The tape is charged by friction and her hand has an induced opposite charge. They therefore attract.
Answer:
ΔTmin = 3.72 °C
Explanation:
With a 16-bit ADC, you get a resolution of 
steps. This means that the ADC will divide the maximum 10V input into 65536 steps:
ΔVmin = 10V / 65536 = 152.59μV
Using the thermocouple sensitiviy we can calculate the smallest temperature change that 152.59μV represents on the ADC:
The centripetal acceleration is 
Explanation:
For an object in uniform circular motion, the centripetal acceleration is given by
where
v is the speed of the object
r is the radius of the circle
The speed of the object is equal to the ratio between the length of the circumference (
) and the period of revolution (T), so it can be rewritten as
Therefore we can rewrite the acceleration as
For the particle in this problem,
r = 2.06 cm = 0.0206 m
While it makes 4 revolutions each second, so the period is
Substituting into the equation, we find the acceleration:
Learn more about centripetal acceleration:
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You have effectively got two capacitors in parallel. The effective capacitance is just the sum of the two.
Cequiv = ε₀A/d₁ + ε₀A/d₂ Take these over a common denominator (d₁d₂)
Cequiv = ε₀d₂A + ε₀d₁A / (d₁d₂) Cequiv = ε₀A( (d₁ + d₂) / (d₁d₂) )
B) It's tempting to just wave your arms and say that when d₁ or d₂ tends to zero C -> ∞, so the minimum will occur in the middle, where d₁ = d₂
But I suppose we ought to kick that idea around a bit.
(d₁ + d₂) is effectively a constant. It's the distance between the two outer plates. Call it D.
C = ε₀AD / d₁d₂ We can also say: d₂ = D - d₁ C = ε₀AD / d₁(D - d₁) C = ε₀AD / d₁D - d₁²
Differentiate with respect to d₁
dC/dd₁ = -ε₀AD(D - 2d₁) / (d₁D - d₁²)² {d2C/dd₁² is positive so it will give us a minimum} For max or min equate to zero.
-ε₀AD(D - 2d₁) / (d₁D - d₁²)² = 0 -ε₀AD(D - 2d₁) = 0 ε₀, A, and D are all non-zero, so (D - 2d₁) = 0 d₁ = ½D
In other words when the middle plate is halfway between the two outer plates, (quelle surprise) so that
d₁ = d₂ = ½D so
Cmin = ε₀AD / (½D)² Cmin = 4ε₀A / D Cmin = 4ε₀A / (d₁ + d₂)
I think that the solar panel would work under a fluorescent or halogen light if the photons are being produced. These types of lights mimic sunlight so it would not work as good as the real thing but it could work. Just not be as powerful.
