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Bumek [7]
3 years ago
14

Describe three objects in your house that use LEDs ​

Engineering
2 answers:
Firlakuza [10]3 years ago
5 0

Answer:

1. Accent Lighting: Often used on stair wells and kitchen islands, LED accent lighting is a cost-effective way to light your home at night – and avoid stubbing your toe in the middle of the night. It’s also effortless to recreate using flexible LED strips or rope lights, which can fit almost anywhere.

2. Faucets & Showerheads: Adding LEDs to a bathroom or kitchen faucet, or a showerhead, produces a neat, futuristic effect especially when paired with streamlined chrome. All you’ll need to do is install a new showerhead to enjoy a shower in bright blue or purple hues. There’s even a designer line of LED sinks and tubs if you care to extend the look to the rest of your bathroom.

3. Wallpapers: LED walls are nothing new; folks have been rigging those up through various methods for years, but now there is LED wallpaper – in a few different styles from high-tech geekery, to more subtle designs. If that’s far too much effort, you can still take part in the trend with the LED wave projector, which drapes your wall in cool blue tones. There are also some lovely ceiling designs that use fibre optics that would probably be quite easy to duplicate with LEDs as well.

4. Centerpieces: Much like adding a tiny, cool plastic LED light disk into a faux candle provides you with a warm flame effect, putting colored LED disks into centerpieces will add atmosphere to the party. Likewise, waterproof LEDs can be added to glasses, and strands of LEDs can be worked into flower arraignments.

5. Planter Pots: Green thumbs don’t get left out when it comes to LEDs – far from it. In fact, LEDs are commonly used to help plants grow. So, an LED planter pot is another perfect combination of style and substance. Studio Shulab’s Lightpot and Rotoluxe both have excellent examples; Firebox has a solar powered version.

Hope it helps

Please mark me as the brainliest

Thank you

Mashutka [201]3 years ago
4 0
Light fixtures lamps and tv i think
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Water at 15°C is to be discharged from a reservoir at a rate of 18 L/s using two horizontal cast iron pipes connected in series
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Answer:

0.245 m^3/s

Explanation:

Flow rate through pipe a is 0.4 m3/s Parallel pipes have a diameter D = 30 cm => r = 15 cm = 0.15 m Length of Pipe a = 1000m Length of Pipe b = 2650m Temperature = 15 degrees Va = V / A = (0.4m3/s) / (3.14 (0.15m)^2) = 5.66 m/s h = (f(LV^2)) / D2g (fa(LaVa^2)) / Da2g = (fb(LbVb^2)) / Da2g and Da = Db; fa = fb LaVa^2 = LbVb^2 => La/Lb = Vb^2/Va^2 Vd^2 = Va^2(La/Lb) => Vb = Va(La/Lb)^(1/2) Vb = 5.66 (1000/2650)^(1/2) => 5.66 x 0.6143 = 3.4769 m/s Vb = 3.4769 m/s V = AVb = 3.14(0.15)^2 x 3.4769 m/s = 0.245 m^3/s

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3 years ago
A fill covering a wide area is to be placed at the surface of this profile. The fill has a total unit weight of 20 kN/m^3 and is
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Answer:

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answer : 81.63 mm

Explanation:

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3 years ago
Decide whether the function is an exponential growth or exponential decay function, and find the constant percentage rate of gro
sasho [114]

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During his military campaign in what is now Germany, Julius Caesar lead his army of 40,000 soldiers to the western bank of the R
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3 years ago
The exhaust steam from a power station turbine is condensed in a condenser operating at 0.0738 bar(abs). The surface of the heat
lozanna [386]

Answer:

Percentage change 5.75 %.

Explanation:Given ;

Given

 Pressure of condenser =0.0738 bar

Surface temperature=20°C

Now from steam table

Properties of steam at 0.0738 bar  

Saturation temperature corresponding to saturation pressure =40°C      

 h_f= 167.5\frac{KJ}{Kg},h_g= 2573.5\frac{KJ}{Kg}

So Δh=2573.5-167.5=2406 KJ/kg

Enthalpy of condensation=2406 KJ/kg

So total heat=Sensible heat of liquid+Enthalpy of condensation

Total\ heat\ =C_p\Delta T+\Delta h

Total heat =4.2(40-20)+2406

Total heat=2,544 KJ/kg

Now film coefficient before inclusion of sensible heat

  h_1=\dfrac{\Delta h}{\Delta T}

  h_1=\dfrac{2406}{20}

h_1=120.3\frac{KJ}{kg-m^2K}

Now film coefficient after inclusion of sensible heat

 h_2=\dfrac{total\ heat}{\Delta T}

 h_2=\dfrac{2,544}{20}

h_2=127.2\frac{KJ}{kg-m^2K}

So\ Percentage\ change=\dfrac{h_2-h_1}{h_1}\times 100

             =\dfrac{127.2-120.3}{120.3}\times 100

                   =5.75 %

So Percentage change 5.75 %.

3 0
3 years ago
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