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?
Answer:
voltage across = 1.6 V
Explanation:
given data
resistance R = 57.61 Ω
capacitance c = 13.13 mF = 13.13 ×
F
inductance L = 196.03 mH = 0.19603 H
fixed rms output Vrms = 23.86 V
to find out
voltage across circuit
solution
we know resonant frequency that is
resonant frequency = 1 / ( 2π√(LC)
put the value
resonant frequency = 1 / ( 2π√(0.19603×13.13 ×)
resonant frequency f = 3.1370 HZ
so current will be at this resonant is
current = Vrms / R
current = 23.86 / 57.61
current = 0.4141 A
and
so voltage across will be
voltage across = current / ( 2π f C )
voltage across = 0.4141 / ( 2π ( 3.1370) 13.13 × )
voltage across = 1.6 V
The distance from the Sun to the Earth is 149,600,000 km.
The answer is C. You divide 4000 kg/s by 115 kg.
