The potential difference across the parallel plate capacitor is 2.26 millivolts
<h3>Capacitance of a parallel plate capacitor</h3>
The capacitance of the parallel plate capacitor is given by C = ε₀A/d where
- ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m,
- A = area of plates and
- d = distance between plates = 4.0 mm = 4.0 × 10⁻³ m.
<h3>Charge on plates</h3>
Also, the surface charge on the capacitor Q = σA where
- σ = charge density = 5.0 pC/m² = 5.0 × 10⁻¹² C/m² and
- a = area of plates.
<h3>
The potential difference across the parallel plate capacitor</h3>
The potential difference across the parallel plate capacitor is V = Q/C
= σA ÷ ε₀A/d
= σd/ε₀
Substituting the values of the variables into the equation, we have
V = σd/ε₀
V = 5.0 × 10⁻¹² C/m² × 4.0 × 10⁻³ m/8.854 × 10⁻¹² F/m
V = 20.0 C/m × 10⁻³/8.854 F/m
V = 2.26 × 10⁻³ Volts
V = 2.26 millivolts
So, the potential difference across the parallel plate capacitor is 2.26 millivolts
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The resistance of the sample is 
Explanation:
The relationship between resistance of a material and temperature is given by the equation

where
is the resistance at the temperature 
is the temperature coefficient of resistance
For the sample of nickel in this problem, we have:
when the temperature is 
While the temperature coefficient of resistance of nickel is

Therefore, the resistance of the sample when its temperature is

is

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Answer:
350 F to 100 F it take approx 87.33 min
Explanation:
given data
oven = 350◦F
cooling rack = 70◦F
time = 30 min
cake = 200◦F
solution
we apply here Newtons law of cooling
= -k(T-Ta)
=
(T(t) -Ta)
=
= -k(T-Ta)
-ky
= -ky
T(t) -Ta = (To -Ta)
T(t) = Ta+ (To -Ta)
put her value for time 30 min and T(t) = 200◦F and To =350◦F and Ta = 70◦F
so here
200 = 70 + ( 350 - 70 ) 
k = 0.025575
so here for T(t) = 100F
100 = 70 + ( 350 - 70 ) 
time = 87.33 min
so here 350 F to 100 F it take approx 87.33 min
There is no <span>radioactive decay</span>