In a parallel circuit, there is only one path for current to take. Please select the best answer from the choices provided T F
1 answer:
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Answer:
a) attractiva, b) dF = , c) F =
, d) F = -1.09 N
Explanation:
a) q1 is negative and the charge of the bar is positive therefore the force is attractive
b) For this exercise we use Coulomb's law, where we assume a card dQ₂ at a distance x
dF =
where k is a constant, Q₁ the charge at the origin, x the distance
c) To find the total force we must integrate from the beginning of the bar at x = d to the end point of the bar x = d + L
∫ dF =
as they indicate that the load on the bar is uniformly distributed, we use the concept of linear density
λ = dQ₂ / dx
DQ₂ = λ dx
we substitute
F =
F = k Q1 λ ()
we evaluate the integral
F = k Q₁ λ
F = k Q₁ λ
we change the linear density by its value
λ = Q2 / L
F =
d) we calculate the magnitude of F
F =9 10⁹ (-4.2 10⁻⁶)
F = -1.09 N
the sign indicates that the force is attractive
Complete Question
The complete question is shown on the first uploaded image
Answer:
The value is
Explanation:
From the question we are told that
The mass of the wheel is m = 6.9 kg
The radius is
The radius of gyration is
The angle is
The force which the massless bar is subjected to
Generally given that the wheels rolls without slipping on the flat stationary ground surface, it implies that point A is the center of rotation.
Generally the moment of inertia about A is mathematically represented as
Here is the moment of inertia about G with respect to the radius of gyration which is mathematically represented as
=>
=>
=>
Generally the torque experienced by the wheel is mathematically represented as
=>
=>
Generally this torque is also mathematically represented as
=>
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