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
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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:
The energy may be carried in the form of (1) radiation, where energy travels in the form of light, and (2) convection, where energy is carried by physical motion of upwelling solar gas.
Explanation:
Answer:
tan is 15 for that triangle
347÷134=2.589552239 meters per second
2.589552239×60= 155.3731343 meters per hour
155.3731343 meters per hour= 0.096544389701642 miles per hour
hopefully this was right.
