Answer:
Minimum capacitance = 200 μF
Explanation:
From image B attached, we can calculate the current flowing through the capacitors.
Thus;
Since V=IR; I = V/R = 5/500 = 0.01 A
Maximum charge in voltage is from 5V to 4.9V. Thus, each capacitor will have 2.5V. Hence, change in voltage(Δv) for each capacitor will be ; Δv = 0.05 V
So minimum capacitance will be determined from;
i(t) = C(dv/dt)
So, C = i(t)(Δt/Δv) = 0.01[0.001/0.05]
C = 0.01 x 0.0002 = 200 x 10^(-6) F = 200 μF
Answer:
P.E = 4.398 Joules.
Explanation:
Given the following data;
Spring constant, k = 71.8N/m
Displacement, x = 0.35m
To find the elastic potential energy;
The elastic potential energy of an object is given by the formula;
Substituting into the equation, we have;
Elastic potential energy = 4.398 Joules.
<em>Therefore, the elastic potential energy of the spring is 4.398 Joules. </em>
Answer:
Given that
m = 5.3 kg
Fx = 2x + 4
We know that work done by force F given as
w= ∫ F. dx
a)
Given that x=1.08 m to x=6.5 m
Fx = 2x + 4
w= ∫ F. dx
w=62.7 J
b)
We know that potential energy given as
∫ dU = -∫F.dx ( w= ∫ F. dx)
ΔU= -62.7 J
c)
We know that form work power energy theorem
Net work = Change in kinetic energy
W= KE₂ - KE₁
62.7 =KE₂ - (1/2)x 5.3 x 3²
KE₂ = 86.55 J
This is the kinetic energy at 6.5m