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svetlana [45]
3 years ago
6

The light energy that falls on a square meter of ground over the course of a typical sunny day is about 20 MJ. The average rate

of electric energy consumption in one house is 1.0 kW
On average, how much energy does one house use during each 24 hr day?
Express your answer using two significant figures.
Physics
1 answer:
Free_Kalibri [48]3 years ago
7 0

Answer:

86MJ

Explanation:

THE TOPIC FROM WHICH THE QUESTION IS OBTAINED IS WORK, ENERGY AND POWER

  • Power is defined as the time rate of doing work.
  • Power =\frac{Energy used/ work done}{time}
  • If the energy use is in joules and the time is in seconds, then power is a measure of watts. 1watt = 1 joule per second.
  • According to the question, we need to find how much energy does one house use during each 24 hour day ( Already given time = 24hours) when the average rate of electric energy consumption in one house is 1.0 kW ( we have been given power = 1kW).
  • Power = 1kW = 1 × 10³W
  • Time = 24hours = 24 ×3600 = 86400seconds.
  • It is important we have our time in seconds so as to be consistent with unit and dimension

Power = \frac{Energy used/ work done}{time}

1 × 10³W =\frac{Energy used}{86400}

  • Energy used =1 × 10³W × 86400
  • Energy used = 86400000J = 86.4 × 10^{6}W  = 86.4MJ  

The question requested the answer in 2 significant figures and the answer is 86MJ

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In the nuclear fission process mass is converted into energy. Determine the total mass that must be converted to energy in one y
brilliants [131]

Answer:

1) Mass that needs to be converted at 100% efficiency is 0.3504 kg

2) Mass that needs to be converted at 30% efficiency is 1.168 kg

Explanation:

By the principle of mass energy equivalence we have

E=mc^{2}

where,

'E' is the energy produced

'm' is the mass consumed

'c' is the velocity of light in free space

Now the energy produced by the reactor in 1 year equals

Energy=Power\times time\\\\\therefore Energy=1\times 10^{9}\times 365\times 24\times 3600\\\\Energy=31.536\times 10^{15}Joules

Thus the mass that is covertred at 100% efficiency is

mass=\frac{Energy}{c^{2}}\\\\mass=\frac{31.536\times 10^{15}}{(3\times 10^{8})^{2}}\\\\mass=\frac{31.536\times 10^{15}}{9\times 10^{16}}\\\\\therefore mass=0.3504kg

Part 2)

At 30% efficiency the mass converted equals

mass|_{30}=\frac{mass|_{100}}{0.3}\\\\mass|_{30}=\frac{0.3504}{0.3}\\\\mass|_{30}=1.168kg

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