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tigry1 [53]
3 years ago
13

1)

Engineering
1 answer:
Kryger [21]3 years ago
7 0

Answer:

b

Explanation:

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Determine the general lighting load for a two-story office building that measures 125 feet by 150 feet.
Irina-Kira [14]

The general lighting load for a two-story office building that measures 125 feet by 150 feet is 112, 500 sq ft.

<h3>What is lighting load?</h3>

Lighting loads are the energy used to power electric lights and they make up nearly a third of US commercial building energy use.

Lighting load = n(LW)

where;

  • L is length of the building
  • W is width of the building
  • n is number of story building

For one story building, = 3

For two story building, n = 6

Lighting load = 6 x 125 x 150 = 112, 500 sq ft.

Learn more about lighting load here: brainly.com/question/14070748

#SPJ12

7 0
2 years ago
An excited electron in an Na atom emits radiation at a wavelength 589 nm and returns to the ground state. If the mean time for t
11Alexandr11 [23.1K]

Answer:   Inherent width in the emission line: 9.20 × 10⁻¹⁵ m or 9.20 fm

                length of the photon emitted: 6.0 m

Explanation:

The emitted wavelength is 589 nm and the transition time is ∆t = 20 ns.

Recall the Heisenberg's uncertainty principle:-

                                 ∆t∆E ≈ h ( Planck's Constant)

The transition time ∆t corresponds to the energy that is ∆E

E=h/t = \frac{(1/2\pi)*6.626*10x^{-34} J.s}{20*10x^{-9} } = 5.273*10x^{-27} J =  3.29* 10^{-8} eV.

The corresponding uncertainty in the emitted frequency ∆v is:

∆v= ∆E/h = (5.273*10^-27 J)/(6.626*10^ J.s)=  7.958 × 10^6 s^-1

To find the corresponding spread in wavelength and hence the line width ∆λ, we can differentiate

                                                    λ = c/v

                                                    dλ/dv = -c/v² = -λ²/c

Therefore,

      ∆λ = (λ²/c)*(∆v) = {(589*10⁻⁹ m)²/(3.0*10⁸ m/s)} * (7.958*10⁶ s⁻¹)

                                 =  9.20 × 10⁻¹⁵ m or 9.20 fm

     The length of the photon (<em>l)</em> is

l = (light velocity) × (emission duration)

  = (3.0 × 10⁸  m/s)(20 × 10⁻⁹ s) = 6.0 m          

                                                   

6 0
3 years ago
Describe the refrigeration cycle &amp; how it transfers heat, including all the components &amp; lines. be sure to discuss why t
Amanda [17]

Heat can be diverted away from the area you want to cool using the refrigeration cycle, also known as a heat pump cycle. This is done by repeatedly compressing and expanding the working refrigerant (air, water, synthetic refrigerants, etc.) to change its pressure.

<h3>What is the cycle of a heat pump?</h3>
  • In order to heat the air within a building, a heat pump warms the air outside by drawing heat from it and adding it to it. The following procedure is used to achieve this: A gas is created when liquid refrigerant absorbs heat from the ambient air in the "evaporator."
  • The pressure buildup from the compressor raises the temperature of the refrigerant, just like it does with everything that is put under pressure. The refrigerant is a heated vapor, ranging in temperature from 120° to 140°F, when it exits the compressor.
  • The chemical refrigerant, once transformed from a gas to a liquid state, has a remarkable capacity to absorb heat. The heat inside your house might be absorbed and released outside when the refrigerant undergoes a state shift as a result of compression and pressurization.  

To learn more about Compressor refer to:

brainly.com/question/29345949

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6 0
1 year ago
3. Ang mga kababaihan noon ay tumutulong sa pakikipaglaban sa mga Espanyol upang makamit ang Kalayaan. Ganito pa rin ba ang mga
likoan [24]

NYC dp my frndsssssssssss

5 0
2 years ago
Calculate the fraction of lattice sites that are Schottky defects for cesium chloride at 573 oC (this temperature is below the m
inn [45]

Answer:

2.9\times 10^{-6}

Explanation:

Q_s = Energy for defect formation = 1.86 eV

T = Temperature = 573^{\circ}\text{C}=573+273.15=846.15\ \text{K}

k = Boltzmann constant = 8.62\times 10^{-5}\ \text{eV/K}

The fraction of lattice sites that are Schottky defects is given by

\dfrac{N_s}{N}=e^{-\dfrac{Q_s}{2kt}}\\\Rightarrow \dfrac{N_s}{N}=e^{-\dfrac{1.86}{2\times 8.62\times 10^{-5}\times 846.15}}\\\Rightarrow \dfrac{N_s}{N}=2.9\times 10^{-6}

The required ratio is 2.9\times 10^{-6}.

6 0
3 years ago
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