Applying Kirchoff's junction rule; the power source and current split when the parallel circuit has two branches.
Discussion:
This forms a basis for the Kirchoff's Junction rule.
In a circuit of resistors; a parallel arrangement of the resistors leaves the current divided in a ratio according to the resistance of the resistors.
However, the case is not the same for a series connection
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Answer:
w = 2w₀ the angular velocity of man doubles
Explanation:
In this exercise, releasing the weights reduces the moment of inertia
I= I₀ / 2
Therefore, since the platform system plus man is isolated, the kinetic moment must be conserved
L₀ = L
I₀ w₀ = I w
I₀ w₀ = I₀ / 2 w
w = 2w₀
therefore the angular velocity of man doubles
Answer:
Explanation:
According to Newton's second law, the tension in the string is equal to the centripetal force, since the mass is under an uniform circular motion:
Here is the centripetal acceleration, which is defined as:
So, replacing:
In this case we have , and . Thus, the tension required to mantain uniform circular motion is:
Answer:
a) E = σ / 2 ε₀ =
Q / 2A ε₀, b) E = 2Q/A ε₀
Explanation:
For this exercise we can use Gauss's Law
Ф = E. dA = / ε₀
Let us define a Gaussian surface as a cylinder with the base parallel to the plane. In this case, the walls of the cylinder and the charged plate have 90 degrees whereby the scalar product is zero, the normal vector at the base of the cylinder and the plate has zero degrees whereby the product is reduced to the algebraic product
Φ = E dA = q_{int} / ε₀ (1)
As they indicate that the plate has an area A, we can use the concept of surface charge density
σ = Q / A
Q = σ A
The flow is to both sides of loaded plate
Φ = 2 E A
Let's replace in equation 1
2E A = σA / ε₀
E = σ / 2 ε₀ =
Q / 2A ε₀
This is in the field at point P.
b) Now we have two plates each with a load Q and 3Q respectively and they ask for the field between them
The electric field is a vector quantity
E = E₁ + E₂
In the gap between the plates the two fields point in the same direction whereby they add
σ₁ = Q / A
E₁ = σ₁ / 2 ε₀
For the plate 2
σ₂ = -3Q / A = -3 σ₁
E₂ = σ₂ / 2 ε₀
E₂ = -3 σ₁ / 2 ε₀
The total field is
E = σ₁ / 2 ε₀ + 3 σ₁ / 2 ε₀
E = σ₁ / 2 ε₀ (1+ 3)
E = 2 σ₁ / ε₀
E = 2Q/A ε₀