A carbon rod with a radius of 2.1 mm is used to make a resistor. The resistivity of this material is 3.2 × 10−5 Ω · m. What leng
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Answer:
Explanation:
Given that:
- Area of the plate of capacitor 1= Area of the plate of capacitor 2=A
- separation distance of capacitor 2,
- separation distance of capacitor 1,
- quantity of charge on capacitor 2,
- quantity of charge on capacitor 1,
We know that the Capacitance of a parallel plate capacitor is directly proportional to the area and inversely proportional to the distance of separation.
Mathematically given as:
.....................................(1)
where:
k = relative permittivity of the dielectric material between the plates= 1 for air
From eq. (1)
For capacitor 2:
For capacitor 1:
We know, potential differences across a capacitor is given by:
..........................................(2)
where, Q = charge on the capacitor plates.
for capacitor 2:
& for capacitor 1:
Answer:
= ( ρ_fluid g A) y
Explanation:
This exercise can be solved in two parts, the first finding the equilibrium force and the second finding the oscillating force
for the first part, let's write Newton's equilibrium equation
B₀ - W = 0
B₀ = W
ρ_fluid g V_fluid = W
the volume of the fluid is the area of the cube times the height it is submerged
V_fluid = A y
For the second part, the body introduces a quantity and below this equilibrium point, the equation is
B - W = m a
ρ_fluid g A (y₀ + y) - W = m a
ρ_fluid g A y + (ρ_fluid g A y₀ -W) = m a
ρ_fluid g A y + (B₀-W) = ma
the part in parentheses is zero since it is the force when it is in equilibrium
ρ_fluid g A y = m a
this equation the net force is
= ( ρ_fluid g A) y
we can see that this force varies linearly the distance and measured from the equilibrium position