Answer:
Knowing we only have one load applied in just one direction we have to use the Hooke's law for one dimension
ex = бx/E
бx = Fx/A = Fx/π
Using both equation and solving for the modulus of elasticity E
E = бx/ex = Fx / πex
E =
Apply the Hooke's law for either y or z direction (circle will change in every direction) we can find the change in radius
ey = (бy - v (бx + бz)) =
бx
= =
Finally
ey = Δr / r
Δr = ey * r = 10 *
Δd = 2Δr =
Explanation:
The potential difference across a and b is 15 v. determine the electrical charge on the 3 μf capacitor will be 45 * C
Capacitance, property of an electric conductor, or set of conductors, that is measured by the amount of separated electric charge that can be stored on it per unit change in electrical potential. Capacitance also implies an associated storage of electrical energy.
Charge (Q) stored in a capacitor is the product of its capacitance (C) and the voltage (V) applied to it. The capacitance of a capacitor should always be a constant, known value. So we can adjust voltage to increase or decrease the cap's charge. More voltage means more charge, less voltage... less charge.
charge = capacitance * voltage
Q = CV
= 3 * * 15 v
= 45 * C
To learn more about capacitance here
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