear powerplant can convert energy from 235U into electricity with an efficiency of 35%, the electrical transmission lines operate at 90% efficiency, and flourescent lights operate at 22% efficiency.

1.) What is the overall efficiency of converting the energy of 235U into flourescent light?

2.) How much energy from 1 kg of 235U is converted into flourescent light?

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Vf=114 m/s.

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Tell the different energy transformations that happen in each of these items
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Goryan [66]

Answer:c=0.213 J/g/K  

Explanation:

Given

sample mass(m_s)=120 gm

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Heat capacity of calorimeter =15.3 J/k

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g Drop the object again and carefully observe its motion after it hits the ground (it should bounce). (Consider only the first b
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Answer:

a) quantity to be measured is the height to which the body rises

b) weighing the body , rule or fixed tape measure

c)   Em₁ = m g h

d) deformation of the body or it is transformed into heat during the crash

Explanation:

In this exercise of falling and rebounding a body, we must know the speed of the body when it reaches the ground, which can be calculated using the conservation of energy, since the height where it was released is known.

a) What quantities must you know to calculate the energy after the bounce?

The quantity to be measured is the height to which the body rises, we assume negligible air resistance.

So let's use the conservation of energy

starting point. Soil

          Em₀ = K = ½ m v²

final point. Higher

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         Em₀ = Em_f

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b) to have the measurements, we begin by weighing the body and calculating its mass, the height was measured with a rule or fixed tape measure and seeing how far the body rises.

c) We use conservation of energy

starting point. Soil

          Em₁ = K = ½ m v²

final point. Higher

          Em_f = U = mg h

         Em₁ = Em_f

         Em₁ = m g h

d) to determine if the energy is conserved, the arrival energy and the output energy must be compared.

There are two possibilities.

* that have been equal therefore energy is conserved

* that have been different (most likely) therefore the energy of the rebound is less than the initial energy, it cannot be stored in the possible deformation of the body or it is transformed into heat during the crash

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