Answer:
1.a) Cmax = 12μF
b) Cmin = 1.09μF
2.a) E1/E2 = 2
b) E1/E2 = 0.5
Explanation:
For the maximum Capacitance, we have to connect them in parallel. In this configuration, Ct = C1 + C2 + C3 = 12μF
For the minumum Capacitance, we have to connect them in parallel. In this configuration, 
= 1.09μF
To calculate the energies:
For the minimum capacitance configuration, the charge is the same on all of the capacitors, so:
For the maximum capacitance configuration, the voltage is the same on all of the capacitors, so:
Answer:
P₀ = 5.76 x 10⁻² MW
Explanation:
given,
efficiency of the power plant = 32%
Power produced by the nuclear fission = 0.18 MW
the power plant output = ?
using formula of efficiency
where P is the power produced in the power plant
P₀ is the power output of the power plant
P₀ = 0.18 x 0.32
P₀ = 0.0576 MW
P₀ = 5.76 x 10⁻² MW
Power plant output is equal to P₀ = 5.76 x 10⁻² MW
Answer:
Part(a): The value of the spring constant is 
.
Part(b): The work done by the variable force that stretches the collagen is 
.
Explanation:
Part(a):
If '
' be the force constant and if due the application of a force '
' on the collagen '
' be it's increase in length, then from Hook's law
Also, Young's modulus of a material is given by
where '
' is the area of the material and '
' is the length.
Comparing equation (
) and (
) we can write
Here, we have to consider only the circular surface of the collagen as force is applied only perpendicular to this surface.
Substituting the given values in equation (
), we have
Part(b):
We know the amount of work done (
) on the collagen is stored as a potential energy (
) within it. Now, the amount of work done by the variable force that stretches the collagen can be written as
Substituting all the values, we can write
