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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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A carriage of 20 kg is pulled with a force of 35 N. How far the carriage will go
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Answer:

2.71 m

Explanation:

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calculate the diameter of a silver wire of length 75cm , which is extended by 1.85mm when a 10kg mass is suspended from it's end
sdas [7]

Answer:0.8\ mm

Explanation:

Given

length of wire l=75\ cm

change in length \Delta l=1.85\ mm

mass of wire m=10\ kg

Young's modulus for silver E=7.9\times 10^{10}\ N/m^2

load on wire F=mg

F=10\times 9.8=98\ kg

change in length is given by

\Delta l=\dfrac{Pl}{AE}

Where A=area of cross-section

A=\dfrac{Pl}{\Delta lE}

A=\dfrac{98\times 0.75}{1.85\times 10^{-3}\times 7.9\times 10^{10}}

A=\dfrac{73.5}{14.615\times 10^{7}}

A=5.029\times 10^{-7}\ m^2

also wire is the shape of cylinder so cross-section is given by

A=\dfrac{\pi d^2}{4}=5.029\times 10^{-7}\ m^2

\Rightarrow d^2=\dfrac{5.029\times 10^{-7}\times 4}{\pi }

\Rightarrow d^2=64.02\times 10^{-8}

\Rightarrow d=8\times 10^{-4}\ m

\Rightarrow d=0.8\ mm

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