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ExtremeBDS [4]
1 year 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:
wel1 year 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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Given Information:

Pendulum 1 mass = m₁ = 0.2 kg

Pendulum 2 mass = m₂ = 0.6 kg

Pendulum 1 length = L₁ = 5 m

Pendulum 2 length = L₂ = 1 m

Required Information:

Affect of mass on the frequency of the pendulum = ?

Answer:

The mass of the ball will not affect the frequency of the pendulum.

Explanation:

The relation between period and frequency of pendulum is given by

f = 1/T

The period of pendulum is given by

T = 2π√(L/g)

Where g is the acceleration due to gravity and L is the length of the string

As you can see the period (and frequency too) of pendulum is independent of the mass of the pendulum. Therefore, the mass of the ball will not affect the frequency of the pendulum.

Bonus:

Pendulum 1:

T₁ = 2π√(L₁/g)

T₁ = 2π√(5/9.8)

T₁ = 4.49 s

f₁ = 1/T₁

f₁ = 1/4.49

f₁ = 0.22 Hz

Pendulum 2:

T₂ = 2π√(L₂/g)

T₂ = 2π√(1/9.8)

T₂ = 2.0 s

f₂ = 1/T₂

f₂ = 1/2.0

f₂ = 0.5 Hz

So we can conclude that the higher length of the string increases the period of the pendulum and decreases the frequency of the pendulum.

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The reciprocal of the total resistance is equal to the sum of the reciprocals of the component resistances:

1/(120.7 Ω) = 1/<em>R₁</em> + 1/(221.0 Ω)

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