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2 answers:
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Answer:
- The difference in length for steel is 2.46 x 10⁻⁴ m
- The difference in length for invar is 1.845 x 10⁻⁵ m
Explanation:
Given;
original length of steel, L₁ = 1.00 m
original length of invar, L₁ = 1.00 m
coefficients of volume expansion for steel, = 3.6 × 10⁻⁵ /°C
coefficients of volume expansion for invar, = 2.7 × 10⁻⁶ /°C
temperature rise in both meter stick, θ = 20.5°C
Difference in length, can be calculated as:
L₂ = L₁ (1 + αθ)
L₂ = L₁ + L₁αθ
L₂ - L₁ = L₁αθ
ΔL = L₁αθ
Where;
ΔL is difference in length
α is linear expansivity =
Difference in length, for steel at 20.5°C:
ΔL = L₁αθ
Given;
L₁ = 1.00 m
θ = 20.5°C
ΔL = 1 x 1.2 x 10⁻⁵ x 20.5 = 2.46 x 10⁻⁴ m
Difference in length, for invar at 20.5°C:
ΔL = L₁αθ
Given;
L₁ = 1.00 m
θ = 20.5°C
ΔL = 1 x 0.9 x 10⁻⁶ x 20.5 = 1.845 x 10⁻⁵ m
Answer:
The answer is below
Explanation:
Given that:
The area of the plates is 6 m by 0.030 m, Therefore the area = 6 m × 0.03 m = 0.18 m²
the relative permittivity of dielectric (εr) is 7.0
Permittivity of free space (εo) = 8.854 × 10^(-12)
capacitance of 100uF
potential difference (V) of 12V
d = separation between plate
The capacitance (C) of a capacitor is given by:
The electric field between plates is given as:
E = V /d