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Answer: E = 1.93×10^-7 J
Explanation: The formulae for getting the energy stored in a capacitor is given as
E =Cv²/2
Where E = energy stored in capacitor (joules), C = capacitance of capacitor (farad), v = applied voltage =150v.
The formulae implies that we need to get the capacitance of the capacitor (C) first before we can get the energy.
From the parameters given to us
Area of plates =A = 87.5cm² = 87.5/10000 = 0.00875m²
Distance between plates = d = 4.50mm = 4.50/1000 = 0.0045m
C = ε0A/d
Where ε0 = permittivity of free space = 8.85×10^-12 F/m
C = 8.85×10^-12 × 0.00875/ 0.0045
C = 7.743×10^-14/0.0045
C = 1.72×10^-11 F.
Recall that E = Cv²/2
E = ( 1.72×10^-11 × 150×150) /2
E = 3.87×10^-7/2
E = 1.93×10^-7 J
Answer:
Explanation:
Let the initial rotational inertia be I and final rotational inertia be I / 6 .
Let the initial angular velocity be ω₁ and final angular velocity be ω₂.
Applying conservation of angular momentum law
I x ω₁ = I / 6 x ω₂
6 ω₁ = ω₂
initial rotational kinetic energy = 1/2 I x ω₁ ²
Final rotational kinetic energy = 1/2 ( I / 6 ) x ω₂ ²
Final rotational kinetic energy / initial rotational kinetic energy
= ( 1 / 6 ) x ω₂ ² / ω₁ ²
= ω₂ ² / 6ω₁ ²
= 36 ω₁ ² / 6ω₁ ²
= 6 .
Answer:
A) Become 1/ 2 of its original value.
B) Become one-half of its original value.
Explanation:
As we know that heat transfer trough the wall of thickness t given as
Where
K=Thermal conductivity of the wall
A= Cross sectional area of the wall
t=Thickness of the wall
ΔT=temperature difference across the wall surfaces.
When the thickness become double ,lets sat t' = 2 t
Then new heat transfer
Therefore the new heat transfer become half of the original.
