A cylindrical container closed of both end has a radius of 7cm and height of 6cm A.)find the total surface area of the container
2 answers:
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The figure mentioned on the question is in the attachment.
Answer: a) = - 38.35N
b) = 30.5 N
c) = 27.45 N
d) a = - 13.16 m/s²
Explanation: A block on an inclined plane has 3 forces acting on it:
- Force due to gravity
= m.g;
- Normal Force due to the plane;
- Force of Friction
= µ.N;
Since the plane is inclined, Normal Force is equal the y-component of the force due to gravity and Force of friction and the x-component of the force due to gravity are opposite forces.
The second attachment ilustrate the forces acting on the block.
Calculating:
A) The magnitude of the x-component of Force due to gravity:
According to the second image:
= P.sinθ
= 5.9.8.sin(36.8)
= - 38.35 N
B) =
= m.g.cosθ
= 5.9.8.cos(36.8)
= 30.5 N
C) = 0.9.30.5
= 27.45 N
D) For the acceleration, use Newton's Law: = m . a
If there is movement, it is only on x-axis, so the net force is:
-
= m.a
- 38.35 - 27.45 = 5a
a = - 13.16 m/s²
The value of acceleration shows there is <u>no</u> <u>movement</u> on the x-axis due to the friction.
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
