Name one way that a mixture can be separated
1 answer:
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Answer:
a) V = - x ( σ / 2ε₀)
c) parallel to the flat sheet of paper
Explanation:
a) For this exercise we use the relationship between the electric field and the electric potential
V = - ∫ E . dx (1)
for which we need the electric field of the sheet of paper, for this we use Gauss's law. Let us use as a Gaussian surface a cylinder with faces parallel to the sheet
Ф = ∫ E . dA = /ε₀
the electric field lines are perpendicular to the sheet, therefore they are parallel to the normal of the area, which reduces the scalar product to the algebraic product
E A = q_{int} /ε₀
area let's use the concept of density
σ = q_{int}/ A
q_{int} = σ A
E = σ /ε₀
as the leaf emits bonnet towards both sides, for only one side the field must be
E = σ / 2ε₀
we substitute in equation 1 and integrate
V = - σ x / 2ε₀
V = - x ( σ / 2ε₀)
if the area of the sheeta is 100 cm² = 10⁻² m²
V = - x (10⁻²/(2 8.85 10⁻¹²) = - x ( 5.6 10⁻¹⁰)
x = 1 cm V = -1 V
x = 2cm V = -2 V
This value is relative to the loaded sheet if we combine our reference system the values are inverted
V ’= V (inf) - V
x = 1 V = 5
x = 2 V = 4
x = 3 V = 3
These surfaces are perpendicular to the electric field lines, so they are parallel to the sheet.
In the attachment we can see a schematic representation of the equipotential surfaces
b) From the equation we can see that the equipotential surfaces are parallel to the sheet and equally spaced
c) parallel to the flat sheet of paper
Answer:
I / = 1.48
, The correct answer is d
Explanation:
the moment of inertia is given by
I = ∫ r² dm
For figures with symmetry it is tabulated. In the case of a thin variation, the moment of inertia with respect to its center of mass is
= 1/12 M L2
There is a widely used theorem, which is the parallel axis theorem, where the moment of inertia of any parallel axis, is the moment of mass inertia plus the moment of inertia of the body taken as a particle
I = + M D²
Let's put these expressions to our case.
As the bar is one meter long its center of mass that this Enel midpoint corresponds to
= 1/12 m L²
= 1/12 m 1.00²
= 8.33 10⁻² m
Let's use the parallel axes theorem for the axis that passes through x = 30 cm. The distance from the enrode masses to the axis is
D = - 0.30
D = 0.50 - 0.30 = 0.20 m
I = + m D²
I = 8.33 10⁻² m + m 0.2²
I = (8.33 10⁻² + 4 10⁻²) m
I = 12.33 10⁻² m
The relationship between these two moments of inertia
I / = 12.333 10⁻² / 8.333 10⁻²
I / = 1.48
The correct answer is d