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trasher [3.6K]
3 years ago
14

Solar panels convert light energy from sunlight into electrical energy. What material is most likely used in solar panels, and w

hy?
A nonmetal is used because it is a semiconductor and can become more conductive when more light shines on it.
A metalloid is used because it is a good conductor but can become less conductive when more light shines on it.
A metalloid is used because it is a semiconductor and can become more conductive when more light shines on it.
A nonmetal is used because it is a poor conductor but can become more conductive when more light shines on it.
Physics
2 answers:
Phantasy [73]3 years ago
8 0

Answer:

Its the first answer :D

Explanation:

klemol [59]3 years ago
3 0

Answer:

A metalloid is used because it is a semiconductor and can become more conductive when more light shines on it

Explanation:

The material used in a solar panel is a metalloid. It can often become conductive when more light shines on it.

Metalloids have properties that straddles between those of metals and non-metals.

In essence, they can be conductive or not under certain conditions.

The most important property they exhibit is that they can become more conductive when more light shines on them. This way more electrons are produced.

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The purpose of the scapula to move during arm elevation is increase the range of elevation of the arm.

<h3>What is the importance of movement of the scapula during arm elevation?</h3>

The scapula is an important bone which is found in the shoulder and back region of the body.

The scapula enables and increases the range of motion of the arm with its motions.

During arm elevation, the scapula undergoes an upward rotational motion.

Therefore, the  purpose of the scapula to move during arm elevation is increase the range of elevation of the arm.

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3 0
1 year ago
The electric field in a certain region of Earth’s atmo- sphere is directed vertically down. At an altitude of 300 m the field ha
I am Lyosha [343]

Answer:

Net charge contained in the cubeq= 3.536×10^-6C

Explanation:

Formular for total flux in a cube is given as:

Total flux= E300Acos(180) + E200Acos(0)

Where A is crossectional area

Total flux= A(E200-E300)

Total flux= q/Eo

q= Eo×total flux

q=(8.84×10^-12)×(100)^2×(100-60)

q= 3.536×10^-6C

6 0
3 years ago
Read 2 more answers
A 0.12-kg metal rod carrying a current of current 4.1 A glides on two horizontal rails separation 6.3 m apart. If the coefficien
Neporo4naja [7]

Answer:

The magnetic field is B  =  8.20 *10^{-3} \  T

Explanation:

From the question we are told that

   The  mass of the metal rod is  m  = 0.12 \ kg

    The current on the rod is  I  = 4.1 \ A

    The distance of separation(equivalent to length of the rod ) is L   = 6.3 \ m

     The coefficient of kinetic friction is \mu_k  =  0.18

      The kinetic frictional force is  F_k  = 0.212 \ N

     The constant speed is v  = 5.1 \ m/s

Generally the magnetic force on the rod is mathematically represented as  

      F  =  B * I  *   L

For  the rod to move with a constant velocity the magnetic force must be equal to the kinetic frictional force so

        F_ k  =  B*  I  *  L

=>      B  =  \frac{F_k}{L  *  I  }

=>       B  =  \frac{0.212}{ 6.3   *  4.1   }

=>       B  =  8.20 *10^{-3} \  T

7 0
3 years ago
Find the net force for <br> 10 N<br> 10 N<br> 25degree<br> 5N<br> 5N
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Answer:

30n

Explanation:

7 0
3 years ago
The speed of a wave is 40 m/s. If the wavelength is 80 centimeters, what is the frequency of the wave?
Vedmedyk [2.9K]
<h2><u>Q</u><u>u</u><u>e</u><u>s</u><u>t</u><u>i</u><u>o</u><u>n</u>:-</h2>

The speed of a wave is 40 m/s. If the wavelength is 80 centimeters, what is the frequency of the wave ?

<h2><u>A</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u>:-</h2>

<h3>Given:-</h3>

Velocity (V) = 40 m/s

Wavelength (\lambda) = 80 cm = 0.8 m

<h3>To Find:-</h3>

The frequency (F) of the wave.

<h2>Solution:-</h2>

We know,

\bf V \: = \: F \: × \: \lambda

40 = F × 0.8

F = \frac{40}{0.8}

F = 50

<h3>The frequency of the wave is <u>5</u><u>0</u><u> </u><u>H</u><u>z</u>. [Answer]</h3>
7 0
3 years ago
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