Answer:
The statement that we can conclude is that the moment of inertial of any body about the centroidal axis is least of all the moment of inertia's of the body about any arbitrary axis in the body.
Explanation:
According to parallel axis theorem we have
Where
is the moment of inertia about any arbitrary axis.
is the moment of inertia about centroidal axis.
A is the area of section of which Moment of area is evaluated
'x' = is the distance between the axis and C.G
thus we conclude that is least of all the moments.
speed of the striking object is given as
now this speed in SI units given as
So here we can use kinematic now to find the height from where it is dropped
so here we will have
now we have
now from above equation
now we know that average height of 1 story building is 3 meter
so approx total number of stories in the building will be
Given Information:
Potential difference = V = 100 V
Capacitance C₁ = 10 mF
Capacitance C₂ = 5 mF
Capacitance C₃ = 4 mF
Required Information:
a. Charge q₃
b. Potential difference V₃
c. Stored energy U₃
d. Charge q ₁
e. Potential difference V₁
f. Stored energy U₁
g. Charge q ₂
h. Potential difference V₂
i. Stored energy U₂
Answer:
a. Charge q₃ = 0.4 C
b. Potential difference V₃ = 100 V
c. Stored energy U₃ = 20 J
d. Charge q ₁ = 0.33 C
e. Potential difference V₁ = 33 V
f. Stored energy U₁ = 5.445 J
g. Charge q ₂ = 0.33 C
h. Potential difference V₂ = 66 V
i. Stored energy U₂ = 10.89 J
Explanation:
Please refer to the circuit attached in the diagram
a. Charge q₃
As we know charge in a capacitor is given by
q₃ = C₃V₃
q₃ = 4x10⁻³*100
q₃ = 0.4 C
b. Potential difference V₃
The potential difference V₃ is same as V
V₃ = 100 V
c. Stored energy U₃
Energy stored in a capacitor is given by
U₃ = ½C₃V₃²
U₃ = ½*4x10⁻³*100²
U₃ = 20 J
d. Charge q ₁
Since capacitor C₁ and C₂ are in series their equivalent capacitance is
Ceq = C₁*C₂/C₁ + C₂
Ceq = 10x10⁻³*5x10⁻³/10x10⁻³ + 5x10⁻³
Ceq = 3.33x10⁻³ F
q ₁ = Ceq*V
q ₁ = 3.33x10⁻³*100
q ₁ = 0.33 C
e. Potential difference V₁
V₁ = q ₁/C₁
V₁ = 0.33/10x10⁻³
V₁ = 33 V
f. Stored energy U₁
U₁ = ½C₁V₁²
U₁ = ½*10x10⁻³*(33)²
U₁ = 5.445 J
g. Charge q ₂
q₂ = Ceq*V
q₂ = 3.33x10⁻³*100
q₂ = 0.33 C
h. Potential difference V₂
V₂ = q ₂/C₂
V₂ = 0.33/5x10⁻³
V₂ = 66 V
i. Stored energy U₂
U₂ = ½C₂V₂²
U₂ = ½*5x10⁻³*(66)²
U₂ = 10.89 J
