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2 years ago

What is the purpose of the antireflective coating on the glass layer of a solar

Physics

2 answers:

bonufazy [111]2 years ago

7 0

Answer:

The answer is D if you have something that is reflective how are you going to absorb any energy.

Explanation:

A P E X

gizmo_the_mogwai [7]2 years ago

3 0

Answer:

I think d. is the answer I'm not conform

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PLEASE PLEASE HELP ME!!!!!!!!!!

marysya [2.9K]

373 kelvin = 99.9 Celsius. Round makes it 100. 373 kelvin also equals 212 Fahrenheit so the correct answer is A.
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8 0

3 years ago

Fan object moves in uniform circular motion in a circle of radius R=200 meters, and the objectes 5.00 seconds to

Leviafan [203]

Answer:

The centripetal acceleration of the object is $31550.72\ m/s^2$ .

Explanation:

We have,

Radius of a circular path is 200 m

It takes 5 seconds to complete 10 revolutions. The angular velocity of the object is given by the rate of change of angular displacement per unit time :

$\omega=\dfrac{\Delta \theta}{\Delta t}\\\\\omega=\dfrac{2\pi \times 10}{5}\\\\\omega=12.56\ rad/s$

The centripetal acceleration of the object is given by :

$a=\omega^2 r\\\\a=(12.56)^2 \times 200\\\\a=31550.72\ m/s^2$

So, the centripetal acceleration of the object is $31550.72\ m/s^2$ .

6 0

3 years ago

Topics related to physics

Bezzdna [24]

Answer:

There are many topics related to physics such as :

Kinematics

Dynamics

Light

Sound

7 0

2 years ago

Roughly how many stars are in the Milky Way Galaxy?

svetlana [45]

Answer:

there are many stars in the milky way galaky

Explanation:

uncountable stars

7 0

3 years ago

Read 2 more answers

An object is placed in front of a convex lens of a length 10cm. What is the nature of the image formed if the object distance is

Lady_Fox [76]

${\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Focal\:length=10\:cm}$

$\:\:\:\:\bullet\:\:\:\sf{Object \ distance = -15\:cm}$

$\\$

${\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Nature \: of \:the\:image}$

$\\$

${\mathfrak{\underline{\purple{\:\:\: Solution:-\:\:\:}}}} \\ \\$

<h3>☯ <u>By using formula of Lens</u> </h3>

$\\$

$\dashrightarrow\:\: {\boxed{\sf{\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}}}}$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{1}{v}-\dfrac{1}{-15}=\dfrac{1}{10}}$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{1}{v}+\dfrac{1}{15}=\dfrac{1}{10}}$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{15}}$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{1}{v} = \dfrac{1}{30}}$

$\\$

$\dashrightarrow\:\: \sf{ v = 30 \ cm}$

$\\$

<h3>☯ <u>Now, Finding the magnification </u></h3>

$\\$

$\dashrightarrow\:\: \sf{ m = \dfrac{-30}{-15}}$

$\\$

$\dashrightarrow\:\: \sf{m = -2}$

$\\$

<h3>☯ <u>Hence</u>, $\\$ </h3>

$\:\:\:\:\star\:\:\:\sf{Image \ distance = 30 \ cm}$

$\:\:\:\:\star\:\:\:\sf{Nature = Real \ \& \ inverted}$

3 0

2 years ago