Answer:0.253Joules
Explanation:
First, we will calculate the force required to stretch the string. According to Hooke's law, the force applied to an elastic material or string is directly proportional to its extension.
F = ke where;
F is the force
k is spring constant = 34N/m
e is the extension = 0.12m
F = 34× 0.12 = 4.08N
To get work done,
Work is said to be done if the force applied to an object cause the body to move a distance from its initial position.
Work done = Force × Distance
Since F = 4.08m, distance = 0.062m
Work done = 4.08 × 0.062
Work done = 0.253Joules
Therefore, work done to stretch the string to an additional 0.062 m distance is 0.253Joules
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Answer:
Ф_cube /Ф_sphere = 3 /π
Explanation:
The electrical flow is
Ф = E A
where E is the electric field and A is the surface area
Let's shut down the electric field with Gauss's law
Фi = ∫ E .dA =
/ ε₀
the Gaussian surface is a sphere so its area is
A = 4 π r²
the charge inside is
q_{int} = Q
we substitute
E 4π r² = Q /ε₀
E = 1 / 4πε₀ Q / r²
To calculate the flow on the two surfaces
* Sphere
Ф = E A
Ф = 1 / 4πε₀ Q / r² (4π r²)
Ф_sphere = Q /ε₀
* Cube
Let's find the side value of the cube inscribed inside the sphere.
In this case the radius of the sphere is half the diagonal of the cube
r = d / 2
We look for the diagonal with the Pythagorean theorem
d² = L² + L² = 2 L²
d = √2 L
we substitute
r = √2 / 2 L
r = L / √2
L = √2 r
now we can calculate the area of the cube that has 6 faces
A = 6 L²
A = 6 (√2 r)²
A = 12 r²
the flow is
Ф = E A
Ф = 1 / 4πε₀ Q/r² (12r²)
Ф_cubo = 3 /πε₀ Q
the relationship of these two flows is
Ф_cube /Ф_sphere = 3 /π