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Ira Lisetskai [31]

3 years ago

15

Consider a household that uses 22.0 kW-hour of electricity per day on average. Most of that electricity is supplied by fossil fu

els. To reduce their carbon footprint, the household wants to install solar panels, which receive on average 376 W/m^2 from the Sun each day.
A) If the solar panels are 18.1% efficient (fraction of solar energy converted into useable electrical energy), what area of solar panels is needed to power the household?

1 answer:

Lelu [443]3 years ago

3 0

Answer:

$A_{net} = 13.469\,m^{2}$

Explanation:

The daily energy consumption of the household is:

$\Delta E_{day} = (22\,kWh)\cdot (\frac{3600000\,J}{1\,kWh} )$

$\Delta E_{day} = 79200000\,J$

The needed area of solar panel to power the household is:

$\Delta E_{day} = \eta\cdot G\cdot A_{net}\cdot \Delta t$

$A_{net} = \frac{\Delta E_{day}}{\eta\cdot G\cdot \Delta t}$

$A_{net} = \frac{79200000\,J}{(0.181)\cdot (376\,\frac{W}{m^{2}} )\cdot (86400\,s)}$

$A_{net} = 13.469\,m^{2}$

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A body which has surface area 5cm² and temperature of 727°C radiates 300J of energy in one minute. Calculate it's emissivity giv

cestrela7 [59]

<h2>Answer: 0.17</h2>

Explanation:

The Stefan-Boltzmann law establishes that a black body (an ideal body that absorbs or emits all the radiation that incides on it) "emits thermal radiation with a total hemispheric emissive power proportional to the fourth power of its temperature":

$P=\sigma A T^{4}$ (1)

Where:

$P=300J/min=5J/s=5W$ is the energy radiated by a blackbody radiator per second, per unit area (in Watts). Knowing $1W=\frac{1Joule}{second}=1\frac{J}{s}$

$\sigma=5.6703(10)^{-8}\frac{W}{m^{2} K^{4}}$ is the Stefan-Boltzmann's constant.

$A=5cm^{2}=0.0005m^{2}$ is the Surface area of the body

$T=727\°C=1000.15K$ is the effective temperature of the body (its surface absolute temperature) in Kelvin.

However, there is no ideal black body (ideal radiator) although the radiation of stars like our Sun is quite close. So, in the case of this body, we will use the Stefan-Boltzmann law for real radiator bodies:

$P=\sigma A \epsilon T^{4}$ (2)

Where $\epsilon$ is the body's emissivity

(the value we want to find)

Isolating $\epsilon$ from (2):

$\epsilon=\frac{P}{\sigma A T^{4}}$ (3)

Solving:

$\epsilon=\frac{5W}{(5.6703(10)^{-8}\frac{W}{m^{2} K^{4}})(0.0005m^{2})(1000.15K)^{4}}$ (4)

Finally:

$\epsilon=0.17$ (5) This is the body's emissivity

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2 years ago

A horizontal uniform bar of mass 2.7 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to t

Jlenok [28]

Answer:

14.36 N

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$T_{1}$ = Tension in string 1

$T_{2}$ = Tension in string 2

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$W_b = m_{b} g = (2.7) (9.8) = 26.46$ N

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weight of the monkey is given as

$W_m = m_{m} g = (1.35) (9.8) = 13.23$ N

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7 0

3 years ago

Determine the mechanical energy of this object a 1-kg ball rolls on the ground at <br> m/s

dedylja [7]

Mechanical energy = potential energy + kinetic energy
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3 years ago

Oduvanchick [21]

Answer:

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Explanation:

This is a uniformly accelerated motion, so we can determine the deceleration of the car by using a suvat equation:

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a is the acceleration

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u = 27.8 m/s

v = 0

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natta225 [31]

Answer:

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2 years ago