Answer:
<h2>
a) Q = 0.759µC</h2><h2>
b) E = 39.5µJ</h2>
Explanation:
a) The charge Q on the positive charge capacitor can be gotten using the formula Q = CV
C = capacitance of the capacitor (in Farads )
V = voltage (in volts) = 100V
C = ∈A/d
∈ = permittivity of free space = 8.85 × 10^-12 F/m
A = cross sectional area = 600 cm²
d= distance between the plates = 0.7cm
C = 8.85 × 10^-12 * 600/0.7
C = 7.59*10^-9Farads
Q = 7.59*10^-9 * 100
Q = 7.59*10^-7Coulombs
Q = 0.759*10^-6C
Q = 0.759µC
b) Energy stored in a capacitor is expressed as E = 1/2CV²
E = 1/2 * 7.59*10^-9 * 100²
E = 0.0000395Joules
E = 39.5*10^-6Joules
E = 39.5µJ
Highest fluid potential energy: answer A
Because the fluid is pushed upwards and potential energy is function of height. Since point A is the highest, there is the highest potential energy.
highest fluid pressure: answer C
This is because it is at the bottom where you have a hydrostatic pressure component
increasing fluid speed: answer B
This is because the section of the pipe is smaller and in order to have the same fluid flow rate the speed must increase
Answer:
I = 0.25 [amp]
Explanation:
To solve this problem we must use ohm's law which tells us that the voltage is equal to the product of the current by the resistance.
V = I*R
where:
V = voltage [Volt]
I = amperage or current [amp]
R = resistance [ohm]
Since all resistors are connected in series, the total resistance will be equal to the arithmetic sum of all resistors.
Rt = 2 + 8 + 14
Rt = 24 [ohm]
Now clearing I for amperage
I = V/Rt
I = 6/24
I = 0.25 [amp].
