Answer:
7.9060 m²
8.57 Volts
5.142×10⁻⁶ Joule
1.2×10⁻⁶ Coulomb
Explanation:
C = Capacitance between plates = 0.14 μF = 0.14×10⁻⁶ F
d = Distance between plates = 0.5 mm = 0.5×10⁻³ m
Q = Charge = 1.2 μC = 1.2×10⁻⁶ C
ε₀ = Permittivity = 8.854×10⁻¹² F/m
Capacitance
∴ Area of each plate is 7.9060 m²
Voltage
∴ Potential difference between the plates if the capacitor is charged to 1.2 μC is 8.57 Volts.
Energy stored
E=0.5CV²
⇒E = 0.5×0.14×10⁻⁶×8.57²
⇒E = 5.142×10⁻⁶ Joule
∴ Stored energy is 5.142×10⁻⁶ Joule
Charge
Q = CV
⇒Q = 0.14×10⁻⁶×8.57
⇒Q = 1.2×10⁻⁶ C
∴ Charge the capacitor carries before a spark occurs between the two plates is 1.2×10⁻⁶ Coulomb
Speed = (distance) / (time)
Speed = (2.3 m) / (3 sec)
Speed = (2.3/3) (m/s)
<em>Speed = 0.766... m/s</em>
The correct answer is
<span>C) either the pressure of the gas, the volume of the gas, or both, will increase.
In fact, the ideal gas law can be written as
</span>
<span>where
p is the gas pressure
V is its volume
n is the number of moles
R is the gas constant
T is the absolute temperature of the gas
We can see that if the temperature T increases, then the term on the right in the equation increases, therefore the term on the left should increase as well. In order for this to be possible, at least one between p and V should increase, or also both of them. Therefore, the correct answer is C.</span>
They can fight the infection but not the disease
