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3 years ago

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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3 years ago

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The basic barometer can be used to measure the height of a building. If the barometric readings at the top and at the bottom of

bixtya [17]

Answer:

h = 287.1 m

Explanation:

the density of mercury \rho =13570 kg/m3

the atmospheric pressure at the top of the building is

$p_t = \rho gh = 13570*908*0.73 = 97.08 kPa$

the atmospheric pressure at bottom

$p_b = \rho gh = 13570*908*0.75 = 100.4 kPa$

$\frac{w_{air}}{A} =p_b -p_t$

we have also

$(\rho gh)_{air} = p_b - p_t$

1.18*9.81*h = (100.4 -97.08)*10^3

h = 287.1 m

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3 years ago

Ammonia contained in a piston-cylinder assembly, initially saturated vapor at 0o F, undergoes an isothermal process during which

Rudik [331]

ANSWERS:

$-P_{2(a)} =15.6lbf/in^2\\-P_{2(b)} =30.146lbf/in^2\\ T_{2(a)} =0^oF\\T_{2(b)} =0^oF\\x_{2(b)} =49.87percent$

Explanation:

Given:

Piston cylinder assembly which mean that the process is constant pressure process P=C.

<u>AMMONIA </u>

state(1)

saturated vapor $x_{1} =1$

The temperature $T_{1} =0^0 F$

Isothermal process $T=C$

$-V_{2} =2V_{1}$ ( double)

$-V_{2} =.5V_{2}$ (reduced by half)

To find the final state by giving the quality in lbf/in we assume the friction is neglected and the system is in equilibrium.

state(1)

using PVT data for saturated ammonia

$-P_{1} =30.416 lbf/in^2\\-v_{1} =v_{g} =9.11ft^3/lb$

then the state exists in the supper heated region.

a) from standard data

$-v_{1(a)} =2v_{1} =18.22ft^3/lb\\-T_{1} =0^oF$

$at\\P_{x} =14lbf/in^2\\-v_{x} =20.289 ft^3/kg$

$at\\P_{y} =16 lbf/in^2\\-v_{y} =17.701ft^3/kg$

assume linear interpolation

$\frac{P_{x}-P_{2(b)} }{P_{x}- P_{y} } =\frac{v_{x}-v_{1(a)} }{v_{x}-v_{y} }$

$P_{1(b)}=P_{x} -(P_{x} -P_{y} )*\frac{v_{x}- v_{1(b)} }{v_{x}-v_{y} }\\ \\P_{1(b)} =14-(14-16)*\frac{20.289-18.22}{20.289-17.701} =15.6lbf/in^2$

$-v_{2(a)} =2v_{1} =4.555ft^3/lb\\v_{g}$

from standard data

$-v_{f} =0.02419ft^3/kg\\-v_{g} =9.11ft^3/kg\\v_{f}$

then the state exist in the wet zone

$-P_{s} =30.146lbf/in^2\\v_{2(a)} =v_{f} +x(v_{g} -v_{f} )$

$x=\frac{v_{2(a)-v_{f} } }{v_{g} -v_{f} } \\x=\frac{4.555-0.02419}{9.11-0.02419} =49.87%$

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3 years ago

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SVEN [57.7K]

Answer:

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2 years ago

Read 2 more answers

Airflow through a long, 0.15-m-square air conditioning duct maintains the outer duct surface temperature at 10°C. If the horizon

Ulleksa [173]

The complete Question is:

Airflow through a long, 0.15-m-square air conditioning duct maintains the outer duct surface temperature at 10°C. If the horizontal duct is uninsulated and exposed to air at 35°C in the crawlspace beneath a home, what is the heat gain per unit length of the duct? Evaluate the properties of air at 300 K. For the sides of the duct, use the more accurate Churchill and Chu correlations for laminar flow on vertical plates.

What is the Rayleigh number for free convection on the outer sides of the duct?

What is the free convection heat transfer coefficient on the outer sides of the duct, in W/m2·K?

What is the Rayleigh number for free convection on the top of the duct?

What is the free convection heat transfer coefficient on the top of the duct, in W/m2·K?

What is the free convection heat transfer coefficient on the bottom of the duct, in W/m2·K?

What is the total heat gain to the duct per unit length, in W/m?

Answers:

- 7709251 or 7.709 ×10⁶

- 4.87

- 965073

- 5.931 W/m² K

- 2.868 W/m² K

- 69.498 W/m

Explanation:

Find the given attachments for complete explanation

4 0

3 years ago

Describe three objects in your house that use LEDs ​

Describe three objects in your house that use LEDs