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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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3 years ago
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A ball dropped from a window strikes the ground 2.76 seconds later. How high is the window above the ground
sweet [91]

Answer:

37.33m

Explanation:

Using the equation of motion

S = ut + 1/2gt^2

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3 years ago
The electric field strength in the space between two closely spaced parallel disks is 1.0 105 N/C. This field is the result of t
balu736 [363]

Answer:

D=2.996\times 10^{-2} m

Explanation:

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Given the electric strength as  1.0\times 10^5 N/C.  transferring 3.9\times 10^9 electrons, the disk's Area can be calculated using the formula:

E=\frac{\eta}{\epsilon_o}=\frac{Q}{A\epsilon_o}\\\\A=\frac{Q}{E\epsilon_o}\\\\=\frac{(3.9\times 10^9)\times (1.6\times10^{-19})}{(1.0\times 10^5 )\times (8.85\times10^{-12})}\\\\A=7.0508\times 10^{-4} \ m^2

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A=\pi(D/2)^2\\\\2\sqrt{\frac{A}{\pi}}=D\\\\=2\sqrt{7.0508\times 10^{-4}/\pi}\\\\D=2.996\times 10^{-2} \ m

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