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.
Answer:
A. We have that radius r = 4.00m intensity I = 8.00 W/m^
total power = power/ Area ( 4πr2)= 8.00 w/m^2( 4π ( 4.00 m)2=1607.68 W
b) I = total power/ 4πr2= 8.00 W/m2 ( 4.00 m/ 9.5 m)2= 1.418 W/m2
c) E = total power x time= 1607 . 68 W x 1s= 1607.68 J
The centrifugal force C = mv^2/r = kq^2/r^2 = P centripetal force. m is the electron mass, q are the proton and electron charges (opposites), and r is the Bohr radius.
Thus 1/2 mv^2/r = 1/2 kq^2/r^2 and KE = 1/2 mv^2 = 1/2 kq^2/r = 1/2 PE
Therefore KE/PE = 1/2, no matter what state the electron is in.