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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 3.18-kg rock is released from rest at a height of 26.6 m. ignore air resistance and determine (a) the kinetic energy at 26.6 m
antiseptic1488 [7]

Answer:

(a) 0 J

(b) 828.96 J

(c) 828.96 J

(d) 828.96 J

(e) 0 J

(f) 828.96 J

Explanation:

Given:

Mass of the rock is, m=3.18\ kg

Initial height of the rock is, h_{i}=26.6\ m

Final height of the rock is, h_f=0\ m

Initial velocity of the rock is, v_i=0\ m/s(Rest)

Acceleration due to gravity is, g=9.8\ m/s^2

(a)

Initial kinetic energy is given as:

K_i=\frac{1}{2}mv_i^2

Plug in the given values and solve for 'K_i'. This gives,

K_i=\frac{1}{2}\times 3.18\times 0^2=0

Therefore, initial kinetic energy is 0 J.

(b)

Initial potential energy is given as:

U_i=mgh_i\\U_i=3.18\times 9.8\times 26.6=828.96\ J

Therefore, initial potential energy is 828.96 J.

(c)

Mechanical energy is equal to the sum of kinetic and potential energy. It is always a constant value. Therefore,

ME=K_i+P_i\\ME=0+828.96=828.96\ J

Therefore, mechanical energy at 26.6 m is 828.96 J.

(d)

Now, at the final height of 0 m, the decrease in potential energy will be equal to the increase in the kinetic energy. In other words, the potential energy at the start will be converted to kinetic energy at the bottom.

Therefore, kinetic energy at the 0 m is given as:

K_f=U_i=828.96\ J

(e)

Potential energy at the final height is given as:

P_f=mgh_f=3.18\times 9.8\times 0=0\ J

Therefore, final potential energy is 0 J.

(f)

Total mechanical energy is a constant and doesn't depend on the height.

So, total mechanical energy at 0 m is same as that at 26.6 m.

Total mechanical energy is 828.96 J.

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