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joja [24]
3 years ago
10

What is the minimum coating thickness (but not zero) that will minimize the reflection at the wavelength of 705 nm where solar c

ells are most efficient?
Physics
1 answer:
DanielleElmas [232]3 years ago
4 0

Answer:

d = 235 nm.

Explanation:

Let the layer be of glass. Refractive index of glass μ = 1.5. Let the required  thickness be d.

For minimum reflection , that is destructive interference in thin films , the condition is

2μd = λ ( for first order)

d = 705 / 2 x 1.5 = 235 nm [ λ =705 nm ]

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PLEASE HELP!!! GIVING BRAINLIEST!! ill also answer questions that you have posted if you answer these correctly!!!! (47pts)
Bingel [31]

\huge\mathfrak\red{answer}

Jake was playing making a paper airplane, after making he kept it on the table and went to have food.

Suddenly his brother saw the plane and threw it in the air, The airplane kept flying for about 3m/s and it hit his mother and due to the force the plane stopped.

{kept in on the table>rest

brother threw the plane>moving

it hit his mother>force that stopped it}

(mark me brainliest if you're satisfied with my answer)

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8 0
3 years ago
HELP RIGHT NOW THE RIGHT ANSWER GETS BRAINLEST!!!
cupoosta [38]
Solar Flares: C
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3 years ago
Read 2 more answers
ASAP pls answer I will mark brainiest if right.
Sergio039 [100]
1) g
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5 0
3 years ago
Water is circulating through a closed system of pipes in a two floor apartment. On the first floor, the water has a gauge pressu
anastassius [24]

Answer:

The value of gauge pressure at outlet = -38557.224 pascal

Explanation:

Apply Bernoulli' s Equation

\frac{P_{1}}{9810} + \frac{V_{1} ^{2}}{19.62} + h_{1} = \frac{P_{2}}{9810} + \frac{V_{2} ^{2}}{19.62} + h_{2} --------------(1)

Where

P_{1} =  Gauge pressure at inlet = 3.70105 pascal

V_{1} = velocity at inlet =  2.4 \frac{m}{sec}

P_{2} = Gauge pressure at outlet = we have to calculate

V_{2} = velocity at outlet = 3.5 \frac{m}{sec}

h_{2} - h_{1} = 3.6 m

Put all the values in equation (1) we get,

⇒ \frac{3.70105}{9810} + \frac{2.4 ^{2}}{19.62} = \frac{P_{2}}{9810} + \frac{3.5 ^{2}}{19.62} + 3.6

⇒ 0.294 = \frac{P_{2}}{9810} + 0.6244 + 3.6

⇒ \frac{P_{2}}{9810} = 0.294 - 0.6244 - 3.6

⇒ \frac{P_{2}}{9810} = - 3.9304

⇒ P_{2} = - 38557.224 pascal

This is the value of gauge pressure at outlet.

3 0
3 years ago
A. 25%<br> B. 20%<br> C. 10%<br> D. 80%<br> please show work :)
Setler79 [48]

Substitute your values into the formula:

W = Work done = 288

Q_{in} = 360

Solve to find e:

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Convert e to a percentage by multiplying by 100.

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<h2>D. 80%</h2>
4 0
3 years ago
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