Normally, when something gets colder, its electrical resistance gets smaller. This is true of component-A in the drawing ... a simple resistor.
The component labeled 'B' has a strange and unusual symbol, and it's not a simple resistor. It's a "thermistor". The word "thermal" always has something to do with heat, and "thermistor" comes from "thermal resistor. These things can be manufactured either way ... using different materials, a thermistor can be manufactured so that its resistance goes UP, or goes DOWN, or doesn'tchange when it gets colder. I'm pretty sure that's what's going on here.
When this circuit gets colder, resistance-A gets smaller, but resistance-B either gets bigger OR doesn't change. Either way, the voltage across B increases. Since the LED is connected directly across B, the current through it depends on that voltage, so the LED gets more current, and becomes brighter, when A and B both get colder.
This circuit could actually be a very useful device. If you took out the LED and put a voltmeter in its place, then the reading on the voltmeter would tell you the temperature of wherever you put the two components A and B.
The electric potential at the origin of the xy coordinate system is negative infinity
<h3>What is the electric field due to the 4.0 μC charge?</h3>
The electric field due to the 4.0 μC charge is E = kq/r² where
- k = electric constant = 9.0 × 10 Nm²/C²,
- q = 4.0 μC = 4.0 × 10 C and
- r = distance of charge from origin = x₁ - 0 = 2.0 m - 0 m = 2.0 m
<h3>What is the electric field due to the -4.0 μC charge?</h3>
The electric field due to the -4.0 μC charge is E = kq'/r² where
- k = electric constant = 9.0 × 10 Nm²/C²,
- q' = -4.0 μC = -4.0 × 10 C and
- r = distance of charge from origin = 0 - x₂ = 0 - (-2.0 m) = 0 m + 2.0 m = 2.0 m
Since both electric fields are equal in magnitude and directed along the negative x-axis, the net electric field at the origin is
E" = E + E'
= -2E
= -2kq/r²
<h3>What is the electric potential at the origin?</h3>
So, the electric potential at the origin is V = -∫₂⁰E".dr
= -∫₂⁰-2kq/r².dr
Since E and dr = dx are parallel and r = x, we have
= -∫₂⁰-2kqdxcos0/x²
= 2kq∫₂⁰dx/x²
= 2kq[-1/x]₂⁰
= -2kq[1/x]₂⁰
= -2kq[1/0 - 1/2]
= -2kq[∞ - 1/2]
= -2kq[∞]
= -∞
So, the electric potential at the origin of the xy coordinate system is negative infinity
Learn more about electric potential here:
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The acceleration of Karla's I IPhone being thrown from Mr. Higley's classroom at 0 m/s will be 29.16 m/s²
<h3>What is acceleration?</h3>
The rate of velocity change concerning time is known as acceleration. According to Newton's second law, the eventual effect of all forces applied to a body is its acceleration.
The pace at which a body's velocity varies is represented by acceleration, which is a vector quantity.
The given data in the problem is given by ;
u is the initial speed = 0 m/sec
v is the final speed= 35 m/sec
t is the time interval= 1.2 second
a is the acceleration=? m/sec²
The formula for acceleration is;
Hence, the acceleration of Karla's iPhone being thrown from Mr. Higley's classroom at 0 m/s will be 29.16 m/s²
To learn more about acceleration, refer to the link;
brainly.com/question/2437624#SPJ2
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